Calc ii complete

Calculus II

This document was written and copyrighted by Paul Dawkins. Use of this document and
its online version is governed by the Terms and Conditions of Use located at
http://tutorial.math.lamar.edu/terms.asp.
The online version of this document is available at http://tutorial.math.lamar.edu. At the
above web site you will find not only the online version of this document but also pdf
versions of each section, chapter and complete set of notes.

Preface
Here are my online notes for my Calculus II course that I teach here at Lamar University.
Despite the fact that these are my “class notes” they should be accessible to anyone
wanting to learn Calculus II or needing a refresher in some of the topics from the class.
These notes do assume that the reader has a good working knowledge of Calculus I topics
including limits, derivatives and integration by substitution.
Here are a couple of warnings to my students who may be here to get a copy of what
happened on a day that you missed
1. Because I wanted to make this a fairly complete set of notes for anyone wanting
to learn calculus I have included some material that I do not usually have time to
cover in class and because this changes from semester to semester it is not noted
here. You will need to find one of your fellow class mates to see if there is
something in these notes that wasn’t covered in class.
2. In general I try to work problems in class that are different from my notes.
However, with Calculus II many of the problems are difficult to make up on the
spur of the moment and so in this class my class work will follow these notes
fairly close as far as worked problems go. With that being said I often don’t have
time in class to work all of these problems and so you will find that some sections
contain problems that weren’t worked in class due to time restrictions.
3. Sometimes questions in class will lead down paths that are not covered here. I try
to anticipate as many of the questions as possible in writing these up, but the
reality is that I can’t anticipate all the questions. Sometimes a very good question
gets asked in class that leads to insights that I’ve not included here. You should
always talk to someone who was in class on the day you missed and compare
these notes to their notes and see what the differences are.
4. This is somewhat related to the previous three items, but is important enough to
merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR
ATTENDING CLASS!! Using these notes as a substitute for class is liable to get
you in trouble. As already noted not everything in these notes is covered in class
and often material or insights not in these notes is covered in class.

© 2005 Paul Dawkins

1

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Calculus II

Integration Techniques
Introduction
In this chapter we are going to be looking at integration techniques. There are a fair
number of them, some easier than others. The point of the chapter is to teach you these
new techniques and so this chapter assumes that you’ve got a fairly good working
knowledge of basic substitutions with integrals. In fact, most integrals involving
“simple” substitutions will not have any of the substitution work shown. It is going to be
assumed that you can verify the substitution portion of the integration yourself.
Also, most of the integrals done in this chapter will be indefinite integrals. It is also
assumed that once you can do the indefinite integrals you can also do the definite
integrals and so to conserve space we concentrate mostly on indefinite integrals. There is
one exception to this and that is the Trig Substitution section and in this case there are
some subtleties involved with definite integrals that we’re going to have to watch out for.
Here is a list of topics that are covered in this chapter.
Integration by Parts – Of all the integration techniques covered in this chapter this is
probably the one that students are most likely to run into down the road in other classes.
Integrals Involving Trig Functions – In this section we look at integrating certain
products and quotients of trig functions.
Trig Substitutions – Here we will look using substitutions involving trig functions an
how they can be used to simplify certain integrals.
Partial Fractions – We will use partial fractions to allow us to do integrals involving
rational functions.
Integrals Involving Roots – We will take a look at a substitution that can, on occasion,
be used with integrals involving roots.
Integrals Involving Quadratics – In this section we are going to look at integrals that
involve quadratics.
Using Integral Tables – Here we look at using Integral Tables as well as relating new
integrals back to integrals that we already know how to do.
Integration Strategy – We give a general set of guidelines for determining how to
evaluate an integral.
Improper Integrals – We will look at integrals with infinite intervals of integration and
integrals with discontinuous integrands in this section.


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Calculus II

Comparison Test for Improper Integrals – Here we will use the Comparison Test to
determine if improper integrals converge or diverge.
Approximating Definite Integrals – There are many ways to approximate the value of a
definite integral. We will look at three of them in this section.

Integration by Parts
Let’s start off with this section with a couple of integrals that we should already be able
to do to get us started. First let’s take a look at the following.
x
x
∫ e dx = e + c
So, that was simple enough. Now, let’s take a look at,

∫ xe

x2

dx

To do this integral we’ll use the following substitution.

u = x2

du = 2 x dx

⇒

x dx =

1
du
2

1 u
1 u
1 x2
∫ e du = 2 e + c = 2 e + c
2
Again, simple enough to do provided you remember how to do substitutions. By the way
make sure that you can do these kinds of substitutions quickly and easily. From this
point on we are going to be doing these kinds of substitutions in our head. If you have to
stop and write these out with every problem you will find that it will take to significantly
longer to do these problems.

∫ xe

x2

dx =

Now, let’s look at the integral that we really want to do.
6x
∫ xe dx
If we just had an x by itself or e6 x by itself we could do the integral easily enough. But,
we don’t have them by themselves, they are instead multiplied together.
There is no substitution that we can use on this integral that will allow us to do the
integral. So, at this point we don’t have the knowledge to do this integral.
To do this integral we will need to use integration by parts so let’s derive the integration
by parts formula. We’ll start with the product rule.
( f g )′ = f ′ g + f g ′
Now, integrate both sides of this.

∫ ( f g )′ dx = ∫ f ′ g + f g ′ dx
The left side is easy enough to integrate and we’ll split up the right side of the integral.
fg = ∫ f ′ g dx + ∫ f g ′ dx


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Calculus II

Note that technically we should have had a constant of integration show up on the left
side after doing the integration. We can drop it at this point since other constants of
integration will be showing up down the road and they would just end up absorbing this
one.
Finally, rewrite the formula as follows and we arrive that the integration by parts formula.
∫ f g ′ dx = fg − ∫ f ′ g dx
This is not the easiest formula to use however. So, let’s do a couple of substitutions.
u = f ( x)
v = g ( x)
du = f ′ ( x ) dx

dv = g ′ ( x ) dx

Both of these are just the standard Calc I substitutions that hopefully you are used to by
now. Don’t get excited by the fact that we are using two substitutions here. They will
work the same way.
Using these substitutions gives us the formula that most people think of as the integration
by parts formula.
∫ u dv = uv − ∫ v du
To use this formula we will need to identify u and dv, compute du and v and then use the
formula. Note as well that computing v is very easy. All we need to do is integrate dv.
v = ∫ dv
So, let’s take a look at the integral we wrote down above.
Example 1 Evaluate the following integral.
6x
∫ xe dx
Solution
So, on some level, the problem here is the x that is in front of the exponential. If that
wasn’t there we could do the integral. Notice as well that anything that we choose for u
will be differentiated and so that seems like choosing u=x will be a good choice since
upon differentiating the x will drop out.

Once u is chosen we know that dv will be everything else that remains.
So, here are the choices for u and dv as well as du and v.
u=x
dv = e6 x dx
1
v = ∫ e6 x dx = e6 x
6

du = dx

The integral is then,


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Calculus II

x
1
dx = e6 x − ⌠ e6 x dx
⎮
⌡6
6
x
1
= e6 x − e6 x + c
6
36
Once we have done the last integral in the problem we will add in the constant of
integration to get our final answer.

∫ xe

6x

Next let’s a look at integration by parts for definite integrals. In this case the formula is,

∫

b
a

b

u dv = uv a − ∫ v du
b

a

b

Note that the uv a in the first term is just the standard integral evaluation notation that
you should be familiar with at this point. All we do is evaluate at b then subtract off the
evaluation at a.

∫

2
−1

xe6 x dx

Solution
This is the same integral that we looked at in the first example so we’ll use the same u
and dv to get,
2
2
x 6x
1 2
6x
xe dx = e
− ∫ e6 x dx
∫ −1
6
6 −1
−1
2

x
1
= e6 x − e6 x
6
36
−1
11
7
= e12 + e−6
36
36

2

−1

Since we need to be able to do the indefinite integral in order to do the definite integral
and doing the definite integral amounts to nothing more than evaluating the indefinite
integral at a couple of points we will concentrate on doing indefinite integrals in the rest
of this section. In fact, through out most of this chapter this will be the case. We will be
doing far more indefinite integrals than definite integrals.
Let’s take a look at some more examples.
⎛t⎞
⌠
⎮ ( 3t + 5 ) cos ⎜ ⎟ dt
⎝4⎠
⌡
Solution
There are two ways to proceed with this example. For many, the first thing that they try
is multiplying the cosine through the parenthesis, splitting up the integral and then doing
integration by parts on the first integral.

5


Calculus II

While that is a perfectly acceptable way of doing the problem it’s more work than we
really need to do. Instead of splitting the integral up let’s instead use the following
choices for u and dv.
⎛t⎞
u = 3t + 5
dv = cos ⎜ ⎟ dt
⎝4⎠
⎛t⎞
du = 3 dt
v = 4sin ⎜ ⎟
⎝4⎠
⌠
⌠ ⎛t⎞
⎛t⎞
⎛t⎞
⎮ ( 3t + 5 ) cos ⎜ ⎟ dt = 4 ( 3t + 5 ) sin ⎜ ⎟ − 12⎮ sin ⎜ ⎟ dt
⎝4⎠
⎝4⎠
⌡
⌡ ⎝4⎠
⎛t⎞
⎛t⎞
= 4 ( 3t + 5 ) sin ⎜ ⎟ + 48cos ⎜ ⎟ + c
⎝4⎠
⎝4⎠
Notice that we pulled any constants out of the integral when we used the integration by
parts formula. We will usually do this in order to simplify the integral a little.
2
∫ w sin (10w) dw
Solution
For this example we’ll use the following choices for u and dv.
u = w2
dv = sin (10w ) dw

du = 2w dw

v=−

1
cos (10 w )
10

2
∫ w sin (10w) dw = −

w2
1
cos (10 w ) + ∫ w cos (10 w ) dw
10
5

In this example, unlike the previous examples, the new integral will also require
integration by parts. For this second integral we will use the following choices.
u=w
dv = cos (10w ) dw

du = dw

v=

1
sin (10w )
10

So, the integral becomes,
2
∫ w sin (10w) dw = −

w2
1⎛ w
1
⎞
cos (10w ) + ⎜ sin (10 w ) − ∫ sin (10 w ) dw ⎟
10
5 ⎝ 10
10
⎠

=−

w2
1⎛ w
1
⎞
cos (10w ) + ⎜ sin (10 w ) +
cos (10w ) ⎟ + c
10
5 ⎝ 10
100
⎠

=−

w2
w
1
cos (10w ) + sin (10w ) +
cos (10w ) + c
10
50
500


6


Calculus II

Be careful with the coefficient on the integral for the second application of integration by
1
parts. Since the integral is multiplied by we need to make sure that the results of
5
1
actually doing the integral is also multiplied by . Forgetting to do this is one of the
5
more common mistakes with integration by parts problems.
As this last example has shown us, we will sometimes need more than one application of
integration by parts to complete a problem. This is something that will happen so don’t
get excited about it when it happens.
In this next example we need to acknowledge an important point about integration
techniques. Some integrals can be done in using several different techniques. That is the
case with the integral in the next example.
Example 5 Evaluate the following integral
∫ x x + 1 dx

(a) Using Integration by Parts.
(b) Using a standard Calculus I substitution.
Solution
(a) First notice that there are no trig functions or exponentials in this integral. While a
good many integration by parts integrals will involve trig functions and/or exponentials
not all of them will so don’t get too locked into the idea of expecting them to show up.

In this case we’ll use the following choices for u and dv.
u=x
dv = x + 1 dx
du = dx

v=

3
2
x + 1) 2
(
3

3
3
2
2
x ( x + 1) 2 − ∫ ( x + 1) 2 dx
3
3
3
5
2
4
= x ( x + 1) 2 − ( x + 1) 2 + c
3
15
(b) Now let’s do the integral with a substitution. We can use the following substitution.
u = x +1
x = u −1
du = dx

∫ x x + 1 dx =

Notice that we’ll actually use the substitution twice, once for the quantity under the
square root and once for the x in front of the square root. The integral is then,


7


Calculus II

∫x

x + 1 dx = ∫ ( u − 1) u du
3

1

= ∫ u 2 − u 2 du
2 5 2 3
= u2 − u2 + c
5
3
5
3
2
2
= ( x + 1) 2 − ( x + 1) 2 + c
5
3
So, we used two different integration techniques in this example and we got two different
answers. The obvious question then should be : Did we do something wrong?
Actually, we didn’t do anything wrong. We need to remember the following fact from
Calculus I.
If f ′ ( x ) = g ′ ( x ) then f ( x ) = g ( x ) + c

In other words, if two functions have the same derivative then they will differ by no more
than a constant. So, how does this apply to the above problem? First define the
following,
f ′( x) = g′ ( x) = x x +1
Then we can compute f(x) and g(x) by integrating as follows,
f ( x ) = ∫ f ′ ( x ) dx
g ( x ) = ∫ g ′ ( x ) dx
We’ll use integration by parts for the first integral and the substitution for the second
integral. Then according to the fact f(x) and g(x) should differ by no more than a
constant. Let’s verify this and see if this is the case. We can verify that they differ my no
more than a constant if we take a look at the difference of the two.
3
5
5
3
4
2
⎛2
⎞ ⎛2
⎞
x ( x + 1) 2 − ( x + 1) 2 ⎟ − ⎜ ( x + 1) 2 − ( x + 1) 2 ⎟
⎜
15
3
⎝3
⎠ ⎝5
⎠
3
4
2
2⎞
⎛2
= ( x + 1) 2 ⎜ x − ( x + 1) − ( x + 1) + ⎟
15
5
3⎠
⎝3
3

= ( x + 1) 2 ( 0 )
=0
So, in this case it turns out the two functions are exactly the same function since the
difference is zero. Note that this won’t always happen. Sometimes the difference will
yield a nonzero constant. For an example of this check out the Constant of Integration
section in my Calculus I notes.


8


Calculus II

So just what have we learned? First, there will, on occasion, be more than one method
for evaluating an integral. Secondly, we saw that different methods will often lead to
different answers. Last, even though the answers are different it can be shown that they
differ by no more than a constant.
When we are faced with an integral the first thing that we’ll need to decide is if there is
more than one way to do the integral. If there is more than one way we’ll then need to
determine which method we should use. The general rule of thumb that I use in my
classes is that you should use the method that you find easiest. This may not be the
method that others find easiest, but that doesn’t make it the wrong method.
One of the more common mistakes with integration by parts is for people to get too
locked into perceived patterns. For instance, all of the previous examples used the basic
pattern of taking u to be the polynomial that sat in front of another function and then
letting dv be the other function. This will not always happen so we need to be careful and
not get locked into any patterns that we think we see.
Let’s take a look at some integrals that don’t fit into the above pattern.
∫ ln x dx
Solution
So, unlike any of the other integral we’ve done to this point there is only a single function
in the integral and no polynomial sitting in front of the logarithm.

The first choice of many people here is to try and fit this into the pattern from above and
make the following choices for u and dv.
u =1
dv = ln x dx
This leads to a real problem however since that means v must be,
v = ∫ ln x dx

In other words, we would need to know the answer ahead of time in order to actually do
the problem. So, this choice simply won’t work.
Therefore, if the logarithm doesn’t belong in the dv it must belong instead in the u. So,
let’s use the following choices instead
u = ln x
dv = dx
1
du = dx
v=x
x


9


Calculus II

⌠1

∫ ln x dx = x ln x − ⎮ x x dx
⌡
= x ln x − ∫ dx
= x ln x − x + c


∫x

5

x 3 + 1 dx

Solution
So, if we again try to use the pattern from the first few examples for this integral our
choices for u and dv would probably be the following.

u = x5
dv = x 3 + 1 dx
However, as with the previous example this won’t work since we can’t easily compute v.
v = ∫ x 3 + 1 dx

This is not an easy integral to do. However, notice that if we had an x2 in the integral
along with the root we could very easily do the integral with a substitution. Also notice
that we do have a lot of x’s floating around in the original integral. So instead of putting
all the x’s (outside of the root) in the u let’s split them up as follows.
u = x3

dv = x 2 x3 + 1 dx

du = 3 x 2 dx

v=

3
2 3
x + 1) 2
(
9

3
3
2 3 3
2
x ( x + 1) 2 − ⌠ x 2 ( x3 + 1) 2 dx
⎮
9
3⌡
3
5
2
4
= x 3 ( x 3 + 1) 2 − ( x 3 + 1) 2 + c
9
45

5
3
∫ x x + 1 dx =

So, in the previous two examples we saw cases that didn’t quite fit into any perceived
pattern that we might have gotten from the first couple of examples. This is always
something that we need to be on the lookout for with integration by parts.
Let’s take a look at another example that also illustrates another integration technique
that sometimes arises out of integration by parts problems.
θ
∫ e cos θ dθ
Solution
Okay, to this point we’ve always picked u in such a way that upon differentiating it
would make that portion go away or at the very least put it the integral into a form that
would make it easier to deal with. In this case no matter which part we make u it will
never go away in the differentiation process.


10


Calculus II

It doesn’t much matter which we choose to be u so we’ll choose in the following way.
Note however that we could choose the other way as well and we’ll get the same result.
u = cos θ
dv = eθ dθ
du = − sin θ dθ

v = eθ


∫e

θ

cos θ dθ = eθ cos θ + ∫ eθ sin θ dθ

So, it looks like we’ll do integration by parts again. Here are our choices this time.
u = sin θ
dv = eθ dθ
du = cos θ dθ

v = eθ

The integral is now,

∫e

θ

cos θ dθ = eθ cos θ + eθ sin θ − ∫ eθ cos θ dθ

Now, at this point it looks like we’re just running in circles. However, notice that we
now have the same integral on both sides and on the right side its got a minus sign in
front of it. This means that we can add the integral to both sides to get,
2∫ eθ cos θ dθ = eθ cos θ + eθ sin θ
All we need to do now is divide by 2 and we’re done. The integral is,
1 θ
θ
θ
∫ e cos θ dθ = 2 ( e cosθ + e sin θ ) + c
Notice that after dividing by the two we add in the constant of integration at that point.
This idea of using integration by parts until you get the same integral on both sides of the
equal sign and then simply solving for the integral is kind of nice to remember. It doesn’t
show up all that often, but when it does it may be the only way to actually do the integral.
We’ve got one more example to do. As we will see some problems could require us to
do integration by parts numerous times and there is a short hand method that will allow
us to do multiple applications of integration by parts quickly and easily.
x
4 2

∫x e

dx

Solution
We start off by choosing u and dv as we always would. However, instead of computing
du and v we put these into the following table. We then differentiate down the column
corresponding to u until we hit zero. In the column corresponding to dv we integrate
once for each entry in the first column. There is also a third column which we will
explain in a bit.


11


Calculus II

Now, multiply along the diagonals show in the table. In front of each product put the
sign in the third column that corresponds to the “u” term for that product. In this case
this would give,
x
x
x
⎛ x⎞
⎛ x⎞
⎛ x⎞
⎛
⎞
⎛
⎞
x 4e 2 dx = ( x 4 ) ⎜ 2e 2 ⎟ − ( 4 x 3 ) ⎜ 4e 2 ⎟ + (12 x 2 ) ⎜ 8e 2 ⎟ − ( 24 x ) ⎜ 16e 2 ⎟ + ( 24 ) ⎜ 32e 2 ⎟
∫
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
x
4 2

x
3 2

x
2 2

x
2

x
2

= 2 x e − 16 x e + 96 x e − 384 xe + 768e + c
We’ve got the integral. This is much easier than writing down all the various u’s and
dv’s that we’d have to do otherwise.
So, in this section we’ve seen how to do integration by parts. In your later math classes
this is liable to be one of the more frequent integration techniques that you’ll encounter.
It is important to not get too locked into patterns that you may think you’ve seen. In most
cases any pattern that you think you’ve seen can (and will be) violated at some point in
time. Be careful!
Also, don’t forget the shorthand method for multiple applications of integration by parts
problems. It can save you a fair amount of work on occasion.

Integrals Involving Trig Functions
In this section we are going to look at quite a few integrals involving trig functions and
some of the techniques we can use to help us evaluate them. Let’s start off with an
integral that we should already be able to do.

∫ cos x sin

5

x dx = ∫ u 5 du

using the substitution u = sin x

1
= sin 6 x + c
6


12


Calculus II

This integral is easy to do with a substitution because the presence of the cosine,
however, what about the following integral.
5
∫ sin x dx
Solution
This integral no longer has the cosine in it that would allow us to use the substitution that
we used above. Therefore, that substitution won’t work and we are going to have to find
another way of doing this integral.

Let’s first notice that we could write the integral as follows,

5
4
2
∫ sin x dx = ∫ sin x sin x dx = ∫ ( sin x ) sin x dx
2

Now recall the trig identity,
cos 2 x + sin 2 x = 1

⇒

sin 2 x = 1 − cos 2 x

With this identity the integral can be written as,

5
2
∫ sin x dx = ∫ (1 − cos x ) sin x dx
2

and we can now use the substitution u = cos x . Doing this gives us,

∫ sin

5

x dx = − ∫ (1 − u 2 ) du
2

= − ∫ 1 − 2u 2 + u 4 du
2
1 ⎞
⎛
= − ⎜ u − u3 + u5 ⎟ + c
3
5 ⎠
⎝
2
1
= − cos x + cos3 x − cos5 x + c
3
5
So, with a little rewriting on the integrand we were able to reduce this to a fairly simple
substitution.

Notice that we were able to do the rewrite that we did in the previous example because
the exponent on the sine was odd. In these cases all that we need to do is strip out one of
the sines. The exponent on the remaining sines will then be even and we can easily
convert the remaining sines to cosines using the identity,
cos 2 x + sin 2 x = 1
(1)
If the exponent on the sines had been even this would have been difficult to do. We
could strip out a sine, but the remaining sines would then have an odd exponent and
while we could convert them to cosines the resulting integral would often be even more
difficult than the original integral in most cases.
Let’s take a look at another example.


13


Calculus II

∫ sin

6

x cos3 x dx

Solution
So, in this case we’ve got both sines and cosines in the problem and in this case the
exponent on the sine is even while the exponent on the cosine is odd. So, we can use a
similar technique in this integral. This time we’ll strip out a cosine and convert the rest to
sines.
6
3
6
2
∫ sin x cos x dx = ∫ sin x cos x cos x dx

= ∫ sin 6 x (1 − sin 2 x ) cos x dx

u = sin x

= ∫ u 6 (1 − u 2 ) du
= ∫ u 6 − u 8 du

1
1
= sin 7 x − sin 9 x + c
7
9
At this point let’s pause for a second to summarize what we’ve learned about integrating
powers of sine and cosine.
n
m
(2)
∫ sin x cos x dx
In (2) if the exponent on the sines (n) is odd we strip out one sine, convert the rest to
cosines using (1) and then use the substitution u = cos x . Likewise, if the exponent on
the cosines (m) is odd we strip out one cosine and convert the rest to sines and the use the
substitution u = sin x .
Of, course if both exponents are odd then we can use either method. However, in these
cases it’s usually easier to convert the term with the smaller exponent.
The one case we haven’t looked at is what happens if both of the exponents are even? In
this case the technique we used in the first couple of examples simply won’t work and in
fact there really isn’t any one set method for doing these integrals. Each integral is
different and in some cases there is more than one way to do the integral.
With that being said most, if not all, of integrals involving products of sines and cosines
in which both exponents are even can be done using one or more of the following
formulas.
1
cos 2 x = (1 + cos ( 2 x ) )
2
1
sin 2 x = (1 − cos ( 2 x ) )
2
1
sin x cos x = sin ( 2 x )
2


14


Calculus II

The first two formulas are the standard half angle formula from a trig class written in a
form that will be more convenient for us to use. The last is the standard double angle
formula for sine, again with a small rewrite.
Let’s take a look at an example.
2
2
∫ sin x cos x dx
Solution
As noted above there are often more than one way to do integrals in which both of the
exponents are even. This integral is an example of that, there are at least two solution
techniques for this problem. We will do both solutions starting with what is probably the
harder of the two, but it’s also the one that many people see first.

Solution 1
In this solution we will use the two half angle formulas above and just substitute them
into the integral.
⌠1
⎛1⎞
2
2
∫ sin x cos x dx = ⎮ 2 (1 − cos ( 2 x ) ) ⎜ 2 ⎟ (1 + cos ( 2 x ) ) dx
⌡
⎝ ⎠
1
= ∫ 1 − cos 2 ( 2 x ) dx
4
So, we still have an integral that can’t be completely done, however notice that we have
managed to reduce the integral down to just one term causing problems (a cosine with an
even power) rather than two terms causing problems.
In fact to eliminate the remaining problem term all that we need to do is reuse the first
half angle formula given above.
1⌠ 1
2
2
∫ sin x cos x dx = 4 ⎮ 1 − 2 (1 + cos ( 4 x ) ) dx
⌡
1 1 1
= ⌠ − cos ( 4 x ) dx
⎮
4⌡ 2 2
1⎛1
1
⎞
= ⎜ x − sin ( 4 x ) ⎟ + c
4⎝2
8
⎠
1
1
= x − sin ( 4 x ) + c
8
32
So, this solution required a total of three trig identities to complete.
Solution 2
In this solution we will use the half angle formula to help simplify the integral as follows.


15


Calculus II

∫ sin

2

x cos 2 x dx = ∫ ( sin x cos x ) dx
2

⌠⎛1
⎞
= ⎮ ⎜ sin ( 2 x ) ⎟ dx
⎠
⌡⎝2
1
= ∫ sin 2 ( 2 x ) dx
4
Now, we use the double angle formula for sine to reduce to an integral that we can do.
1
2
2
∫ sin x cos x dx = 8 ∫ 1 − cos ( 4 x ) dx
1
1
= x − sin ( 4 x ) + c
8
32
This method required only two trig identities to complete.
2

Notice that the difference between these two methods is more one of “messiness”. The
second method is not appreciably easier (other than needing one less trig identity) it is
just not as messy and that will often translate into an “easier” process.
In the previous example we saw two different solution methods that gave the same
answer. Note that this will not always happen. In fact, more often than not we will get
different answers. However, as we discussed in the Integration by Parts section, the two
answers will differ by no more than a constant.
In general when we have products of sine and cosine in which both exponents are even
we will need to use a series of half angle and/or double angle formulas to reduce the
integral into a form that we can integrate. Also, the larger the exponents the more we’ll
need to use these formulas and hence the messier the problem.
Sometimes in the process of reducing integrals in which both exponents are even we will
run across products of sine and cosine in which the arguments are different. These will
require one of the following formulas to reduce the products to integrals that we can do.
1
⎡sin (α − β ) + sin (α + β ) ⎤
⎦
2⎣
1
sin α sin β = ⎡cos (α − β ) − cos (α + β ) ⎤
⎦
2⎣
1
cos α cos β = ⎡cos (α − β ) + cos (α + β ) ⎤
⎦
2⎣

sin α cos β =

Let’s take a look at an example of one of these kinds of integrals.
∫ cos (15x ) cos ( 4 x ) dx
Solution


16


Calculus II

This integral requires the last formula listed above.
1
∫ cos (15x ) cos ( 4 x ) dx = 2 ∫ cos (11x ) + cos (19 x ) dx
1⎛ 1
1
⎞
= ⎜ sin (11x ) + sin (19 x ) ⎟ + c
2 ⎝ 11
19
⎠
Okay, at this point we’ve covered pretty much all the possible cases involving products
of sine and cosine. It’s now time to look at integrals that involve products of secants and
tangents.
This time, let’s do a little analysis of the possibilities before we just jump into examples.
The general integral will be,
n
m
(3)
∫ sec x tan x dx
The first thing to notice is that we can easily convert even powers of secants to tangents
and even powers of tangents to secants by using a formula similar to (1). In fact, the
formula can be derived from (1) so let’s do that.
sin 2 x + cos 2 x = 1
sin 2 x cos 2 x
1
+
=
2
2
cos x cos x cos 2 x
tan 2 x + 1 = sec 2 x

(4)

Now, we’re going to want to deal with (3) similarly to how we dealt with (2). We’ll want
to eventually use one of the following substitutions.
u = tan x
du = sec 2 x dx
u = sec x
du = sec x tan x dx
So, if we use the substitution u = tan x we will need two secants left for the substitution
to work. This means that if the exponent on the secant (n) is even we can strip two out
and then convert the remaining secants to tangents using (4).
Next, if we want to use the substitution u = sec x we will need one secant and one tangent
left over in order to use the substitution. This means that if the exponent on the tangent
(m) is odd we can strip one out along with one of the secants of course. The tangent will
then have an even exponent and so we can use (4) to convert the rest to tangents to
secants. Note that this method does require that we have at least one secant in the
integral as well. If there aren’t any secants then we’ll need to do something different.
If the exponent on the secant is even and the exponent on the tangent is odd then we can
use either case. Again, it will be easier to convert the term with the smallest exponent.


17


Calculus II

Let’s take a look at a couple of examples.
9
5
∫ sec x tan x dx
Solution
First note that since the exponent on the secant isn’t even we can’t use the substitution
u = tan x . However, the exponent on the tangent is odd and we’ve got a secant in the
integral and so we will be able to use the substitution u = sec x . This means striping out
a single tangent (along with a secant) and converting the remaining tangents to secants
using (4).

Here’s the work for this integral.
9
5
8
4
∫ sec x tan x dx = ∫ sec x tan x tan x sec x dx
= ∫ sec8 x ( sec 2 x − 1) tan x sec x dx
2

u = sec x

= ∫ u 8 ( u 2 − 1) du
2

= ∫ u12 − 2u10 + u 8 du
=

1
2
1
sec13 x − sec11 x + sec9 x + c
13
11
9

4
6
∫ sec x tan x dx
Solution
So, in this example the exponent on the tangent is even so the substitution u = sec x
won’t work. The exponent on the secant is even and so we can use the substitution
u = tan x for this integral. That means that we need to strip out two secants and convert
the rest to tangents.

Here is the work for this integral.
4
6
2
6
2
∫ sec x tan x dx = ∫ sec x tan x sec x dx

= ∫ ( tan 2 x + 1) tan 6 x sec 2 x dx

u = tan x

= ∫ ( u 2 + 1) u 6 du
= ∫ u 8 + u 6 du

1
1
= tan 9 x + tan 7 x + c
9
7
Both of the previous examples fit very nicely into the patterns discussed above and so
were not all that difficult to work. However, there are a couple of exceptions to the


18


Calculus II

patterns above and in these cases there is no single method that will work for every
problem. Each integral will be different and may require different solution methods.
Let’s first take a look at a couple of integrals that have odd exponents on the tangents, but
no secants. In these cases we can’t use the substitution u = sec x since it requires there to
be at least one secant in the integral.
∫ tan x dx
Solution
This integral is nothing more than a Calculus I substitution.
⌠ sin x
u = cos x
∫ tan x dx = ⎮ cos x dx
⌡
1
= −⌠ du
⎮
⌡u
= − ln cos x + c
r ln x = ln x r
= ln cos x

−1

+c

ln sec x + c

3
∫ tan x dx
Solution
The trick to this one is do the following manipulation of the integrand.
3
2
∫ tan x dx = ∫ tan x tan x dx

= ∫ tan x ( sec 2 x − 1) dx

= ∫ tan x sec 2 x dx − ∫ tan x dx

We can now use the substitution u = tan x on the first integral and the results from the
previous example to on the second integral.

∫ tan

3

x dx =

1
tan 2 x − ln sec x + c
2

Note that all odd powers of tangent (with the exception of the first power) can be
integrated using the same method we used in the previous example. For instance,

∫ tan

5

x dx = ∫ tan 3 x ( sec2 x − 1) dx = ∫ tan 3 x sec 2 x dx − ∫ tan 3 x dx

So, a quick substitution ( u = tan x ) will give us the first integral and the second integral
will always be the previous odd power.


19


Calculus II

Now let’s take a look at a couple of examples in which the exponent on the secant is odd
and the exponent on the tangent is even. In these cases the substitutions used above
won’t work.
∫ sec x dx
Solution
This one isn’t too bad once you see what you’ve got to do. By itself the integral can’t be
done. However, if we manipulate the integrand as follows we can do it.
⌠ sec x ( sec x + tan x )
dx
sec x dx = ⎮
∫
sec x + tan x
⌡

=⌠
⎮
⌡

sec 2 x + tan x sec x
dx
sec x + tan x

In this form we can do the integral using the substitution u = sec x + tan x . Doing this
gives,
∫ sec x dx = ln sec x + tan x + c
3
∫ sec x dx
Solution
This one is different from any of the other integrals that we’ve done in this section. The
first step to doing this integral is to perform integration by parts using the following
choices for u and dv.
u = sec x
dv = sec 2 x dx
du = sec x tan x dx
v = tan x

Note that using integration by parts on this problem is not an obvious choice, but it does
work very nicely here. The integral is then,
3
2
∫ sec x dx = sec x tan x − ∫ sec x tan x dx
Now the new integral also has an odd exponent on the secant and an even exponent on
the tangent and so the previous examples of products of secants and tangents still won’t
do us any good.
To do this integral we’ll first write the tangents in the integral in terms of secants.
3
2
∫ sec x dx = sec x tan x − ∫ sec x ( sec x − 1) dx
= sec x tan x − ∫ sec3 x dx + ∫ sec x dx
Now, we can use the results from the previous example to do the second integral and
notice that the first integral is exactly the integral we’re being asked to evaluate with a


20


Calculus II

minus sign in front. So, add it to both sides to get,
2 ∫ sec3 x dx = sec x tan x + ln sec x + tan x
Finally divide by a two and we’re done.
1
3
∫ sec x dx = 2 ( sec x tan x + ln sec x + tan x ) + c
The two integrals in the last two examples will arise on occasion in some of the work that
we’ll be doing in later sections and chapters so it wouldn’t be a bad idea to make sure
you’ve got them written down somewhere.
Now that we’ve looked at products of secants and tangents let’s also acknowledge that
because we can relate cosecants and cotangents by
1 + cot 2 x = csc 2 x
all of the work that we did for products of secants and tangents will also work for
products of cosecants and cotangents. I’ll leave it to you to verify that.
There is one final topic to be discussed in this section before moving on.
To this point we’ve looked only at products of sines and cosines and products of secants
and tangents. However, the methods used to do these integrals can also be used on some
quotients involving sines and cosines and quotients involving secants and tangents (and
hence quotients involving cosecants and cotangents).
Let’s take a quick look at an example of this.
7
⌠ sin x dx
⎮
⌡ cos 4 x
Solution
If this were a product of sines and cosines we would know what to do. We would strip
out a sine (since the exponent on the sine is odd) and convert the rest to cosines. We’ll
the same thing will work in this case.
7
6
⌠ sin x dx = ⌠ sin x sin x dx
⎮
⎮
⌡ cos 4 x
⌡ cos 4 x
2
⌠ ( sin x )
sin x dx
=⎮
4
⌡ cos x
3

2
⌠ (1 − cos x )
sin x dx
=⎮
4
⌡ cos x
3

At this point all we need to do is use the substitution u = cos x and we’re done.


21


Calculus II

7
⌠ sin x dx = −⌠ (1 − u
⎮
⎮
4
⌡ cos 4 x
⌡ u

)

2 3

du

= − ∫ u −4 − 3u −2 + 3 − u 2 du
1
1 ⎞
⎛ 1 1
= − ⎜ − 3 + 3 + 3u − u 3 ⎟ + c
3 ⎠
u
⎝ 3u
1
3
1
=
−
− 3cos x + cos3 x + c
3
3cos x cos x
3

So, under the right circumstances, we can use the ideas developed to help us deal with
products of trig functions to deal with quotients of trig functions. The natural question
then, is just what are the right circumstances?
First notice that if the quotient had been reversed,
4
⌠ cos x dx
⎮
⌡ sin 7 x
we wouldn’t have been able to strip out a sine.
4
4
⌠ cos x dx = ⌠ cos x 1 dx
⎮
⎮
⌡ sin 7 x
⌡ sin 6 x sin x
In this case the “stripped out” sine remains in the denominator and it won’t do us any
good for the substitution u = cos x since this substitution requires a sine in the numerator
of the quotient. Also note that, while we could convert the sines to cosines, the resulting
integral would still be a fairly difficult integral.

So, we can use the methods we applied to products of trig functions to quotients of trig
functions provided the term that needs parts stripped out in is the numerator of the
quotient.

Trig Substitutions
As we have done in the last couple of sections, let’s start off with a couple of integrals
that we should already be able to do with a standard substitution.
3
1
x 25 x 2 − 4 dx = ( 25 x 2 − 4 ) 2 + c
∫
75
1
x
⌠
25 x 2 − 4 + c
dx =
⎮
2
25
⌡ 25 x − 4
Both of these used the substitution u = 25 x 2 − 4 .
However, let’s take a look at the following integral.


22


Calculus II

⌠ 25 x 2 − 4
dx
⎮
x
⌡
Solution
In this case the substitution u = 25 x 2 − 4 will not work and so we’re going to have to do
something different for this integral.

It would be nice if we could get rid of the square root somehow. The following
substitution will do that for us.
2
x = sec θ
5
Do not worry about where this came from at this point. As we work the problem you will
see that it works and that if we have a similar type of square root in the problem we can
always use a similar substitution.
Notice that this is not the standard substitution we are used to working with. To this
point we’ve used substitution that were in the form u = f ( x ) . In this case we are going
to explicitly give a substitution for x. The substitution will work in pretty much the same
manner however. Before we actually do the substitution however let’s verify the claim
that this will allow us to get rid of the square root.
⎛ 4 ⎞
25 x 2 − 4 = 25 ⎜ ⎟ sec 2 θ − 4 = 4 ( sec 2 θ − 1) = 2 sec 2 θ − 1
⎝ 25 ⎠

To get rid of the square root all we need to do is recall the relationship,
tan 2 θ + 1 = sec 2 θ
⇒
sec 2 θ − 1 = tan 2 θ
Using this fact the square root becomes,
25 x 2 − 4 = 2 tan 2 θ = 2 tan θ

Note the presence of the absolute value bars there. These are important. Recall that
x2 = x

There should always be absolute value bars at this stage. If we knew that tan θ was
always positive or always negative we could eliminate the absolute value bars using,
if x ≥ 0
⎧x
x =⎨
⎩− x if x < 0
Without limits we won’t be able to determine this, however, we will need to eliminate
them in order to do the integral. Therefore, since we doing an indefinite integral we will
assume that tan θ will be positive and so we can drop the absolute value bars. This
gives,
25 x 2 − 4 = 2 tan θ


23


Calculus II

So, we were able to eliminate the square root using this substitution. Let’s go ahead and
put the substitution into the integral and see what we get. In doing the substitution don’t
forget that well also need to substitute for the dx. This is easy enough to get from the
substitution.
2
2
x = sec θ
⇒
dx = sec θ tan θ dθ
5
5
Using this substitution the integral becomes,
⌠ 25 x 2 − 4
⌠ 2 tan θ
dx = ⎮ 2
⎮
x
⌡ 5 sec θ
⌡

⎛2
⎞
⎜ sec θ tan θ ⎟ dθ
⎝5
⎠

= 2 ∫ tan 2 θ dθ

So, with this substitution we were able to reduce the given integral to an integral
involving trig functions and we saw how to do these problems in the previous section.
Let’s finish the integral.

⌠ 25 x 2 − 4
dx = 2∫ sec 2 θ − 1 dθ
⎮
x
⌡
= 2 ( tan θ − θ ) + c
So, we’ve got an answer for the integral. Unfortunately the answer isn’t given in x’s as it
should be. So, we need to write our answer in terms of x. We can do this with some right
triangle trig. From our original substitution we have,
5 x hypotenuse
sec θ =
=
2
adjacent
This gives the following right triangle.

From this we can see that,
25 x 2 − 4
2
We can deal with the θ in one of any variety of ways. From our substitution we can see
that,
⎛ 5x ⎞
θ = sec−1 ⎜ ⎟
⎝ 2 ⎠
While this is a perfectly acceptable method of dealing with the θ we can use any of the
tan θ =


24


Calculus II

possible six inverse trig functions and since sine and cosine are the two trig functions
most people are familiar with we will usually use the inverse sine or inverse cosine. In
this case we’ll use the inverse cosine.
⎛ 2 ⎞
θ = cos −1 ⎜ ⎟
⎝ 5x ⎠
So, with all of this the integral becomes,
⎛ 25 x 2 − 4
⌠ 25 x 2 − 4
⎛ 2 ⎞⎞
dx = 2 ⎜
− cos −1 ⎜ ⎟ ⎟ + c
⎮
⎜
x
2
⎝ 5x ⎠ ⎟
⌡
⎝
⎠
⎛ 2 ⎞
= 25 x 2 − 4 − 2 cos −1 ⎜ ⎟ + c
⎝ 5x ⎠
We now have the answer back in terms of x.
Wow! That was a lot of work. Most of these won’t take as long to work however. This
first one needed lot’s of explanation since it was the first one. The remaining examples
won’t need quite as much explanation and so won’t take as long to work.
However, before we move onto more problems let’s first address the issue of definite
integrals and how the process differs in these cases.
4
5

⌠
⎮2
⌡

25 x 2 − 4
dx
x

5

Solution
The limits here won’t change the substitution so that will remain the same.
2
x = sec θ
5
Using this substitution the square root still reduces down to,
25 x 2 − 4 = 2 tan θ

However, unlike the previous example we can’t just drop the absolute value bars. In this
case we’ve got limits on the integral and so we can use the limits as well as the
substitution to determine the range of θ that we’re in. Once we’ve got that we can
determine how to drop the absolute value bars.
Here’s the limits of θ .
2
x=
5
4
x=
5


⇒
⇒

2
=
5
4
=
5

2
sec θ
5
2
sec θ
5

25

⇒

θ =0

⇒

θ=

π
3


Calculus II

In the range of 0 ≤ θ ≤

π

tangent is positive and so in this case we can just drop the
3
absolute value bars. So, let’s do the substitution. Note that the work is identical to the
previous example and so most of it is left out.
4

π
⌠ 5 25 x 2 − 4
dx = 2∫ 3 sec 2 θ − 1 dθ
⎮2
0
x
⌡
5

π 3

= 2 ( tan θ − θ ) 0

2π
3
Note that because of the limits we didn’t need to resort to a right triangle to complete the
problem.

=2 3−

Let’s take a look at a different set of limits for this integral.
−

2

⌠ 5 25 x 2 − 4
dx
⎮ 4
x
⌡−
5

Solution
Again, the substitution and square root are the same as the first two examples.
2
x = sec θ
25 x 2 − 4 = 2 tan θ
5

Let’s next see the limits θ for this problem.
2
2 2
x=−
⇒
− = sec θ
⇒
θ =π
5
5 5
4
4 2
2π
x=−
θ=
⇒
− = sec θ
⇒
5
5 5
3
Note that in determining the value of θ we used the smallest positive value. Now in the
2π
≤ θ ≤ π tangent is negative and so in this case we can drop the absolute
range of
3
value bars, but will need to add in a minus sign upon doing so. In other words,
25 x 2 − 4 = −2 tan θ

So, the only change this will make in the integration process is to put a minus sign in
front of the integral. The integral is then,


26


Calculus II

−

2

π
⌠ 5 25 x 2 − 4
dx = −2 ∫ 2π sec 2 θ − 1 dθ
⎮ 4
x
⌡−
3
5

π

= −2 ( tan θ − θ ) 2π 3
=

2π
−2 3
3

In the last two examples we saw that we have to be very careful with definite integrals.
We need to make sure that we determine the limits on θ and whether or not this will
mean that we can drop the absolute value bars or if we need to add in a minus sign when
we drop them.
Before moving on to the next example let’s get the general form for the substitution that
we used in the previous set of examples.
a
b2 x2 − a 2
⇒
x = sec θ
b
Let’s work a new and different type of example.
1
⌠
dx
⎮ 4
⌡ x 9 − x2
Solution
Now, the square root in this problem looks to be (almost) the same as the previous ones
so let’s try the same type of substitution and see if it will work here as well.
x = 3sec θ

Using this substitution the square root becomes,
9 − x 2 = 9 − 9sec 2 θ = 3 1 − sec 2 θ = 3 − tan 2 θ

So, this will be trouble. Using this substitution we will get complex values and we don’t
want that. So, using secant for the substitution won’t work.
However, the following substitution (and differential) will work.
x = 3sin θ
dx = 3cos θ dθ
With this substitution the square root is,
9 − x 2 = 3 1 − sin 2 θ = 3 cos 2 θ = 3 cos θ = 3cos θ

We were able to drop the absolute value bars because we are doing an indefinite integral
and so we’ll assume that everything is positive.
The integral is now,


27


Calculus II

1
1
⌠
⌠
dx = ⎮
3cos θ dθ
⎮ 4
4
2
⌡ x 9− x
⌡ 81sin θ ( 3cos θ )
1
1
= ⌠ 4 dθ
⎮
81 ⌡ sin θ
1
= ∫ csc 4 θ dθ
81
In the previous section we saw how to deal with integrals in which the exponent on the
secant was even and since cosecants behave an awful lot like secants we should be able to
do something similar with this.
Here is the integral.
1
1
⌠
dx = ∫ csc2 θ csc2 θ dθ
⎮ 4
81
⌡ x 9 − x2
1
= ∫ ( cot 2 θ + 1) csc 2 θ dθ
81
1
= − ∫ u 2 + 1 du
81
1 ⎛1
⎞
= − ⎜ cot 3 θ + cot θ ⎟ + c
81 ⎝ 3
⎠

u = cot θ

Now we need to go back to x’s using a right triangle. Here is the right triangle for this
problem and trig functions for this problem.
sin θ =

x
3

cot θ =

9 − x2
x


⎛
1
1 1 ⎛ 9 − x2
⌠
dx = − ⎜ ⎜
⎮ 4
x
81 ⎜ 3 ⎜
⌡ x 9 − x2
⎝ ⎝

3

⎞
9 − x2
+
⎟
⎟
x
⎠

⎞
⎟+c
⎟
⎠

3

=−

(9 − x2 ) 2
243x 3

−

9 − x2
+c
81x

Here’s the general form for this type of square root.

28


Calculus II

a 2 − b2 x2

⇒

x=

a
sin θ
b

There is one final case that we need to look at. The next integral will also contain
something that we need to make sure we can deal with.
1

⌠6
x5
⎮
dx
3
⎮
2
2
⌡ 0 ( 36 x + 1)
Solution
First, notice that there really is a square root in this problem even though it isn’t explicitly
written out.

( 36 x

2

+ 1)

3
2

3

1
⎛
⎞
= ⎜ ( 36 x 2 + 1) 2 ⎟ =
⎝
⎠

(

36 x 2 + 1

)

3

This square root is not in the form we saw in the previous examples. Here we will use
the substitution.
1
1
x = tan θ
dx = sec2 θ dθ
6
6
With this substitution the denominator becomes,

(

) (
3

36 x 2 + 1 =

) (
3

tan 2 θ + 1 =

sec 2 θ

) = secθ
3

3

Now, because we have limits we’ll need to convert them to θ so we can determine how
to drop the absolute value bars.
1
x=0
0 = tan θ
⇒
⇒
θ =0
6
1
1 1
π
x=
θ=
⇒
= tan θ
⇒
6
6 6
4
In this range secant is positive and so we can drop the absolute value bars.
Here is the integral,
1

π

⌠6
x5
⌠4
⎮
dx = ⎮
3
⎮
⌡0
⌡ 0 ( 36 x 2 + 1) 2

tan 5 θ
sec3 θ

1
7776

⎛1 2 ⎞
⎜ sec θ ⎟ dθ
⎝6
⎠

π

1 ⌠ 4 tan 5 θ
dθ
=
⎮
46656 ⌡ 0 sec θ
There are several ways to proceed from this point. Normally with an odd exponent on


29


Calculus II

the tangent we would strip on of them out and convert to secants. However, that would
require that we also have a secant in the numerator which we don’t have. Therefore, it
seems like the best way to do this one would be to convert the integrand to sines and
cosines.
1

π
⌠6
1 ⌠ 4 sin 5 θ
x5
⎮
dx =
dθ
⎮
3
⎮
46656 ⌡ 0 cos 4 θ
2
2
⌡ 0 ( 36 x + 1)
π

1 ⌠ 4 (1 − cos θ )
=
sin θ dθ
⎮
46656 ⌡ 0
cos 4 θ
2

2

We can now use the substitution u = cos θ and we might as well convert the limits as
well.
θ =0
u = cos 0 = 1

θ=

π

u = cos

4

π
4

=

2
2

1

2
⌠6
1
x5
−4
−2
2
⎮
dx = −
3
∫1 u − 2u + 1 du
⎮
46656
⌡ 0 ( 36 x 2 + 1) 2
2

1 ⎛ 1 2
⎞ 2
=−
⎜− 3 + +u⎟
46656 ⎝ 3u u
⎠1

=

1
11 2
−
17496 279936

The general form for this final type of square root is
a 2 + b2 x2

⇒

x=

a
tan θ
b

We have a couple of final examples to work in this section. Not all trig substitutions will
just jump right out at us. Sometimes we need to do a little work on the integrand first to
get it into the correct form.
x
⌠
dx
⎮
⌡ 2x2 − 4x − 7
Solution
In this case the quantity under the root doesn’t obviously fit into any of the cases we
looked at above and in fact isn’t in the any of the forms we saw in the previous examples.
Note however that if we complete the square on the quadratic we can make it look


30


Calculus II

somewhat like the above integrals.
Remember that completing the square requires a coefficient of one in front of the x2.
Once we have that we take half the coefficient of the x, square it, and then add and
subtract it to the quantity. Here is the completing the square for this problem.
7⎞
7⎞
9⎞
2
2
⎛
⎛
⎛
2 ⎜ x 2 − 2 x − ⎟ = 2 ⎜ x 2 − 2 x + 1 − 1 − ⎟ = 2 ⎜ ( x − 1) − ⎟ = 2 ( x − 1) − 9
2⎠
2⎠
2⎠
⎝
⎝
⎝
So, the root becomes,
2 x 2 − 4 x − 7 = 2 ( x − 1) − 9
2

This looks like a secant substitution except we don’t just have an x that is squared. That
is okay, it will work the same way.
3
3
3
x −1 =
x = 1+
dx =
sec θ
sec θ
sec θ tan θ dθ
2
2
2
The root reduces to,
2 x 2 − 4 x − 7 = 2 ( x − 1) − 9 = 9sec 2 θ − 9 = 3 tan 2 θ = 3 tan θ = 3 tan θ
2

Note we could drop the absolute value bars since we are doing an indefinite integral.
Here is the integral.
3
⌠ 1 + 2 sec θ ⎛ 3
x
⎞
⌠
sec θ tan θ ⎟ dθ
dx = ⎮
⎮
⎜
⌡ 2 x2 − 4 x − 7
⎠
⌡ 3 tan θ ⎝ 2
1
3
=⌠
sec θ + sec 2 θ dθ
⎮
2
⌡ 2
1
3
=
ln sec θ + tan θ + tan θ + c
2
2
And here is the right triangle for this problem.
2 ( x − 1)
sec θ =
3

2x2 − 4x − 7
tan θ =
3



31


Calculus II

1
x
⌠
ln
dx =
⎮
2
2
⌡ 2x − 4x − 7

2 ( x − 1)
2 x2 − 4 x − 7
2x2 − 4x − 7
+
+
+c
3
3
2


∫e

4x

1 + e 2 x dx

Solution
This doesn’t look to be anything like the other problems in this section. However it is.
To see this we first need to notice that,

e2 x = ( e x )

2

With this we can use the following substitution.
e x = tan θ
e x dx = sec 2 θ dθ
Remember that to compute the differential all we do is differentiate both sides and then
tack on dx or dθ onto the appropriate side.
With this substitution the square root becomes,

1 + e 2 x = 1 + ( e x ) = 1 + tan 2 θ = sec 2 θ = sec θ = sec θ
2

Again, we can drop the absolute value bars because we are doing an indefinite integral.
Here’s the integral.

∫e

4x

1 + e 2 x dx = ∫ e3 x e x 1 + e 2 x dx
= ∫ (ex )

3

1 + e 2 x ( e x ) dx

= ∫ tan 3 θ ( sec θ ) ( sec 2 θ ) dθ

= ∫ ( sec 2 θ − 1) sec 2 θ sec θ tan θ dθ

u = sec θ

= ∫ u 4 − u 2 du
1
1
= sec5 θ − sec3 θ + c
5
3
Here is the right triangle for this integral.
tan θ =


ex
1

sec θ =

32

1 + e2 x
= 1 + e2 x
1


Calculus II

4x
2x
∫ e 1 + e dx =

5
3
1
(1 + e2 x ) 2 − 1 (1 + e2 x ) 2 + c
5
3

So, as we’ve seen in the final two examples in this section some integrals that look
nothing like the first few examples can in fact be turned into a trig substitution problem
with a little work.
Before leaving this section let’s summarize all three cases in one place.
a
a 2 − b2 x 2
⇒
x = sin θ
b
a
b2 x2 − a 2
⇒
x = sec θ
b
a
⇒
a 2 + b2 x2
x = tan θ
b

Partial Fractions
In this section we are going to take a look a integrals of rational expressions of
polynomials and once again let’s start this section out with an integral that we can already
do so we can contrast it with the integrals that we’ll be doing in this section.
⌠ 2 x − 1 dx = ⌠ 1 du
using u = x 2 − x + 6 and du = ( 2 x − 1) dx
⎮ 2
⎮
⌡ x −x+6
⌡u
= ln x 2 − x + 6 + c

So, if the numerator is the derivative of the denominator (or a constant multiple of the
derivative of the denominator) doing this kind of integral is fairly simple. However,
often the numerator isn’t the derivative of the denominator (or a constant multiple). For
example, consider the following integral.
⌠ 3 x + 11 dx
⎮ 2
⌡ x − x−6


33


Calculus II

In this case the numerator is definitely not the derivative of the denominator nor is it a
constant multiple of the derivative of the denominator. Therefore, the simple substitution
that we used above won’t work.
However, if we notice that

3 x + 11
4
1
=
−
2
x − x−6 x−3 x+ 2

then the integral can be done.
⌠ 3x + 11 dx = ⌠ 4 − 1 dx
⎮ 2
⎮
⌡ x − x−6
⌡ x−3 x+ 2
= 4 ln x − 3 − ln x + 2 + c
This process of taking a rational expression and decomposing it into simpler rational
expressions that we can add or subtract to get the original rational expression is called
partial fraction decomposition. Many integrals involving rational expressions can be
done if we first do partial fractions on the integrand.
So, let’s do a quick review of partial fractions. We’ll start with a rational expression in
the form,
P ( x)
f ( x) =
Q ( x)
where both P(x) and Q(x) are polynomials and the degree of P(x) is smaller than the
degree of Q(x). Recall that the degree of a polynomial is the largest exponent in the
polynomial. Partial fractions can only be done if the degree of the numerator is strictly
less than the degree of the denominator. That is important to remember.
So, once we’ve determined that partial fractions can be done we factor the denominator
as completely as possible. Then for each factor in the denominator we can use the
following table to determine the term(s) we pick up in the partial fraction decomposition.
Factor in
denominator
ax + b

( ax + b )

k

ax 2 + bx + c

( ax 2 + bx + c )


Term in partial
fraction decomposition
A
ax + b
Ak
A1
A2
+
+ +
2
k
ax + b ( ax + b )
( ax + b )

Ax + B
ax + bx + c
A1 x + B1
A2 x + B2
+
+
ax 2 + bx + c ( ax 2 + bx + c )2
2

k

34

+

Ak x + Bk

( ax

2

+ bx + c )

k


Calculus II

Notice that the first and third cases are really special cases of the second and fourth cases
respectively.
There are several methods for determining the coefficients for each term and we will go
over each of those in the following examples.
Let’s start with actually doing the integral above.
⌠ 3 x + 11 dx
⎮ 2
⌡ x − x−6
Solution
The first step is to factor the denominator as much as possible and get the form of the
partial fraction decomposition.
3 x + 11
A
B
=
+
( x − 3)( x + 2 ) x − 3 x + 2
The next step is to actually add the right side back up.
A ( x + 2 ) + B ( x − 3)
3 x + 11
=
( x − 3)( x + 2 )
( x − 3)( x + 2 )

Now, we need to choose A and B so that the numerators of these two are equal for every
x. So, the next step is to set numerators equal.
3 x + 11 = A ( x + 2 ) + B ( x − 3)
Note that in most problems we will go straight from the general form of the
decomposition to this step and not bother with actually adding the terms back up. The
only point to adding the terms is to get the numerator and we can get that without actually
writing down the results of the addition.
At this point we have one of two ways to proceed. One way will always work, but is
often more work. The other, while it won’t always work, is often quicker when it does
work. In this case both will work and so we’ll use the quicker way for this example.
We’ll take a look at the other method in a later example.
What we’re going to do here is to notice that the numerators must be equal for any x that
we would choose to use. In particular the numerators must be equal for x=-2 and x=3.
So, let’s plug these in and see what we get.
x = −2
5 = A ( 0 ) + B ( −5 )
⇒
B = −1

x=3

20 = A ( 5 ) + B ( 0 )

⇒

A=4

So, by carefully picking the x’s we got the unknown constants to quickly drop out. Note
that these are the values we claimed they would be above.
At this point there really isn’t a whole lot to do other than the integral.


35


Calculus II

⌠ 3x + 11 dx = ⌠ 4 − 1 dx
⎮ 2
⎮
⌡ x − x−6
⌡ x−3 x+ 2
4
1
=⌠
dx − ⌠
dx
⎮
⎮
⌡ x −3
⌡ x+2
= 4 ln x − 3 − ln x + 2 + c
Recall that to do this integral we first split it up into two integrals and then used the
substitutions,
u = x−3
v = x+2
on the integrals to get the final answer.
Before moving onto the next example a couple of quick notes are in order here. First,
many of the integrals in partial fractions problems come down to the type of integrals
seen above. Make sure that you can do those integrals.
There is also another integral that often shows up in these kinds of problems so we may
as well give the formula for it here since we are already on the subject.
⌠ 1 dx = 1 tan −1 ⎛ x ⎞ + c
⎮ 2
⎜ ⎟
⌡ x + a2
a
⎝a⎠
It will be an example or two before we use this so don’t forget about it.
Now, let’s work some more examples.
x2 + 4
⌠
dx
⎮ 3
⌡ 3x + 4 x 2 − 4 x
Solution
We won’t be putting as much detail into this solution as we did in the previous example.
The first thing is to factor the denominator and get the form of the partial fraction
decomposition.
x2 + 4
A
B
C
= +
+
x ( x + 2 )( 3x − 2 ) x x + 2 3x − 2

The next step is to set numerators equal. If you need to actually add the right side
together to get the numerator for that side then you should do so, however, it will
definitely make the problem quicker if you can do this step in your head.
x 2 + 4 = A ( x + 2 )( 3 x − 2 ) + Bx ( 3 x − 2 ) + Cx ( x + 2 )
As with the previous example it looks like we can just pick a few values of x and find the
constants so let’s do that.


36


Calculus II

x=0

4 = A ( 2 )( −2 )

⇒

A = −1

x = −2

8 = B ( −2 )( −8 )

⇒

40
⎛ 2 ⎞⎛ 8 ⎞
= C ⎜ ⎟⎜ ⎟
9
⎝ 3 ⎠⎝ 3 ⎠

⇒

1
2
40 5
=
C=
16 2

x=

2
3

B=

Note that unlike the first example most of the coefficients here are fractions. That is not
unusual so don’t get excited about it when it happens.
Now, let’s do the integral.
5
1
x2 + 4
1
⌠
dx = ⌠ − + 2 + 2 dx
⎮
⎮ 3
2
⌡ x x + 2 3x − 2
⌡ 3x + 4 x − 4 x
1
5
= − ln x + ln x + 2 + ln 3x − 2 + c
2
6
Again, as noted above, integrals that generate natural logarithms are very common in
these problems so make sure you can do them.
⌠ x 2 − 29 x + 5
dx
⎮
2
2
⌡ ( x − 4 ) ( x + 3)
Solution
This time the denominator is already factored so let’s just jump right to the partial
fraction decomposition.
x 2 − 29 x + 5
A
B
Cx + D
=
+
+ 2
2
2
2
( x − 4 ) ( x + 3) x − 4 ( x − 4 ) x + 3

Setting numerators gives,
2
x 2 − 29 x + 5 = A ( x − 4 ) ( x 2 + 3) + B ( x 2 + 3) + ( Cx + D )( x − 4 )
In this case we aren’t going to be able to just pick values of x that will give us all the
constants. Therefore, we will need to work this the second (and often longer) way. The
first step is to multiply out the right side and collect all the like terms together. Doing
this gives,
x 2 − 29 x + 5 = ( A + C ) x 3 + ( −4 A + B − 8C + D ) x 2 + ( 3 A + 16C − 8 D ) x − 12 A + 3B + 16 D
Now we need to choose A, B, C, and D so that these two are equal. In other words we
will need to set the coefficients of like powers of x equal. This will give a system of
equations that can be solved.


37


Calculus II

x3 :
x2 :
x1 :
x0 :

⎫
⎪
−4 A + B − 8C + D = 1 ⎪
⎬
3 A + 16C − 8 D = −29 ⎪
−12 A + 3B + 16 D = 5 ⎪
⎭
A+C = 0

⇒

A = 1, B = −5, C = −1, D = 2

Note that we used x0 to represent the constants. Also note that these systems can often be
quite large and have a fair amount of work involved in solving them. The best way to
deal with these is to use some form of computer aided solving techniques.
Now, let’s take a look at the integral.

⌠ x 2 − 29 x + 5
⌠ 1
5
−x + 2
−
+ 2
dx = ⎮
dx
⎮
2
2
2
x +3
⌡ x − 4 ( x − 4)
⌡ ( x − 4 ) ( x + 3)
⌠ 1
5
2
x
=⎮
−
− 2
+ 2
dx
2
x +3 x +3
⌡ x − 4 ( x − 4)

= ln x − 4 +

5
1
2
⎛ x ⎞
− ln x 2 + 3 +
tan −1 ⎜
⎟+c
x−4 2
3
⎝ 3⎠

In order to take care of the third term we needed to split it up into two separate terms.
Once we’ve done this we can do all the integrals in the problem. The first two use the
substitution u = x − 4 , the third uses the substitution v = x 2 + 3 and the fourth term uses
the formula given above for inverse tangents.
⌠ x 3 + 10 x 2 + 3x + 36
dx
⎮
2
⎮
2
⌡ ( x − 1) ( x + 4 )
Solution
Let’s first get the general form of the partial fraction decomposition.
x 3 + 10 x 2 + 3x + 36
A
Bx + C Dx + E
=
+ 2
+
2
x − 1 x + 4 ( x 2 + 4 )2
( x − 1) ( x 2 + 4 )

Now, set numerators equal, expand the right side and collect like terms.

x3 + 10 x 2 + 3 x + 36 = A ( x 2 + 4 ) + ( Bx + C )( x − 1) ( x 2 + 4 ) + ( Dx + E )( x − 1)
2

= ( A + B ) x 4 + ( C − B ) x3 + (8 A + 4 B − C + D ) x 2 +

( −4 B + 4C − D + E ) x + 16 A − 4C − E
Setting coefficient equal gives the following system.


38


Calculus II

x4 :
x3 :
x2 :
1

x :
x0 :

A+ B = 0 ⎫
⎪
C − B =1 ⎪
⎪
8 A + 4 B − C + D = 10 ⎬ ⇒
−4 B + 4C − D + E = 3 ⎪
⎪
16 A − 4C − E = 36 ⎪
⎭

A = 2, B = −2, C = −1, D = 1, E = 0

Don’t get excited if some of the coefficients end up being zero. It happens on occasion.
Here’s the integral.
⌠ x 3 + 10 x 2 + 3x + 36
⌠ 2
−2 x − 1
x
+ 2
+
dx = ⎮
dx
⎮
2
⎮
x − 1 x + 4 ( x 2 + 4 )2
x − 1) ( x 2 + 4 )
⌡
⌡ (

⌠ 2
2x
1
x
dx
=⎮
− 2
− 2
+
2
x − 1 x + 4 x + 4 ( x + 4 )2
⌡
1
⎛ x⎞ 1 1
= 2 ln x − 1 − ln x 2 + 4 − tan −1 ⎜ ⎟ −
+c
2
2
⎝2⎠ 2 x +4
To this point we’ve only looked at rational expressions where the degree of the numerator
was strictly less that the degree of the denominator. Of course not all rational expressions
will fit into this form and so we need to take a look at a couple of examples where this
isn’t the case.
4
3
2
⌠ x − 5 x + 6 x − 18 dx
⎮
x3 − 3x 2
⌡
Solution
So, in this case the degree of the numerator is 4 and the degree of the numerator is 3.
Therefore, partial fractions can’t be done on this rational expression.

To fix this up we’ll need to do long division on this to get it into a form that we can deal
with. Here is the work for that.
x−2
x3 − 3x 2 x 4 − 5 x 3 + 6 x 2 − 18

− ( x 4 − 3x3 )
− 2 x 3 + 6 x 2 − 18
− ( −2 x3 + 6 x 2 )
− 18


39


Calculus II

So, from the long division we see that,
x 4 − 5 x 3 + 6 x 2 − 18
18
= x−2− 3
3
2
x − 3x
x − 3x 2
and the integral becomes,
4
3
2
⌠ x − 5 x + 6 x − 18 dx = ⌠ x − 2 − 18 dx
⎮
⎮
⌡
x3 − 3x 2
x3 − 3x 2
⌡
18
= ∫ x − 2 dx − ⌠ 3
dx
⎮
⌡ x − 3x 2
The first integral we can do easily enough and the second integral is now in a form that
allows us to do partial fractions. So, let’s get the general form of the partial fractions for
the second integrand.
18
A B
C
= + 2+
2
x ( x − 3) x x
x −3
Setting numerators equal gives us,
18 = Ax ( x − 3) + B ( x − 3) + Cx 2
Now, there is a variation of the method we used in the first couple of examples that will
work here. There are a couple of values of x that will allow us to quickly get two of the
three constants, but there is no value of x that will just hand us the third.
What we’ll do in this example is pick x’s to get the two constants that we can easily get
and then we’ll just pick another value of x that will be easy to work with (i.e. it won’t
give large/messy numbers anywhere) and then we’ll use the fact that we also know the
other two constants to find the third.
x=0
18 = B ( −3)
⇒ B = −6
x=3

18 = C ( 9 )

⇒

C=2

x =1

18 = A ( −2 ) + B ( −2 ) + C = −2 A + 14

⇒

A = −2

4
3
2
⌠ x − 5 x + 6 x − 18 dx = x − 2 dx − ⌠ − 2 − 6 + 2 dx
⎮
⎮
∫
⌡ x x2 x − 3
x3 − 3x 2
⌡
1
6
= x 2 − 2 x + 2 ln x − − 2 ln x − 3 + c
x
2
In the previous example there were actually two different ways of dealing with the x2 in
the denominator. One is to treat is as a quadratic which would give the following term in
the decomposition
Ax + B
x2
and the other is to treat it as a linear term in the following way,


40


Calculus II

x2 = ( x − 0)

2

which gives the following two terms in the decomposition,
A B
+
x x2
We used the second way of thinking about it in our example. Notice however that the
two will give identical partial fraction decompositions. So, why talk about this? Simple.
This will work for x2, but what about x3 or x4? In these cases we really will need to use
the second way of thinking about these kinds of terms.
x3 ⇒

A B C
+ +
x x 2 x3

x4 ⇒

A B C D
+ + +
x x 2 x3 x 4

Let’s take a look at one more example.
2
⌠ x dx
⎮ 2
⌡ x −1
Solution
In this case the numerator and denominator have the same degree. As with the last
example we’ll need to do long division to get this into the correct form. I’ll leave the
details of that to you to check.
2
⌠ x dx = ⌠ 1 + 1 dx = dx + ⌠ 1 dx
⎮
⎮ 2
∫ ⎮ x2 − 1
⌡
⌡
x2 −1
⌡ x −1

So, we’ll need to partial fraction the second integral. Here’s the decomposition.
1
A
B
=
+
( x − 1)( x + 1) x − 1 x + 1
Setting numerator equal gives,
1 = A ( x + 1) + B ( x − 1)
Picking value of x gives us the following coefficients.
x = −1

1 = B ( −2 )

⇒

B=−

x =1

1 = A ( 2)

⇒

A=

1
2

1
2

2
⌠ x dx = dx + ⌠ 1 − 1 dx
2
2
⎮ 2
∫ ⎮ x −1 x +1
⌡
⌡ x −1
1
1
= x + ln x − 1 − ln x + 1 + c
2
2


41


Calculus II

Integrals Involving Roots
In this section we’re going to look at an integration technique that can be useful for some
integrals with roots in them. We’ve already seen some integrals with roots in them.
Some can be done quickly with a simple Calculus I substitution and some can be done
with trig substitutions.
However, not all integrals with roots will allow us to do either of these methods. Let’s
look at a couple of examples to see another technique that can be used on occasion to
help with these integrals.
⌠ x+2
dx
⎮3
⌡ x −3
Solution
Sometimes when faced with an integral that contains a root we can use the following
substitution to simplify the integral into a form that can be easily worked with.
u = 3 x−3

So, instead of letting u be the stuff under the radical as we often did in Calculus I we let u
be the whole radical. Now, there will be a little more work here since we will also need
to know what x is so we can substitute in for that in the numerator and so we can compute
the differential, dx. This is easy enough to get however. Just solve the substitution for x.
x = u3 + 3
dx = 3u 2 du
Using this substitution the integral is now,
3
⌠ ( u + 3) + 2 2
3u du = ∫ 3u 4 + 15u du
⎮
u
⌡
3
15
= u5 + u2 + c
5
2
5
2
3
15
= ( x − 3) 3 + ( x − 3) 3 + c
5
2
So, sometimes, when an integral contains the root

n

g ( x ) the substitution,

u = n g ( x)

can be used to simplify the integral into a form that we can deal with.
Let’s take a look at another example real quick.
2
⌠
dx
⎮
⌡ x − 3 x + 10


42


Calculus II

Solution
We’ll do the same thing we did in the previous example. Here’s the substitution,
u = x + 10
x = u 2 − 10
dx = 2u du

With this substitution the integral is,
2
2
4u
⌠
dx = ⌠ 2
du
( 2u ) du = ⌠ 2
⎮
⎮
⎮
⌡ u − 10 − 3u
⌡ u − 3u − 10
⌡ x − 3 x + 10
This integral can now be done with partial fractions.
4u
A
B
=
+
( u − 5)( u + 2 ) u − 5 u + 2
Setting numerators equal gives,
4u = A ( u + 2 ) + B ( u − 5 )
Picking value of u gives the coefficients.
u = −2

−8 = B ( −7 )

u =5

20 = A ( 7 )

8
7
20
A=
7

B=

20
8
2
⌠
dx = ⌠ 7 + 7 du
⎮
⎮
⌡ u −5 u + 2
⌡ x − 3 x + 10
20
8
= ln u − 5 + ln u + 2 + c
7
7
20
8
= ln x + 10 − 5 + ln x + 10 + 2 + c
7
7
So, we’ve seen a nice method to eliminate roots from the integral and put into a form that
we can deal with. Note however, that this won’t always work and sometimes the new
integral will be just as difficult to do.

Integrals Involving Quadratics
To this point we’ve seen quite a few integrals that involve quadratics. A couple of
examples are,
1
⌠ 1
⎛x⎞
⌠ x dx = 1 ln x 2 ± a + c
dx = tan −1 ⎜ ⎟
⎮ 2
⎮ 2
2
⌡ x ±a
2
a
⎝a⎠
⌡ x +a
We also saw that integrals involving
done with a trig substitution.


b2 x2 − a 2 ,

43

a 2 − b 2 x 2 and

a 2 + b 2 x 2 could be


Calculus II

Notice however that all of these integrals were missing an x term. They all consist of a
quadratic term and a constant.
Some integrals involving general quadratics are easy enough to do. For instance, the
following integral can be done with a quick substitution.
2x + 3
1 1
⌠
dx = ⌠ du
du = 4 ( 2 x + 3) dx )
⎮ 2
⎮
( u = 4 x 2 + 12 x − 1
⌡ 4 x + 12 x − 1
4⌡ u
1
= ln 4 x 2 + 12 x − 1 + c
4
Some integrals with quadratics can be done with partial fractions. For instance,
⌠ 10 x − 6 dx = ⌠ 4 − 2 dx = 4 ln x + 5 − 2 ln 3 x + 1 + c
⎮ 2
⎮
⌡ 3 x + 16 x + 5
⌡ x + 5 3x + 1
3
Unfortunately, these methods won’t work on a lot of integrals. A simple substitution will
only work if the numerator is a constant multiple of the derivative of the denominator and
partial fractions will only work if the denominator can be factored.
This section is how to deal with integrals involving quadratics when the techniques that
we’ve looked at to this point simply won’t work.
Back in the Trig Substitution section we saw how to deal with square roots that had a
general quadratic in them. Let’s take a quick look at another one like that since the idea
involved in doing that kind integral is exactly what we are going to need for the other
integrals in this section.

∫

x 2 + 4 x + 5 dx

Solution
Recall from the Trig Substitution section that in order to do a trig substitution here we
first needed to complete the square on the quadratic. This gives,
2
x2 + 4 x + 5 = x2 + 4 x + 4 − 4 + 5 = ( x + 2) + 1

After completing the square the integral becomes,

∫

x 2 + 4 x + 5 dx = ∫

( x + 2)

2

+ 1 dx

Upon doing this we can identify the trig substitution that we need. Here it is,
x + 2 = tan θ
x = tan θ − 2
dx = sec2 θ dθ

( x + 2)

2

+ 1 = tan 2 θ + 1 = sec 2 θ = sec θ = sec θ

Recall that since we are doing an indefinite integral we can drop the absolute value bars.


44


Calculus II

∫

x 2 + 4 x + 5 dx = ∫ sec3 θ dθ
=

1
( secθ tan θ + ln secθ + tan θ ) + c
2

We can finish the integral out with the following right triangle.
tan θ =

∫

x+2
1

x 2 + 4 x + 5 dx =

x2 + 4 x + 5
= x2 + 4x + 5
1

sec θ =

1
2

(( x + 2)

)

x 2 + 4 x + 5 + ln x + 2 + x 2 + 4 x + 5 + c

So, by completing the square we were able to take an integral that had a general quadratic
in it and convert it into a form that allowed use a known integration technique.
Let’s do a quick review of completing the square before proceeding. Here is the general
completing the square formula that we’ll use.
2

2

2

b⎞
b2
⎛b⎞ ⎛b⎞
⎛
x + bx + c = x + bx + ⎜ ⎟ − ⎜ ⎟ + c = ⎜ x + ⎟ + c −
2⎠
4
⎝2⎠ ⎝2⎠
⎝
2

2

This will always take a general quadratic and write it in terms of a squared term and a
constant term.
Recall as well that in order to do this we must have a coefficient of one in front of the x2.
If not we’ll need to factor out the coefficient before completing the square. In other
words,
⎛
⎞
⎜
⎟
b
c
ax 2 + bx + c = a ⎜ x 2 + x + ⎟
⎜
a
a⎟
⎜ complete the ⎟
⎝ square on this! ⎠
Now, let’s see how completing the square can be used to do integrals that we aren’t able
to do at this point.


45


Calculus II

1
⌠
dx
⎮ 2
⌡ 2 x − 3x + 2

Solution
Okay, this doesn’t factor so partial fractions just won’t work on this. Likewise, since the
numerator is a “1” we can’t use the substitution u = 2 x 2 − 3 x + 8 . So, let’s see what
happens if we complete the square on the denominator.
3
⎛
⎞
2 x 2 − 3 x + 2 = 2 ⎜ x 2 − x + 1⎟
2
⎝
⎠
3
9 9
⎛
⎞
= 2 ⎜ x 2 − x + − + 1⎟
2
16 16 ⎠
⎝
2
⎛⎛
3⎞
7⎞
2⎜⎜ x − ⎟ + ⎟
⎜⎝
4 ⎠ 16 ⎟
⎝
⎠

With this the integral is,
1
1⌠
1
⌠
dx = ⎮
dx
⎮ 2
7
⌡ 2 x − 3x + 2
2 ⌡ ( x − 3 )2 + 16
4
Now this may not seem like all that great of a change. However, notice that we can now
use the following substitution.
3
u = x−
du = dx
4
and the integral is now,
1
1⌠ 1
⌠
dx = ⎮ 2 7 du
⎮ 2
⌡ 2 x − 3x + 2
2 ⌡ u + 16
We can now see that this is an inverse tangent! So, using the formula from above we get,
1
1 ⎛ 4 ⎞ −1 ⎛ 4u ⎞
⌠
dx = ⎜
tan ⎜
⎮ 2
⎟+c
⌡ 2 x − 3x + 2
2⎝ 7 ⎟
⎠
⎝ 7⎠
2
⎛ 4x − 3 ⎞
tan −1 ⎜
=
⎟+c
7
⎝ 7 ⎠
3x − 1
⌠
dx
⎮ 2
⌡ x + 10 x + 28
Solution
This example is a little different from the previous one. In this case we do have an x in
the numerator however the numerator still isn’t a multiple of the derivative of the
denominator and so a simple Calculus I substitution won’t work.

So, let’s again complete the square on the denominator and see what we get,


46


Calculus II

x 2 + 10 x + 28 = x 2 + 10 x + 25 − 25 + 28 = ( x + 5 ) + 3
2

Upon completing the square the integral becomes,
⌠ 3x − 1
3x − 1
⌠
dx = ⎮
dx
⎮ 2
2
⌡ x + 10 x + 28
⌡ ( x + 5) + 3
At this point we can use the same type of substitution that we did in the previous
example. The only real difference is that we’ll need to make sure that we plug the
substitution back into the numerator as well.
u = x+5
x = u −5
dx = du
3x − 1
⌠ 3 ( u − 5) − 1
⌠
dx = ⎮
du
⎮ 2
2
⌡ x + 10 x + 28
⌡ u +3
3u
16
=⌠ 2
− 2
du
⎮
⌡ u +3 u +3
3
16
⎛ u ⎞
tan −1 ⎜
= ln u 2 + 3 −
⎟+c
2
3
⎝ 3⎠
3
16
⎛ x+5⎞
2
= ln ( x + 5 ) + 3 −
tan −1 ⎜
⎟+c
2
3
⎝ 3 ⎠
So, in general when dealing with an integral in the form,

⌠ Ax + B dx
⎮ 2
⌡ ax + bx + c

(1)

where the denominator doesn’t factor we complete the square on the denominator and
then do a substitution that will yield an inverse tangent and/or a logarithm depending on
the exact form of the numerator.
Let’s now take a look at a couple of integrals that are in the same general form as (1)
except the denominator will also be raised to a power. In other words, let’s look at
integrals in the form,
⌠
Ax + B
(2)
dx
⎮
n
ax 2 + bx + c )
⌡(
⌠
x
dx
⎮ 2
3
x − 6 x + 11)
⌡(
Solution
For the most part this integral will work the same as the previous two with one exception


47


Calculus II

that will occur down the road. So, let’s start by completing the square on the quadratic in
the denominator.
2
x 2 − 6 x + 11 = x 2 − 6 x + 9 − 9 + 11 = ( x − 3) + 2

⌠
⌠
x
x
dx = ⎮
dx
⎮ 2
3
3
2
⎮⎡
⎤
x − 6 x + 11)
⌡(
⌡ ⎣( x − 3 ) + 2 ⎦
Now, we will use the same substitution that we’ve used to this point in the previous two
examples.
u = x −3
x = u +3
dx = du
⌠
⌠ u +3
x
dx = ⎮
du
⎮ 2
3
3
x − 6 x + 11)
u2 + 2)
⌡(
⌡(
⌠
⌠
3
u
du + ⎮
du
=⎮
3
3
u2 + 2)
u2 + 2)
⌡(
⌡(
Now, here is where the differences start cropping up. The first integral can be done with
the substitution v = u 2 + 2 and isn’t too difficult. The second integral however, can’t be
done with the substitution used on the first integral and it isn’t an inverse tangent.
It turns out that a trig substitution will work nicely on the second integral and it will be
the same as we did when we had square roots in the problem.
u = 2 tan θ
du = 2 sec2 θ dθ

With these two substitutions the integrals become,


48


Calculus II

⌠
⌠
1 1
3
x
dx = ⌠ 3 dv + ⎮
⎮
⎮ 2
3
3
2
2⌡ v
⌡ ( x − 6 x + 11)
⌡ ( 2 tan θ + 2 )

(

)

2 sec 2 θ dθ

=−

1 1 ⌠ 3 2 sec 2 θ
dθ
+⎮
4 v 2 ⎮ 8 ( tan 2 θ + 1)3
⌡

=−

1
1
3 2 ⌠ sec 2 θ
dθ
+
⎮
4 ( u 2 + 2 )2
8 ⎮ ( sec 2 θ )3
⌡

=−

1
4

=−

1
4

2

+

3 2⌠ 1
dθ
⎮
8 ⌡ sec 4 θ

2

+

3 2
4
∫ cos θ dθ
8

1

( ( x − 3) + 2 )
2

1

( ( x − 3) + 2 )
2

Okay, at this point we’ve got two options for the remaining integral. We can either use
the ideas we learned in the section about integrals involving trig integrals or we could use
the following formula.
1
m −1
m
m −1
m−2
∫ cos θ dθ = m sin θ cos θ + m ∫ cos θ dθ
Let’s use this formula to do the integral.
1
3
4
3
2
∫ cos θ dθ = 4 sin θ cos θ + 4 ∫ cos θ dθ
1
3⎛1
1
⎞
= sin θ cos3 θ + ⎜ sin θ cos θ + ∫ cos 0 θ dθ ⎟
4
4⎝2
2
⎠
1
3
3
= sin θ cos3 θ + sin θ cos θ + θ
4
8
8
Next, let’s use the following right triangle to get this back to x’s.
u
x−3
x−3
tan θ =
=
sin θ =
cos θ =
2
2
2
( x − 3) + 2


49

cos 0 θ = 1!

2

( x − 3)

2

+2


Calculus II

The cosine integral is then,

∫ cos

4

θ dθ =

1 2 2 ( x − 3)
3 2 ( x − 3) 3 −1 ⎛ x − 3 ⎞
+
+ tan ⎜
⎟
2
4 ( x − 3) 2 + 2
8 ( x − 3) 2 + 2 8
⎝ 2 ⎠

(

2
2

=

)

x −3

( ( x − 3) + 2 )
2

All told then the original integral is,
⌠
x
1
1
dx = −
⎮ 2
3
4 ( x − 3) 2 + 2
⌡ ( x − 6 x + 11)

(

⎛
3 2⎜ 2
8 ⎜ 2
⎜
⎝
=

1
8

+

+

2

( ( x − 3)

( ( x − 3) + 2 )

x −3
3 2
3
⎛ x−3⎞
+ tan −1 ⎜
⎟
2
8 ( x − 3) + 2 8
⎝ 2 ⎠

x −3

3 x − 11
2

)

2

2

+

2

+2

)

2

⎞
x−3
3 2
3 −1 ⎛ x − 3 ⎞ ⎟
+
+ tan ⎜
⎟⎟
8 ( x − 3) 2 + 2 8
⎝ 2 ⎠⎟
⎠

x−3
9
9 2
⎛ x−3⎞
+
tan −1 ⎜
⎟+c
2
32 ( x − 3) + 2 64
⎝ 2 ⎠

It’s a long and messy answer, but there it is.
⌠
x −3
dx
⎮
2
4 − 2x − x2 )
⌡(
Solution
As with the other problems we’ll first complete the square on the denominator.

(

)

4 − 2 x − x 2 = − ( x 2 + 2 x − 4 ) = − ( x 2 + 2 x + 1 − 1 − 4 ) = − ( x + 1) − 5 = 5 − ( x + 1)
2

2

The integral is,

⌠
⌠
x−3
x −3
dx = ⎮
dx
⎮
2 2
2 2
⎮⎡
⌡ ( 4 − 2x − x )
⌡ ⎣5 − ( x + 1) ⎤
⎦
Now, let’s do the substitution.
u = x +1
and the integral is now,


x = u −1

50

dx = du


Calculus II

⌠
⌠ u−4
x−3
dx = ⎮
du
⎮
2
2
4 − 2 x − x2 )
5 − u2 )
⌡(
⌡(
⌠
⌠
u
4
=⎮
du − ⎮
du
2
2
2 2
⌡ (5 − u )
⌡ (5 − u )
In the first integral we’ll use the substitution
v = 5 − u2
and in the second integral we’ll use the following trig substitution
u = 5 sin θ
du = 5 cos θ dθ
Using these substitutions the integral becomes,
⌠
⌠
1 1
4
x −3
dx = − ⌠ 2 dv − ⎮
⎮
⎮
2
2
2
2
2⌡ v
⌡ ( 4 − 2x − x )
⌡ ( 5 − 5sin θ )

(

)

5 cos θ dθ

=

11 4 5⌠
cos θ
−
dθ
⎮
2 v 25 ⌡ (1 − sin 2 θ )2

=

1 1 4 5 ⌠ cos θ
−
dθ
⎮
2 v 25 ⌡ cos 4 θ

=

11 4 5
3
−
∫ sec θ dθ
2 v 25

=

11 2 5
−
( secθ tan θ + ln secθ + tan θ ) + c
2 v 25

We’ll need the following right triangle to finish this integral out.
u
x +1
5
x +1
sin θ =
=
sec θ =
tan θ =
2
2
5
5
5 − ( x + 1)
5 − ( x + 1)

So, going back to x’s the integral becomes,


51


Calculus II

⎛
⌠
x−3
1 1
2 5 ⎜ 5 ( x + 1)
−
+ ln
dx =
⎮
2 2
2 5 − u2
25 ⎜ 5 − ( x + 1)2
⎜
⌡ ( 4 − 2x − x )
⎝
=

1
4x −1
2 5
+
ln
2
10 5 − ( x + 1)
25

5
5 − ( x + 1)

x +1+ 5
5 − ( x + 1)

2

2

+

x +1
5 − ( x + 1)

2

⎞
⎟+c
⎟
⎟
⎠

+c

Often the following formula is needed when using the trig substitution that we used in the
previous example.

∫ sec

m

θ dθ =

1
m−2
m−2
tan θ sec m − 2 θ +
∫ sec θ dθ
m −1
m −1

Note that we’ll only need the two trig substitutions that we used here. The third trig
substitution that we used will not be needed here.

Using Integral Tables

Note : Of all the notes that I’ve written up for download, this is the one section that is
tied to the book that we are currently using here at Lamar University. In this section we
discuss using tables of integrals to help us with some integrals. However, I haven’t had
the time to construct a table of my own and so I will be using the tables given in
Stewart’s Calculus (5th edition). As soon as I get around to writing my own table I’ll post
it online and make any appropriate changes to this section.

So, with that out of the way let’s get on with this section.
This section is entitled Using Integral Tables and we will be using integral tables.
However, at some level, this isn’t really the point of this section. To a certain extent the
real subject of this section is how to take advantage of known integrals to do integrals
that may not look like anything the ones that we do know how to do or are given in a
table of integrals.
For the most part we’ll be doing this by using substitution to put integrals into a form that
we can deal with. However, not all of the integrals will require a substitution. For some
integrals all that we need to do is a little rewriting of the integrand to get into a form that
we can deal with.
We’ve already related a new integral to one we could deal with least once. In the last
example in the Trig Substitution section we looked at the following integral.


52


Calculus II

∫e

1 + e 2 x dx

4x

At first glance this looks nothing like a trig substitution problem. However, with the
substitution u = e x we could turn the integral into,

∫u

1 + u 2 du

3

which definitely is a trig substitution problem ( u = tan θ ). We actually did this process
in a single step by using e x = tan θ , but the point is that with a substitution we were able
to convert an integral into a form that we could deal with.
So, let’s work a couple examples using substitutions and tables.

⌠ 7 + 9x 2
dx
⎮
x2
⌡
Solution
So, the first thing we should do is go to the tables and see if there is anything in the tables
that is close to this. In the tables in Stewart we find the following integral,

(

)

a2 + u2
⌠ a2 + u 2
du = −
+ ln u + a 2 + u 2 + c
⎮
u2
u
⌡

This is nearly what we’ve got in our integral. The only real difference is that we’ve got a
coefficient in front of the x2 and the formula doesn’t. This is easily enough dealt with.
All we need to do is the following manipulation on the integrand.
2
2
⌠ 9( 7 + x )
9
⌠3 7+x
⌠
⌠ 7 + 9 x2
dx = ⎮
dx = ⎮ 9 2
dx = 3⎮
⎮
2
2
x
x
x
⌡
⌡
⌡
⌡

+ x2
x2

dx

7
.
3

So, we can now use the formula with a =
⎛
⌠ 7 + 9 x2
dx = 3 ⎜ −
⎮
⎜
x2
⌡
⎝

7
9

7
9

+ x2
x

⎛
⎞⎞
7
+ ln ⎜ x +
+ x2 ⎟ ⎟ + c
⎜
⎟⎟
9
⎝
⎠⎠

cos x
⌠
dx
⎮
⌡ sin x 3sin x − 9
Solution
Going through our tables we aren’t going to find anything that looks like this in them.
However, notice that with the substitution u = sin x we can rewrite the integral as,
cos x
1
⌠
⌠
dx = ⎮
du
⎮
⌡ u 3u − 9


53


Calculus II

and this is in the tables.
1
1
⌠
du =
ln
⎮
⌡ u a + bu
a
=

a + bu − a
+c
a + bu + a

⎛ a + bu
2
tan −1 ⎜
⎜
−a
−a
⎝

⎞
⎟+c
⎟
⎠

if a > 0
if a < 0

Notice that this is a formula that will depend upon the value of a. This will happen on
occasion. In our case we have a=-9 and b=3 so we’ll use the second formula.
⎛ 3u − 9 ⎞
cos x
2
⌠
dx =
tan −1 ⎜
⎟+c
⎮
⎜ − ( −9 ) ⎟
− ( −9 )
⎝
⎠
=

⎛ 3sin x − 9 ⎞
2
tan −1 ⎜
⎟+c
⎜
⎟
3
9
⎝
⎠

This final example uses a type of formula known as a reduction formula.
⌠ 4⎛ x⎞
⎮ cot ⎜ ⎟ dx
⌡
⎝2⎠
Solution
x
We’ll first need to use the substitution u = since none of the formulas in our tables
2
have that in them. Doing this gives,
⌠ 4⎛x⎞
4
⎮ cot ⎜ ⎟ dx = 2∫ cot u du
2⎠
⌡
⎝

To help us with this integral we’ll use the following formula.
−1
n
n −1
n−2
∫ cot u du = n − 1 cot u − ∫ cot u du
Formulas like this are called reduction formulas. Reduction formulas generally don’t
explicitly give the integral. Instead they reduce the integral to an easier one. In fact they
often reduce the integral to a different version of itself!
For our integral we’ll use n=4.
⌠ 4⎛ x⎞
⎛ 1 3
⎞
2
⎮ cot ⎜ ⎟ dx = 2 ⎜ − cot u − ∫ cot u du ⎟
⌡
⎝2⎠
⎝ 3
⎠
At this stage we can either reuse the reduction formula with n=2 or use the formula
2
∫ cot u du = − cot u − u + c


54


Calculus II

We’ll reuse the reduction formula with n=2 so we can address something that happens on
occasion.
⎛ 1 3
⌠ 4⎛ x⎞
⎛ 1
⎞⎞
0
cot 0 u = 1!
⎮ cot ⎜ ⎟ dx = 2 ⎜ − cot u − ⎜ − cot u − ∫ cot u du ⎟ ⎟
⌡
⎝2⎠
⎝ 1
⎠⎠
⎝ 3
2
= − cot 3 u + 2 cot u + 2∫ du
3
2
= − cot 3 u + 2 cot u + 2u + c
3
2
⎛x⎞
⎛ x⎞
= − cot 3 ⎜ ⎟ + 2 cot ⎜ ⎟ + x + c
3
⎝2⎠
⎝2⎠
Don’t forget that a 0 = 1 . Often people forget that and then get stuck on the final integral!
There really wasn’t a lot to this section. Just don’t forget that sometimes a simple
substitution or rewrite of an integral can take it from undoable to doable.

Integration Strategy
We’ve now seen a fair number of different integration techniques and so we should
probably pause at this point and talk a little bit about a strategy to use for determining the
correct technique to use when faced with an integral.
There are a couple of points that need to be made about this strategy. First, it isn’t a hard
and fast set of rules for determining the method that should be used. It is really nothing
more than a general set of guidelines that will help us to identify techniques that may
work. Some integrals can be done in more than one way and so depending on the path
you take through the strategy you may end up with a different technique than somebody
else who also went through this strategy.
Second, while the strategy is presented as a way to identify the technique that could be
used on an integral also keep in mind that, for many integrals, it can also automatically
exclude certain techniques as well. When going through the strategy keep two lists in
mind. The first list is integration techniques that simply won’t work and the second list is
techniques that look like they might work. After going through the strategy and the
second list has only one entry then that is the technique to use. If, on the other hand,
there are more than one possible technique to use we will then have to decide on which is
liable to be the best for us to use. Unfortunately there is no way to teach which technique
is the best as that usually depends upon the person and which technique they find to be
the easiest.
Third, don’t forget that many integrals can be evaluated in multiple ways and so more
than one technique may be used on it. This has already been mentioned in each of the
previous points, but is important enough to warrant a separate mention. Sometimes one


55


Calculus II

technique will be significantly easier than the others and so don’t just stop at the first
technique that appears to work. Always identify all possible techniques and then go back
and determine which you feel will be the easiest for you to use.
Next, it’s entirely possible that you will need to use more than one method to completely
do an integral. For instance a substitution may lead to using integration by parts or
partial fractions integral.
Finally, in my class I will accept any valid integration technique as a solution. As already
noted there is often more than one way to do an integral and just because I find one
technique to be the easiest doesn’t mean that you will as well. So, in my class, there is no
one right way of doing an integral. You may use any integration technique that I’ve
taught you in this class or you learned in Calculus I to integrals in this class. In other
words, always take the approach that you find to be the easiest.
Note that this final point is more geared towards my class and it’s completely possible
that your instructor may not agree with this and so be careful in applying this point if you
aren’t in my class.
Okay, let’s get on with the strategy.

1. Simplify the integrand, if possible. This step is very important in the integration
process. Many integrals can be taken from impossible or very difficult to very
easy with a little simplification or manipulation. Don’t forget basic trig and
algebraic identities as these can often be used to simplify the integral.
We used this idea when we were looking at integrals involving trig functions. For
example consider the following integral.
2
∫ cos x dx
This integral can’t be done as is however, simply by recalling the identity,
1
cos 2 x = (1 + cos ( 2 x ) )
2
the integral becomes very easy to do.
Note that this example also shows that simplification does not necessarily mean
that we’ll write the integrand in a “simpler” form. It only means that we’ll write
the integrand into a form that we can deal with and this is often longer and/or
“messier” than the original integral.
2. See if a “simple” substitution will work. Look to see if a simple substitution
can be used instead of the often more complicated methods from Calculus II. For
example consider both if the following integrals.
2
⌠ x dx
⎮ 2
∫ x x − 1 dx
⌡ x −1


56


Calculus II

The first integral can be done with partial fractions and the second could be done
with a trig substitution.
However, both could also be evaluated using the substitution u = x 2 − 1 and the
work involved in the substitution would be significantly less than the work
involved in either partial fractions or trig substitution.
So, always look for quick, simple substitutions before moving on to the more
complicated Calculus II techniques.
3. Identify the type of integral. Note that any integral may fall into more than one
of these types. Because of this fact it’s usually best to go all the way through the
list and identify all possible types since one may be easier than the other and it’s
entirely possible that the easier type is listed lower in the list.
a. Is the integrand a rational expression (i.e is the integrand a polynomial
divided by a polynomial)? If so, then partial fractions may work on the
integral.
b. Is the integrand a polynomial times a trig function, exponential, or
logarithm? If so, then integration by parts may work.
c. Is the integrand a product of sines and cosines, secant and tangents, or
cosecants and cotangents? If so, then the topics from the second section
may work.
Likewise, don’t forget that some quotients involving these functions can
also be done using these techniques.
d. Does the integrand involve b 2 x 2 + a 2 , b 2 x 2 − a 2 , or a 2 − b 2 x 2 ? If
so, then a trig substitution might work nicely.
e. Does the integrand have roots other than those listed above in it? If so,
then the substitution u = n g ( x ) might work.
f. Does the integrand have a quadratic in it? If so, then completing the
square on the quadratic might put it into a form that we can deal with.
4. Can we relate the integral to an integral we already know how to do? In
other words, can we use a substitution or manipulation to write the integrand into
a form that does fit into the forms we’ve looked at previously in this chapter.
A typical example here is the following integral.

∫ cos x

1 + sin 2 x dx

This integral doesn’t obviously fit into any of the forms we looked at in this
chapter. However, with the substitution u = sin x we can reduce the integral to
the form,

∫

1 + u 2 du

which is a trig substitution problem.


57


Calculus II

5. Do we need to use multiple techniques? In this step we need to ask ourselves if
it is possible that we’ll need to use multiple techniques. The example in the
previous part is a good example. Using a substitution didn’t allow us to actually
do the integral. All it did was put the integral and put it into a form that we could
use a different technique on.
Don’t ever get locked into the idea that an integral will only require one step to
completely evaluate it. Many will require more than one step.
6. Try again. If everything that you’ve tried to this point doesn’t work then go back
through the process and try again. This time try a technique that that you didn’t
use the first time around.

As noted above this strategy is not a hard and fast set of rules. It is only intended to
guide you through the process of best determining how to do any given integral. Note as
well that the only place Calculus II actually arises is in the third step. Steps 1, 2 and 4
involve nothing more than manipulation of the integrand either through direct
manipulation of the integrand or by using a substitution. The last two steps are simply
ideas to think about in going through this strategy.
Many students go through this process and concentrate almost exclusively on Step 3
(after all this is Calculus II, so it’s easy to see why they might do that….) to the exclusion
of the other steps. One very large consequence of that exclusion is that often a simple
manipulation or substitution is overlooked that could make the integral very easy to do.
Before moving on to the next section we should work a couple of quick problems
illustrating a couple of not so obvious simplifications/manipulations and a not so obvious
substitution.
⌠ tan x dx
⎮
⌡ sec 4 x
Solution
This integral almost falls into the form given in 3c. It is a quotient of tangent and secant
and we know that sometimes we can use the same methods for products of tangents and
secants on quotients.

The process from that section tells us that if we have even powers of secant to strip two
of them off and convert the rest to tangents. That won’t work here. We can split two
secants off, but they would be in the denominator and they won’t do us any good there.
Remember that the point of splitting them off is so they would be there for the
substitution u = tan x . That requires them to be in the numerator.
So, that won’t work and so we’ll have to find another solution method.


58


Calculus II

There are in fact two solution methods to this integral depending on how you want to go
about it. We’ll take a look at both.
Solution 1
In this solution method we could just convert everything to sines and cosines and see if
that gives us an integral we can deal with.
⌠ tan x dx = ⌠ sin x cos 4 x dx
⎮
⎮
⌡ sec 4 x
⌡ cos x
= ∫ sin x cos3 x dx

u = cos x

= − ∫ u 3 du

1
= − cos 4 x + c
4
Note that just converting to sines and cosines won’t always work and if it does it won’t
always work this nicely. Often there will be a lot more work that would need to be done
to complete the integral.
Solution 2
This solution method goes back to dealing with secants and tangents. Let’s notice that if
we had a secant in the numerator we could just use u = sec x as a substitution and it
would be a fairly quick and simple substitution to use. We don’t have a secant in the
numerator. However we could very easily get a secant in the numerator simply by
multiplying the numerator and denominator by secant.
⌠ tan x dx = ⌠ tan x sec x dx
u = sec x
⎮
⎮
⌡ sec 4 x
⌡ sec5 x
1
= ⌠ 5 du
⎮
⌡u
1 1
=−
+c
4 sec 4 x
1
= − cos 4 x + c
4
In the previous example we saw two “simplifications” that allowed us to do the integral.
The first was using identities to rewrite the integral into terms we could deal with and the
second involved multiplying the numerator and the denominator by something to again
put the integral into terms we could deal with.
Using identities to rewrite an integral is an important “simplification” and we should not
forget about it. Integrals can often be greatly simplified or at least put into a form that
can be dealt with by using an identity.
The second “simplification” is not used as often, but does show up on occasion so again,
it’s best to not forget about it. In fact, let’s take another look at an example in which
multiplying the numerator and denominator by something will allow us to do an integral.


59


Calculus II

1
⌠
dx
⎮
⌡ 1 + sin x

Solution
This is an integral in which if we just concentrate on the third step we won’t get
anywhere. This integral doesn’t appear to be any of the kinds of integrals that we worked
in this chapter.

We can do the integral however, if we do the following,
1
1 1 − sin x
⌠
dx = ⌠
dx
⎮
⎮
⌡ 1 + sin x
⌡ 1 + sin x 1 − sin x
1 − sin x
=⌠
dx
⎮
⌡ 1 − sin 2 x
This does not appear to have done anything for us. However, if we now remember the
first “simplification” we looked at above we will notice that we can use an identity to
rewrite the denominator. Once we do that we can further reduce the integral into
something we can deal with.
1
1 − sin x
⌠
dx = ⌠
dx
⎮
⎮
⌡ 1 + sin x
⌡ cos 2 x
1
sin x 1
dx
=⌠
−
⎮
2
⌡ cos x cos x cos x
= ∫ sec 2 x − tan x sec x dx
= tan x − sec x + c

So, we’ve seen once again that multiplying the numerator and denominator by something
can put the integral into a form that we can integrate. Notice as well that this example
also showed that “simplifications” do not necessarily put an integral into a simpler form.
They only put the integral into a form that is easier to integrate.
Let’s now take a quick look at an example of a substitution that is not so obvious.
∫ cos x dx

( )

Solution
We introduced this example saying that the substitution was not so obvious. However,
this is really an integral that falls into the form given by 3e in our strategy above.
However, many people miss that form and so don’t think about it. So, let’s try the
following substitution.
u= x
x = u2
dx = 2u du


60


Calculus II

With this substitution the integral becomes,
∫ cos x dx = 2∫ u cos u du

( )

This is now an integration by parts integral. Remember that often we will need to use
more than one technique to completely do the integral. This is a fairly simple integration
by parts problem so I’ll leave the remainder of the details to you to check.

∫ cos ( x ) dx = 2 ( cos ( x ) +

x sin

( x )) + c

Before leaving this section we should also point out that there are integrals out there in
the world that just can’t be done in terms of functions that we know. Some examples of
these are.
2
⌠ sin ( x )
x
e − x dx
cos ( x 2 ) dx
dx
⎮
∫
∫
∫ cos ( e ) dx
⌡ x
That doesn’t mean that these integrals can’t be done at some level. If you go to a
computer algebra system such as Maple and have it do these integrals here is what it will
return the following.

∫e

− x2

dx =

π
2

∫ cos ( x ) dx =
2

erf ( x )
⎛ 2⎞
FresnelC ⎜ x
⎜ π ⎟
⎟
2
⎝
⎠

π

⌠ sin ( x )
dx = Si ( x )
⎮
⌡ x

∫ cos ( e ) dx = Ci ( e )
x

x

So it appears that these integrals can in fact be done. However this is a little misleading.
Here are the definitions of each of the functions given above.
Error Function

erf ( x ) =

2

π

∫

x

0

e −t dt
2

The Sine Integral
x
⌠ sin t dt
Si ( x ) = ⎮
⌡0 t

The Fresnel Cosine Integral

⌠
⎛π ⎞
FresnelC ( x ) = ⎮ cos ⎜ t 2 ⎟ dt
⌡0
⎝2 ⎠
x

The Cosine Integral


61


Calculus II

cos t − 1
Ci ( x ) = γ + ln ( x ) + ⌠
dt
⎮
⌡0
t
Where γ is the Euler-Mascheroni constant.
x

Note that the first three are simply defined in terms of themselves and so when we say we
can integrate them all we are really doing is renaming the integral. The fourth one is a
little different and yet it is still defined in terms of an integral that can’t be done in
practice.
It will be possible to integrate every integral given in this class, but it is important to note
that there are integrals that just can’t be done. We should also note that after we look at
Series we will be able to write down series representations of each of the integrals above.

Improper Integrals
In this section we need to take a look at a couple of different kinds of integrals. Both of
these are examples of integrals that are called Improper Integrals.
Let’s start with the first kind of improper integrals that we’re going to take a look at.

Infinite Interval
In this kind of integrals we are going to take a look at integrals that in which one or both
of the limits of integration are infinity. In these cases the interval of integration is said to
be over an infinite interval.
Let’s take a look at an example that will also show us how we are going to deal with
these integrals.
∞
⌠ 1 dx
⎮ 2
⌡1 x

Solution
This is an innocent enough looking integral. However, because infinity is not a real
number we can’t just integrate as normal and then “plug in” the infinity to get an answer.

To see how we’re going to do this integral let’s think of this as an area problem. So
1
instead of asking what the integral is, let’s instead ask what the area under f ( x ) = 2 on
x
the interval [1, ∞ ) is.
We still aren’t able to do this, however, let’s step back a little and instead ask what the
area under f(x) is on the interval [1,t ] were t > 1 and t is finite. This is a problem that we


62


Calculus II

can do.
t

t
1
1
1
At = ⌠ 2 dx = −
= 1−
⎮
⌡1 x
x1
t

Now, we can get the area under f(x) on [1, ∞ ) simply by taking the limit of At as t goes to
infinity.

⎛ 1⎞
A = lim At = lim ⎜ 1 − ⎟ = 1
t →∞
t →∞
⎝ t⎠
This is then how we well do the integral it self.
∞
t
⌠ 1 dx = lim ⌠ 1 dx
⎮ 2
⎮ 2
t →∞ ⌡ x
⌡1 x
1
t

⎛ 1⎞
= lim ⎜ − ⎟
t →∞
⎝ x ⎠1
⎛ 1⎞
= lim ⎜ 1 − ⎟
t →∞
⎝ t⎠
=1
So, this is how we will deal with these kinds of integrals in general. We will replace the
infinity with a variable (usually t), do the integral and then take the limit of the result as t
goes to infinity.
On a side note, notice that the area under a curve on an infinite interval was not infinity
as we might have suspected it to be. In fact, it was a surprisingly small number. Of
course this won’t always be the case, but it is important enough to point out that not all
areas on an infinite interval will yield infinite areas.
Let’s now get some definitions out of the way. We will call these integrals convergent if
the associated limit exists and is a finite number (i.e. it’s not plus or minus infinity) and
divergent if the associated limits either doesn’t exist or is (plus or minus) infinity.
Let’s now formalize up the method for dealing with infinite intervals. There are
essentially three cases that we’ll need to look at.

1. If

∫ f ( x ) dx
t

a

exists for every t > a then,

∫

∞
a

f ( x ) dx = lim ∫ f ( x ) dx
t

t →∞

a

provided the limit exists and is finite.


63


Calculus II

2. If

∫ f ( x ) dx
b

t

exists for every t < b then,

∫ ∞ f ( x ) dx = lim ∫ f ( x ) dx
b

b

−

t →−∞

t

provided the limits exists and is finite.
3. If

c

∫∞
−

f ( x ) dx and

∫

∞
c

f ( x ) dx are both convergent then,

∞

c

∞

−

−

c

∫ ∞ f ( x ) dx = ∫ ∞ f ( x ) dx + ∫

f ( x ) dx

Where c is any number. Note as well that this requires BOTH of the integrals to
be convergent in order for this integral to also be convergent. If either of the two
integrals is divergent then so is this integral.

Let’s take a look at a couple more examples.
Example 2 Determine if the follow integral is convergent or divergent and if it’s
convergent find it’s value.
∞
⌠ 1 dx
⎮
⌡1 x
Solution
So, the first thing we do is convert the integral to a limit.
∞
t
⌠ 1 dx = lim ⌠ 1 dx
⎮
⎮
t →∞ ⌡ x
⌡1 x
1
Now, do the integral and the limit.
∞
⌠ 1 dx = lim ln ( x ) t
⎮
1
t →∞
⌡1 x
= lim ( ln ( t ) − ln1)
t →∞

=∞

So, the limit is infinite and so the integral is divergent.
If we go back to thinking in terms of area notice that the area under g ( x ) =

1
on the
x

1
which was
x2
quite small. There really isn’t all that much difference between these two functions and
yet there is a large difference in the area under them. We can actually extend this out to
the following fact.
interval [1, ∞ ) is infinite. This is in contrast to the area under f ( x ) =

Fact
If a>0 then


64


Calculus II

∞
⌠ 1 dx
⎮ p
⌡a x
is convergent if p > 1 and divergent if p ≤ 1 .

One thing to note about this fact is that it’s in essence saying that if an integrand goes to
zero fast enough then the integral will converge. How fast is fast enough? If we use this
1
fact as a guide it looks like integrands that go to zero faster than goes to zero will
x
probably converge.
Example 3 Determine if the following integral is convergent or divergent. If it is
convergent find its value.
0
1
⌠
dx
⎮
⌡ −∞ 3 − x
Solution
There really isn’t much to do with these problems once you know how to do them. We’ll
convert the integral to a limit/integral pair, evaluate the integral and then the limit.
0
0
1
1
⌠
⌠
dx = lim ⎮
dx
⎮
t →−∞ ⌡
⌡ −∞ 3 − x
3− x
t
= lim −2 3 − x
t →−∞

(

0
t

= lim −2 3 + 2 3 − t
t →−∞

)

= −2 3 + ∞
=∞

So, the limit is infinite and so this integral is divergent.

∫

∞
−∞

2

xe − x dx

Solution
In this case we’ve got infinities in both limits and so we’ll need to split the integral up
into two separate integrals. We can split the integral up at any point, so let’s choose a=0
since this will be a convenient point for the evaluate process. The integral is then,

∫

∞
−∞

xe − x dx = ∫
2

0
−∞

∞

xe − x dx + ∫ xe− x dx
2

2

0

We’ve now got to look at each of the individual limits.


65


Calculus II

∫

0

0

xe − x dx = lim ∫ xe − x dx
2

−∞

t →−∞

2

t

0

⎛ 1 2⎞
= lim ⎜ − e − x ⎟
t →−∞
⎝ 2
⎠t

⎛ 1 1 2⎞
= lim ⎜ − + e − t ⎟
t →−∞
⎝ 2 2
⎠
1
=−
2
So, the first integral is convergent. Note that this does NOT mean that the second
integral will also be convergent. So, let’s take a look at that one.

∫

∞

t

xe − x dx = lim ∫ xe − x dx
2

t →∞

0

2

0

t

⎛ 1 2⎞
= lim ⎜ − e − x ⎟
t →∞
⎝ 2
⎠0
⎛ 1 2 1⎞
= lim ⎜ − e −t + ⎟
t →∞
2⎠
⎝ 2
1
=
2
This integral is convergent and so since they are both convergent the integral we were
actually asked to deal with is also convergent and its value is,
∞
0
∞
2
2
2
1 1
xe − x dx = ∫ xe − x dx + ∫ xe − x dx = − + = 0
∫ −∞
0
−∞
2 2

∫

∞
−2

sin x dx

Solution
First convert to a limit.

∫

∞
−2

t

sin x dx = lim ∫ sin x dx
t →∞

−2

= lim ( − cos x ) −2
t

t →∞

= lim ( cos 2 − cos t )
t →∞

This limit doesn’t exist and so the integral is divergent.
In most examples in a Calculus II class that are worked over infinite intervals the limit
either exists or is infinite. However, there are limits that don’t exist, as the previous
example showed, so don’t forget about those.

Discontinuous Integrand

66


Calculus II

We now need to look at the second type of improper integrals that we’ll be looking at in
this section. These are integrals that have discontinuous integrands. The process here is
basically the same with one on subtle difference. Here are the general cases that we’ll
look at for these integrals.
1. If f(x) is continuous on the interval [ a, b ) and not continuous at x=b then,

∫

b
a

f ( x ) dx = lim ∫ f ( x ) dx
−
t

t →b

a

provided the limit exists and is finite. Note as well that we do need to use a left
hand limit here since the interval of integration is entirely on the left side of the
upper limit.
2. If f(x) is continuous on the interval ( a, b ] and not continuous at x=a then,

∫

b
a

f ( x ) dx = lim ∫ f ( x ) dx
+
b

t →a

t

provided the limit exists and is finite. In this case we need to use a right hand
limit here since the interval of integration is entirely on the right side of the lower
limit.
3. If f(x) is not continuous at x=c where a < c < b and

∫ f ( x ) dx
c

a

and

∫

b
c

f ( x ) dx

are both convergent then,

∫

b
a

f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
c

b

a

c

As with the infinite interval case this requires BOTH of the integrals to be
convergent in order for this integral to also be convergent. If either of the two
integrals is divergent then so is this integral.

∫

4. If f(x) is not continuous at x=a and x=b and if

c
a

f ( x ) dx and

∫

b
c

f ( x ) dx are

both convergent then,

∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
b

c

b

a

a

c

Where c is any number. Again, this requires BOTH of the integrals to be
convergent in order for this integral to also be convergent.

Note that the limits in these cases really do need to be right or left handed limits. Since
we will be working inside the interval of integration we will need to make sure that we
stay inside that interval. This means that we’ll used one-sided limits to make sure we
stay inside the interval.
Let’s do a couple of examples of these kinds of integrals.


67


Calculus II

3
1
⌠
dx
⎮
⌡0 3 − x
Solution
The problem point is the upper limit so we are in the first case above.
3
t
1
1
⌠
⌠
dx = lim ⎮
dx
⎮
t →3− ⌡
⌡0 3 − x
3− x
0

(
= lim ( 2

= lim −2 3 − x
−
t →3

t →3−

)

t

0

3 − 2 3−t

)

=2 3
The limit exists and is finite and so the integral converges and the integrals value is 2 3 .
3
⌠ 1 dx
⎮ 3
⌡ −2 x
Solution
This integrand is not continuous at x=0 and so we’ll need to split the integral up at that
point.
3
0
3
⌠ 1 dx = ⌠ 1 dx + ⌠ 1 dx
⎮ 3
⎮ 3
⎮ 3
⌡ −2 x
⌡ −2 x
⌡0 x

Now we need to look at each of these integrals and see if they are convergent.
0
t
⌠ 1 dx = lim ⌠ 1 dx
⎮ 3
⎮ 3
t → 0− ⌡ − 2 x
⌡ −2 x
t

⎛ 1 ⎞
= lim ⎜ − 2 ⎟
−
t →0 ⎝
2 x ⎠ −2
⎛ 1 1⎞
= lim ⎜ − 2 + ⎟
−
t →0 ⎝
2t 8 ⎠
= −∞
At this point we’re done. One of the integrals is divergent that means the integral that we
were asked to look at is divergent. We don’t even need to bother with the second
integral.
Before leaving this section lets note that we can also have integrals that involve both of
these cases. Consider the following integral.

68


Calculus II

∞
⌠ 1 dx
⎮ 2
⌡0 x

Solution
This is an integral over an infinite interval that also contains a discontinuous integrand.
To do this integral we’ll need to split it up into two integrals. We can split it up
anywhere, but pick a value that will be convenient for evaluation purposes.
∞
1
∞
⌠ 1 dx = ⌠ 1 dx + ⌠ 1 dx
⎮ 2
⎮ 2
⎮ 2
⌡0 x
⌡0 x
⌡1 x

In order for the integral in the example to be convergent we will need BOTH of these to
be convergent. If one or both are divergent then the whole integral will also be divergent.
We know that the second integral is convergent by the fact given in the infinite interval
portion above. So, all we need to do is check the first integral.
1
1
⌠ 1 dx = lim ⌠ 1 dx
⎮ 2
⎮ 2
t → 0+ ⌡ t x
⌡0 x
1

⎛ 1⎞
= lim ⎜ − ⎟
+
t →0 ⎝
x⎠t

1⎞
⎛
= lim ⎜ −1 + ⎟
+
t →0 ⎝
t⎠
=∞
So, the first integral is divergent and so the whole integral is divergent.

Comparison Test for Improper Integrals
Now that we’ve seen how to actually compute improper integrals we need to address one
more topic about them. Often we aren’t concerned with the actual value of these
integrals. Instead we might only be interested in whether the integral is convergent or
divergent. Also, there will be some integrals that we simply won’t be able to integrate
and yet we would still like to know if they converge or diverge.
To deal with this we’ve got a test for convergence or divergence that we can use to help
us answer the question of convergence for an improper integral.
We will give this test only for a sub-case of the infinite interval integral, however
versions of the test exist for the other sub-cases of the infinite interval integrals as well as
integrals with discontinuous integrands.
Comparison Test


69


Calculus II

If f ( x ) ≥ g ( x ) ≥ 0 on the interval [ a, ∞ ) then,
1. If
2. If

∫

∞

a

f ( x ) dx converges then so does

∞

∫ g ( x ) dx .
a

∞

∞

a

a

∫ g ( x ) dx diverges then so does ∫

f ( x ) dx .

Note that if you think in terms of area the Comparison Test makes a lot of sense. If f(x) is
larger than g(x) then the area under f(x) must also be larger than the area under g(x).

∫

So, if the area under the larger function is finite (i.e.
under the smaller function must also be finite (i.e.

∞

a

f ( x ) dx converges) then the area

∞

∫ g ( x ) dx converges).
a

Likewise, if the area under the smaller function is infinite (i.e.
the area under the larger function must also be infinite (i.e.

∫

∞

a

∞

∫ g ( x ) dx diverges) then
a

f ( x ) dx diverges).

Be careful not to misuse this test. If the smaller function converges there is no reason to
believe that the larger will also converge (after all infinity is larger than a finite
number…) and if the larger function diverges there is no reason to believe that the
smaller function will also converge.
Let’s work a couple of example using the comparison test. Note that all we’ll be able to
do is determine the convergence of the integral. We won’t be able to determine the value
of the integrals and so won’t even bother with that.
Example 1 Determine if the following integral is convergent or divergent.
∞
2
⌠ cos x dx
⎮
⌡ 2 x2
Solution
Let’s take a second and think about how the Comparison Test works. If this integral is
convergent then we’ll need to find a larger function that also converges on the same
interval. Likewise, if this integral is divergent then we’ll need to find a smaller function
that also diverges.

So, it seems like it would be nice to have some idea as to whether the integral converges
or diverges ahead of time so we will know whether we will need to look for a larger (and
convergent) function or a smaller (and divergent) function.
To get the guess for this function let’s notice that the numerator is nice and bounded and
simply won’t get too large. Therefore, it seems likely that the denominator will
determine the convergence/divergence of this integral and we know that


70


Calculus II

∞

⌠ 1 dx
⎮ 2
⌡2 x
converges since p=2>1 by the fact in the previous section. So let’s guess that this
integral will converge.
So we now know that we need to find a function that is larger than
cos 2 x
x2
and also converges. Making a fraction larger is actually a fairly simple process. We can
either make the numerator larger or we can make the denominator smaller. In this case
can’t do a lot about the denominator. However we can use the fact that 0 ≤ cos 2 x ≤ 1 to
make the numerator larger (i.e. we’ll replace the cosine with something we know to be
larger, namely 1). So,
cos 2 x 1
≤ 2
x2
x
Now, as we’ve already noted
∞
⌠ 1 dx
⎮ 2
⌡2 x
converges and so by the Comparison Test we know that
∞
2
⌠ cos x dx
⎮
⌡ 2 x2
must also converge.
∞
1
⌠
dx
⎮
⌡3 x + ex
Solution
Let’s first take a guess about the convergence of this integral. As noted after the fact in
the last section about
∞
⌠ 1 dx
⎮ p
⌡a x
1
if the integrand goes to zero faster than then the integral will probably converge.
x
Now, we’ve got an exponential in the denominator which is approaching infinity much
faster than the x and so it looks like this integral should probably converge.

So, we need a larger function that will also converge. In this case we can’t really make
the numerator larger and so we’ll need to make the denominator smaller in order to make
the function larger as a whole. We will need to be careful however. There are two ways
to do this and only one, in this case only one, of them will work for us.
First, notice that since the lower limit of integration is 3 we can say that x ≥ 3 > 0 and we


71


Calculus II

know that exponentials are always positive. So, the denominator is the sum of two
positive terms and if we were to drop one of them the denominator would get smaller.
This would in turn make the function larger.
The question then is which one to drop? Let’s first drop the exponential. Doing this
gives,
1
1
<
x
x+e
x
This is a problem however, since
∞
⌠ 1 dx
⎮
⌡3 x
diverges by the fact. We’ve got a larger function that is divergent. This doesn’t say
anything about the smaller function. Therefore, we chose the wrong one to drop.
Let’s try it again and this time let’s drop the x.
1
1
< x = e− x
x
e
x+e
Also,

∫

∞

3

t

e − x dx = lim ∫ e − x dx
t →∞

3

= lim ( −e − t + e −3 )
t →∞

= e −3
So,

∫

∞

3

e − x dx is convergent. Therefore, by the Comparison test
∞
1
⌠
dx
⎮
⌡3 x + ex

is also convergent.
∞
1
⌠
dx
⎮
⌡ 3 x − e− x
Solution
This is very similar to the previous example with a couple of very important differences.
First, notice that the exponential now goes to zero as x increases instead of growing larger
as it did in the previous example (because of the negative in the exponent). Also note
that the exponential is now subtracted off the x instead of added onto it.

The fact that the exponential goes to zero means that this time the x in the denominator
will probably dominate the term and that means that the integral probably diverges. We
will therefore need to find a smaller function that also diverges.
Making fractions smaller is pretty much the same as making fractions larger. In this case
we’ll need to either make the numerator smaller or the denominator larger.


72


Calculus II

This is where the second change will come into play. As before we know that both x and
the exponential are positive. However, this time since we are subtracting the exponential
from the x if we were to drop the exponential the denominator will become larger and so
the fraction will become smaller. In other words,
1
1
>
−x
x−e
x
and we know that
∞
⌠ 1 dx
⎮
⌡3 x
diverges and so by the Comparison Test we know that
∞
1
⌠
dx
⎮
⌡ 3 x − e− x
must also diverge.
∞
4
⌠ 1 + 3sin ( 2x )
dx
⎮
x
⌡1
Solution
First notice that as with the first example, the numerator in this function is going to be
bounded since the sine is never larger than 1. Therefore, since the exponent on the
denominator is less than 1 we can guess that the integral will probably diverge. We will
need a smaller function that also diverges.

We know that 0 ≤ sin 4 ( 2 x ) ≤ 1 . In particular, this term is positive and so if we drop it
from the numerator the numerator will get smaller. This gives,
1 + 3sin 4 ( 2 x )
1
>
x
x
and
∞
⌠ 1
dx
⎮
⌡1 x
diverges so by the Comparison Test
∞
4
⌠ 1 + 3sin ( 2x )
dx
⎮
x
⌡1
also diverges.
Okay, we’ve seen a few examples of the Comparison Test now. However, most of them
worked pretty much the same way. All the functions were rational and all we did for
most of them was add or subtract something from the numerator or denominator to get
what we want.


73


Calculus II

Let’s take a look at an example that works a little differently so we don’t get too locked
into these ideas.
∞ −x
⌠ e dx
⎮
⌡1 x
Solution
Normally, the presence of just an x in the denominator would lead us to guess divergent
for this integral. However, the exponential in the numerator will approach zero so fast
that instead we’ll need to guess that this integral converges.

To get a larger function we’ll use the fact that we know from the limits of integration that
x>1. This means that if we just replace the x in the denominator with 1 (which is always
smaller than x) we will make the denominator smaller and so the function will get larger.
e− x e− x
<
= e− x
x
1
and we can show that

∫

∞

1

e − x dx

converges. In fact, we’ve already done this for a lower limit of 3 and changing that to a 1
won’t change the convergence of the integral. Therefore, by the Comparison Test
∞ −x
⌠ e dx
⎮
⌡1 x
also converges.
We should also really work an example that doesn’t involve a rational function since
there is no reason to assume that we’ll always be working with rational functions.

∫

∞

1

2

e − x dx

Solution
We know that exponentials with negative exponents die down to zero very fast so it
makes sense to guess that this integral will be convergent. We need a larger function, but
this time we don’t have a fraction to work with so we’ll need to do something different.

We’ll take advantage of the fact that e − x is a decreasing function. This means that
x1 > x2
⇒
e − x1 < e − x2
In other words, plug in a larger number and the function gets smaller.
From the limits of integration we know that x>1 and this means that if we square x we
will get larger. Or,
x2 > x
provided x > 1
Note that we can only say this since x>1. This won’t be true if x<1! We can now use the

74


Calculus II

fact that e − x is a decreasing function to get,
2

e− x < e− x
2

So, e− x is a larger function than e− x and we know that

∫

∞

1

e − x dx

converges so by the Comparison Test we also know that

∫

∞

1

2

e− x dx

is convergent.
The last two examples made use of the fact that x>1. Let’s take a look at an example to
see how do we would have to go about these if the lower limit had been smaller than 1.
∞ −x
⌠ e dx
⎮1
⌡ x
2

Solution

1
in the denominator and get a larger function
2
for all x in the interval of integration as we did in Example 5. Remember that we need a
function (that is also convergent) that is always larger than the given function..
In this case we can’t just replace x with

1
3
plug in x = into the denominator and
2
4
1
compare this to what we would have if we plug in x = .
2
−x
−x
e
4
e
= e− x < 2e− x =
34 3
12
1
So, for x’s in the range ≤ x < 1 we won’t get a larger function as required for use in the
2
Comparison Test.
To see why we can’t just replace x with

However, this isn’t the problem it might at first appear to be. We can always write the
integral as follows,
∞ −x
1 −x
∞ −x
⌠ e dx = ⌠ e dx + ⌠ e dx
⎮1
⎮1
⎮
⌡ x
⌡ x
⌡1 x
2

2

∞

e− x
dx
= 0.34039 + ⌠
⎮
⌡1 x

We used Maple to get the value of the first integral. Now, if the second integral


75


Calculus II

converges it will have a finite value and so the sum of two finite values will also be finite
and so the original integral will converge. Likewise, if the second integral diverges it
will either be infinite or not have a value at all and adding a finite number onto this will
not all of a sudden make it finite or exist and so the original integral will diverge.
Therefore, this integral will converge or diverge depending only on the convergence of
the second integral.
As we saw in Example 5 the second integral does converge and so the whole integral
must also converge.
As we saw in this example, if we need to, we can split the integral up into one that
doesn’t involve any problems and can be computed and one that may contain problem
what we can use the Comparison Test on to determine its converge.

Approximating Definite Integrals
In this chapter we’ve spent quite a bit to time on computing the values of integrals.
However, not all integrals can be computed. A perfect example is the following definite
integral.

∫

2
0

2

e x dx

We now need to talk a little bit about estimating values of definite integrals. We will
look at three different methods, although one should already be familiar to you from your
Calculus I days.
We will develop all three methods for estimating

∫ f ( x ) dx
b

a

by thinking of the integral as an area problem and using known shapes to estimate the
area under the curve.
Let’s get first develop the methods and then we’ll try to estimate the integral shown
above.

Midpoint Rule
This is the rule that you should be somewhat familiar to you. We will divide the interval
[ a, b ] into n subintervals of equal width,
b−a
n
We will denote each of the intervals as follows,
[ x0 , x1 ] , [ x1 , x2 ] ,… , [ xn−1 , xn ] where x0 = a and xn = b

Δx =


76


Calculus II

Then for each interval let xi* be the midpoint of the interval. We then sketch in

rectangles for each subinterval with a height of f ( xi* ) . Here is a graph showing the set
up using n=6.

We can easily find the area for each of these rectangles and so for a general n we get that,

∫ f ( x ) dx ≈ Δx f ( x ) + Δx f ( x ) +
b

*
1

a

*
2

*
+ Δx f ( xn )

Or, upon factoring out a Δx we get the general Mid Point Rule.

∫ f ( x ) dx ≈ Δx ⎡ f ( x ) + f ( x ) +
⎣
b

a

*
1

*
2

*
+ f ( xn ) ⎤
⎦

Trapezoid Rule
For this rule we will do the same set up as for the Midpoint Rule. We will break up the
interval [ a, b ] into n subintervals of width,
b−a
n
Then on each subinterval we will approximate the function with a straight line that is
equal to the function values at either endpoint of the interval. Here is a sketch of this
case for n=6.

Δx =


77


Calculus II

Each of these objects is a trapezoid (hence the rules name…) and as we can see some of
them do a very good job of approximating the actual area under the curve and others
don’t do such a good job.
The area of the trapezoid in the interval [ xi −1 , xi ] is given by,

Δx
( f ( xi−1 ) + f ( xi ) )
2
So, if we use n subintervals the integral is approximately,
b
Δx
Δx
∫ a f ( x ) dx ≈ 2 ( f ( x0 ) + f ( x1 ) ) + 2 ( f ( x1 ) + f ( x2 ) ) +
Ai =

+

Δx
( f ( xn−1 ) + f ( xn ) )
2

Upon doing a little simplification we arrive at the general Trapezoid Rule.

Δx
⎣
∫ f ( x ) dx ≈ 2 ⎡ f ( x ) + 2 f ( x ) + 2 f ( x ) +
b

a

0

1

2

+ 2 f ( xn −1 ) + f ( xn ) ⎤
⎦

Note that all the function evaluations, with the exception of the first and last, are
multiplied by 2.

Simpson’s Rule
This is the final method we’re going to take a look at and in this case we will again divide
up the interval [ a, b ] into n subintervals. However unlike the previous two methods we
need to require that n be even. The reason for this will be evident in a bit The width of
each subinterval is,
b−a
Δx =
n


78


Calculus II

In the Trapezoid Rule we approximated the curve with a straight line. For Simpson’s
Rule we are going to approximate the function with a quadratic and we’re going to
require that the quadratic agree with three of the points from our subintervals. Below is a
sketch of this using n=6. Each of the approximations is colored differently so we can see
how they actually work.

Notice that each approximation actually covers two of the subintervals. This is the
reason for requiring n to be even. It can be shown that the area under the approximation
on the intervals [ xi −1 , xi ] and [ xi , xi +1 ] is,
Ai =

Δx
( f ( xi−1 ) + 4 f ( xi ) + f ( xi+1 ) )
3

If we use n subintervals the integral is then approximately,
b
Δx
Δx
∫ a f ( x ) dx ≈ 3 ( f ( x0 ) + 4 f ( x1 ) + f ( x2 ) ) + 3 ( f ( x2 ) + 4 f ( x3 ) + f ( x4 ) )
Δx
+ +
( f ( xn−2 ) + 4 f ( xn−1 ) + f ( xn ) )
3
Upon simplifying we arrive at the general Simpson’s Rule.

Δx
⎣
∫ f ( x ) dx ≈ 3 ⎡ f ( x ) + 4 f ( x ) + 2 f ( x ) +
b

a

0

1

2

+ 2 f ( xn − 2 ) + 4 f ( xn −1 ) + f ( xn ) ⎤
⎦

In this case notice that all the function evaluations at points with odd subscripts are
multiplied by 4 and all the function evaluations at points with even subscripts (except for
the first and last) are multiplied by 2. If you can remember this, this a fairly easy rule to
remember.
Okay, it’s time to work an example and see how these rules work.


79


Calculus II

Example 1 Using n=4 and all three rules to approximate the value of the following
integral.

∫

2
0

2

e x dx

Solution
First, for reference purposes, Maple gives the following value for this integral.

∫

2
0

2

e x dx = 16.45262776

In each case the width of the subintervals will be,
2−0 1
Δx =
=
4
2
and so the subintervals will be,
[0, 0.5] , [ 0.5, 1] , [1, 1.5] , [1.5, 2]
Let’s go through each of the methods.
Midpoint Rule

∫

2
0

2

e x dx ≈

)

(

1 ( 0.25)2 ( 0.75)2 (1.25)2 (1.75)2
e
+e
+e
+e
= 14.48561253
2

Remember that we evaluate at the midpoints of each of the subintervals here! The
Midpoint Rule has an error of 1.96701523.
Trapezoid Rule

∫

2
0

2

e x dx ≈

)

(

2
2
2
2
1 2 ( 0 )2
0.5
1
1.5
2
e + 2e( ) + 2e( ) + 2e( ) + e( ) = 20.64455905
2

The Trapezoid Rule has an error of 4.19193129
Simpson’s Rule

∫

2
0

2

e x dx ≈

)

(

2
2
2
2
1 2 ( 0 )2
e + 4e( 0.5) + 2e(1) + 4e(1.5) + e( 2) = 17.35362645
3

The Simpson’s Rule has an error of 0.90099869.
None of the estimations in the previous example are all that good. The best
approximation in this case is from the Simpson’s Rule and yet it’s still had an error of
almost 1. To get a better estimation we would need to use a larger n. So, for
completeness sake here are the estimates for some larger value of n.

n
8
16

Midpoint
Trapezoid
Simpson’s
Approx.
Error
Approx.
Error
Approx.
Error
15.9056767 0.5469511 17.5650858 1.1124580 16.5385947 0.0859669
16.3118539 0.1407739 16.7353812 0.2827535 16.4588131 0.0061853


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Calculus II

32 16.4171709 0.0354568 16.5236176 0.0709898 16.4530297 0.0004019
64 16.4437469 0.0088809 16.4703942 0.0177665 16.4526531 0.0000254
128 16.4504065 0.0022212 16.4570706 0.0044428 16.4526294 0.0000016
In this case we where able to determine the error for each estimate because we could get
our hands on the exact value. Often this won’t be the case and so we’d next like to look
at error bounds for each estimate.
These bounds will give the largest possible error in the estimate, but it should also be
pointed out that the actual error may be significantly smaller than the bound. The bound
is only there so we can say that we know the actual error will be less than the bound.
So, suppose that f ′′ ( x ) ≤ K and f ( 4) ( x ) ≤ M for a ≤ x ≤ b then if EM, ET, and ES are
the actual errors for the Midpoint, Trapezoid and Simpson’s Rule we have the following
bounds,
EM

K (b − a )
≤
24n 2

3

K (b − a )
ET ≤
12n 2

3

M (b − a )
ES ≤
180n 4

5

Example 2 Determine the error bounds for the estimations in the last example.
Solution
We already know that n=4, a=0, and b=2 so we just need to compute K (the largest value
of the second derivative) and M (the largest value of the fourth derivative). This means
that we’ll need the second and fourth derivative of f(x).
2
f ′′ ( x ) = 2e x (1 + 2 x 2 )

f ( 4) ( x ) = 4e x ( 3 + 12 x 2 + 4 x 4 )
2

Here is a graph of the second derivative.

Here is a graph of the fourth derivative.


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Calculus II

So, from these graphs it’s clear that the largest value of both of these are at x=2. So,
⇒
f ′′ ( 2 ) = 982.76667
K = 983
f ( 4) ( 2 ) = 25115.14901

⇒
We rounded to make the computations simpler.

M = 25116

Here are the bounds for each rule.
EM ≤
ET ≤
ES ≤

983 ( 2 − 0 )
24 ( 4 )

= 20.4791666667

2

983 ( 2 − 0 )
12 ( 4 )

3

3

2

25116 ( 2 − 0 )
180 ( 4 )

= 40.9583333333
5

4

= 4.36041666667

In each case we can see that the errors are significantly smaller than the actual bounds.

Applications of Integrals
Introduction
In this section we’re going to take a look at some of applications of integration. It should
be noted as well that these applications are presented here, as opposed to Calculus I,
simply because many of the integrals that arise from these applications tend to require
techniques that we discussed in the previous chapter.
Here is a list of applications that we’ll be taking a look at in this chapter.

Arc Length – We’ll determine the length of a curve in this section.
Surface Area – In this section we’ll determine the surface area of a solid of revolution.


82


Calculus II

Center of Mass – Here we will determine the center of mass or centroid of a thin plate.
Hydrostatic Pressure and Force – We’ll determine the hydrostatic pressure and force
on a vertical plate submerged in water.
Probability – Here we will look at probability density functions and computing the mean
of a probability density function.

Arc Length
In this section we are going to look at computing the arc length of a function. Because
it’s easy enough to derive the formulas that we’ll use in this section we will derive one of
them and leave the other to you to derive.
We want to determine the length of the continuous function y = f ( x ) on the interval

[ a, b ] .

Initially we’ll need to estimate the length of the curve. We’ll do this by dividing

the interval up into n equal subintervals each of width Δx and we’ll denote the point on
the curve at each point by Pi. We can then approximate the curve by a series of straight
lines connecting the points. Here is a sketch of this situation for n=9.

Now denote the length of each of these line segments by Pi −1 Pi and the length of the
curve will then be approximately,
n

L ≈ ∑ Pi −1 Pi
i =1

and we can get the exact length by taking n larger and larger. In other words, the exact
length will be,
n

L = lim ∑ Pi −1 Pi
n →∞


i =1

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Calculus II

Now, let’s get a better grasp on the length of each of these line segments. First, on each
segment let’s define Δyi = yi − yi −1 = f ( xi ) − f ( xi −1 ) . We can then compute directly the
length of the line segments as follows.
Pi −1 Pi =

( xi − xi −1 ) + ( yi − yi −1 )
2

2

= Δx 2 + Δyi2

By the Mean Value Theorem we know that on the interval [ xi −1 , xi ] there is a point xi* so
that,
f ( xi ) − f ( xi −1 ) = f ′ ( xi* ) ( xi − xi −1 )

Δyi = f ′ ( xi* ) Δx
Therefore, the length can now be written as,
Pi −1 Pi =

( xi − xi −1 ) + ( yi − yi −1 )
2

2

= Δx 2 + ⎡ f ′ ( xi* ) ⎤ Δx 2
⎣
⎦
2

= 1 + ⎡ f ′ ( xi* ) ⎤
⎣
⎦
The exact length of the curve is then,

2

Δx

n

L = lim ∑ Pi −1 Pi
n →∞

i =1

⎡
⎤
= lim ∑ 1 + ⎣ f ( xi* ) ⎦
n →∞
i =1
n

2

Δx

However, using the definition of the definite integral, this is nothing more than,
b

L = ⌠ 1 + ⎡ f ′ ( x ) ⎤ dx
⎣
⎦
⌡a
2

A slightly more convenient notation (in my opinion anyway) is the following.
b

2
⌠
⎛ dy ⎞
L = ⎮ 1 + ⎜ ⎟ dx
⎝ dx ⎠
⌡a

In a similar fashion we can also derive a formula for x = h ( y ) on [ c, d ] . This formula is,
d

2
⌠
⌠ 1 + ⎡ h′ ( y ) ⎤ dy = ⎮ 1 + ⎛ dx ⎞ dy
L=
⎜ ⎟
⎣
⎦
⎮
⌡c
⎝ dy ⎠
⌡c
Again, the second form is probably a little more convenient.
d

2

Note the difference in the derivative under the square root! Don’t get too confused. With
one we differentiate with respect to x and with the other we differentiate with respect to y.

84


Calculus II

One way to keep the two straight is to notice that the differential in the “denominator” of
the derivative will match up with the differential in the integral. This is one of the
reasons why the second form is a little more convenient.
Before we work any examples we need to make a small change in notation. Instead of
having two formulas for the arc length of a function we are going to reduce it, in part, to a
single formula.
From this point on we are going to use the following formula for the length of the curve.
L = ∫ ds
where,
2

⎛ dy ⎞
ds = 1 + ⎜ ⎟ dx
⎝ dx ⎠
2

⎛ dx ⎞
ds = 1 + ⎜ ⎟ dy
⎝ dy ⎠

if y = f ( x ) , a ≤ x ≤ b
if x = h ( y ) , c ≤ y ≤ d

Note that no limits where put on the integral as the limits will depend upon the ds that
we’re using. Using the first ds will require x limits of integration and using the second ds
will require y limits of integration.
Thinking of the arc length formula as a single integral with different ways to define ds
will be convenient when we run across arc lengths in future sections. Also, this ds
notation will be a nice notation for the next section as well.
Now that we’ve derived the arc length formula let’s work some examples.
Example 1 Determine the length of y = ln ( sec x ) between 0 ≤ x ≤

π
4

.

Solution
In this case we’ll need to use the first ds since the function is in the form y = f ( x ) . So,

let’s get the derivative out of the way.
2

⎛ dy ⎞
2
⎜ ⎟ = tan x
⎝ dx ⎠

dy sec x tan x
=
= tan x
sec x
dx

Let’s also get the root out of the way since there is often simplification that can be done
and there’s no reason to do that inside the integral.
2

⎛ dy ⎞
1 + ⎜ ⎟ = 1 + tan 2 x = sec 2 x = sec x = sec x
⎝ dx ⎠
Note that we could drop the absolute value bars here since secant is positive in the range
given.


85


Calculus II

The arc length is then,
π

L = ∫ 4 sec x dx
0

= ln sec x + tan x
= ln
Example 2 Determine the length of x =

(

)

π
4
0

2 +1

3
2
( y − 1) 2 between 1 ≤ y ≤ 4 .
3

Solution
There is a very common mistake that students make in problems of this type. Many
students see that the function is in the form x = h ( y ) and they immediately decide that it

will be too difficult to work with it in that form so they solve for y to get the function into
the form y = f ( x ) . While that can be done here it will lead to a messier integral for us
to deal with.
Sometimes it’s just easier to work with functions in the form x = h ( y ) . In fact, if you
can work with functions in the form y = f ( x ) then you can work with functions in the
form x = h ( y ) . There really isn’t a difference between the two so don’t get excited
about functions in the form x = h ( y ) .
Let’s compute the derivative and the root.
2

⎛ dx ⎞
1+ ⎜ ⎟ = 1+ y −1 = y
⇒
⎝ dy ⎠
As you can see keeping the function in the form x = h ( y ) is going to lead to a very easy
integral. To see what would happen if we tried to work with the function in the form
y = f ( x ) see the next example.
1
dx
= ( y − 1) 2
dy

Let’s get the length.
L=∫

4

1

y dy
4

2 3
= y2
3 1

=

14
3

As noted in the last example we really do have a choice as to which ds we use. Provided
we can get the function in the form required for a particular ds we can use it. However,


86


Calculus II

as also noted above, there will often be a significant difference in difficulty in the
resulting integrals. Let’s take a quick look at what would happen in the previous example
if we did put the function into the form y = f ( x ) .
Example 3 Redo the previous example using the function in the form y = f ( x ) instead.
Solution
In this case the function and its derivative would be,
2

⎛ 3x ⎞ 3
y = ⎜ ⎟ +1
⎝ 2 ⎠
The root in the arc length formula would then be.

dy ⎛ 3 x ⎞
=⎜ ⎟
dx ⎝ 2 ⎠

2

1
3

2

2

2

1
⎛ dy ⎞
=
1+ ⎜ ⎟ = 1+
2
⎝ dx ⎠
3x 3
( )

−

( 32x ) 3 + 1 = ( 32x ) 3 + 1
2

1

( 32x ) 3

( 32x ) 3

All the simplification work above was just to put the root into a form that will allow us to
do the integral.
Now, before we write down the integral we’ll also need to determine the limits. This
particular ds requires x limits of integration and we’ve got y limits. They are easy enough
to get however. Since we know x as a function of y all we need to do is plug in the
original y limits of integration and get the x limits of integration. Doing this gives,
2 3
0 ≤ x ≤ ( 3) 2
3
Not easy limits to deal with, but there they are.
Let’s now write down the integral that will give the length.
2

3

( 3) 2

2
⌠3
( 32x ) 3 + 1
L=⎮
dx
1
⎮
3x 3
⎮
(2)
⌡0
That’s a really unpleasant looking integral. It can be evaluated however using the
following substitution.
2

−

⎛ 3x ⎞ 3
u = ⎜ ⎟ +1
⎝ 2 ⎠
x=0

x=

2
( 3)
3

1

⇒
3
2

⎛ 3x ⎞ 3
du = ⎜ ⎟ dx
⎝ 2 ⎠
u =1

⇒

u=4



87


Calculus II

L=∫

4

1

u du
4

2 3
= u2
3 1

14
3
So, we got the same answer as in the previous example. Although that shouldn’t really
be all that surprising since we were dealing with the same curve.

=

From a technical standpoint the integral in the previous example was not that difficult. It
was just a Calculus I substitution. However, from a practical standpoint the integral was
significantly more difficult than the integral we used in Example 2. So, the moral of the
story here is that we can use either formula (provided we can get the function in the
correct form of course) however one will often be significantly easier to actually
evaluate.
Okay, let’s work one more example.
Example 4 Determine the length of x =

1 2
1
y for 0 ≤ x ≤ . Assume that y is positive.
2
2

Solution
We’ll use the second ds for this one as the function is already in the correct form for that
one. Also, the other ds would again lead to a particularly difficult integral. The
derivative and root will then be,

dx
=y
dy

2

⎛ dx ⎞
1+ ⎜ ⎟ = 1+ y2
⎝ dy ⎠

⇒

Before writing down the length notice that we were given x limits and we will need y
limits for this ds. With the assumption that y is positive there are easy enough to get.
They are,
0 ≤ y ≤1
The integral for the arc length is then,
L=∫

1
0

1 + y 2 dy

This integral will require the following trig substitution.
y = tan θ
dy = sec 2 θ dθ
y=0
⇒
0 = tan θ
⇒
θ =0
y =1

⇒

1 = tan θ

⇒

θ=

π
4

1 + y 2 = 1 + tan 2 θ = sec 2 θ = sec θ = sec θ


88


Calculus II

The length is then,
π

L = ∫ 4 sec3 θ dθ
0

π

1
= ( sec θ tan θ + ln sec θ + tan θ
2
1
=
2

(

(

2 + ln 1 + 2

)

4
0

))

The first couple of examples ended up being fairly simple Calculus I substitutions.
However, as this last example had shown we will often end up with trig substitutions as
well for these integrals.

Surface Area
In this section we are going to look once again at solids of revolution. We first looked at
them back in Calculus I when we found the volume of the solid of revolution. In this
section we want to find the surface area of this region.
So, for the purposes of the derivation of the formula, let’s look at rotating the continuous
function y = f ( x ) in the interval [ a, b ] about the x-axis. Below is a sketch of a function
and the solid of revolution we get by rotating the function about the x-axis.

We can derive a formula for the surface area much as we derived the formula for arc
length. We’ll start by dividing the integral into n equal subintervals of width Δx . On
each subinterval we will approximate the function with a straight line that agrees with the
function at the endpoints of the each interval. Here is a sketch of that for our
representative function using n=4.


89


Calculus II

Now, rotate the approximations about the x-axis and we get the following solid.

The approximation on each interval gives a distinct portion of the solid and to make this
clear each portion is colored differently. Each of these portions are called frustums and
we know how to find the surface area of frustums.
The surface area of a frustum is given by,
A = 2π rl
where,
1
r = ( r1 + r2 )
r1 = radius of right end
2
r2 = radius of left end
and l is the length of the slant of the frustum.
For the frustum on the interval [ xi −1 , xi ] we have,
r1 = f ( xi )
r2 = f ( xi −1 )
l = Pi −1 Pi


( length of the line segment connecting Pi
90

and Pi −1 )


Calculus II

and we know from the previous section that,
Pi −1 Pi = 1 + ⎡ f ′ ( xi* ) ⎤
⎣
⎦

2

Δx where xi* is some point in [ xi −1 , xi ]

Before writing down the formula for the surface area we are going to assume that Δx is
“small” and since f(x) is continuous we can then assume that,
f ( xi ) ≈ f ( xi* )
and
f ( xi −1 ) ≈ f ( xi* )
So, the surface area of the frustum on the interval [ xi −1 , xi ] is approximately,

⎛ f ( xi ) + f ( xi −1 ) ⎞
Ai = 2π ⎜
⎟ Pi −1 Pi
2
⎝
⎠
≈ 2π f ( xi* ) 1 + ⎡ f ′ ( xi* ) ⎤
⎣
⎦

2

Δx

The surface area of the whole solid is then approximately,
S ≈ ∑ 2π f ( xi* ) 1 + ⎡ f ′ ( xi* ) ⎤
⎣
⎦
i =1
n

2

Δx

and we can get the exact surface area by taking the limit as n goes to infinity.
S = lim ∑ 2π f ( xi* ) 1 + ⎡ f ′ ( xi* ) ⎤
⎣
⎦
n →∞
i =1
n

b

2

Δx

= ⌠ 2π f ( x ) 1 + ⎡ f ′ ( x ) ⎤ dx
⎣
⎦
⌡a
2

If we wanted to we could also derive a similar formula for rotating x = h ( y ) on [ c, d ]
about the y-axis. This would give the following formula.
d

S = ⌠ 2π h ( y ) 1 + ⎡ h′ ( y ) ⎤ dy
⎣
⎦
⌡c
2

These are not the “standard” formulas however. Notice that the roots in both of these
formulas are nothing more than the two ds’s we used in the previous section. Also, we
will replace f(x) with y and h(y) with x. Doing this gives the following two formulas for
the surface area.
S = ∫ 2π y ds

rotation about x − axis

S = ∫ 2π x ds

rotation about y − axis

where,


91


Calculus II

2

⎛ dy ⎞
ds = 1 + ⎜ ⎟ dx
⎝ dx ⎠
2

⎛ dx ⎞
ds = 1 + ⎜ ⎟ dy
⎝ dy ⎠

if y = f ( x ) , a ≤ x ≤ b
if x = h ( y ) , c ≤ y ≤ d

There are a couple of things to note about these formulas. First, notice that the variable
in the integral itself is always the opposite variable from the one we’re rotating about.
Second, we are allowed to use either ds in either formula. This means that there are, in
some way, four formulas here. We will choose the ds based upon which is the most
convenient for a given function and problem.
Now let’s work a couple of examples.
Example 1 Determine the surface area of the solid obtained by rotating y = 9 − x 2 ,
−2 ≤ x ≤ 2 about the x-axis.
Solution
The formula that we’ll be using here is,
S = ∫ 2π y ds

since we are rotating about the x-axis and we’ll use the first ds in this case because our
function is in the correct form for that ds and we won’t gain anything by solving it for x.
Let’s first get the derivative and the root taken care of.
1
dy 1
2 −2
= ( 9 − x ) ( −2 x ) = −
dx 2

x
1
2 2

(9 − x )

2

x2
9
3
⎛ dy ⎞
=
=
1+ ⎜ ⎟ = 1+
2
2
9− x
9− x
⎝ dx ⎠
9 − x2
Here’s the integral for the surface area,
2

3
⌠
S = ⎮ 2π y
dx
⌡ −2
9 − x2
There is a problem however. The dx means that we shouldn’t have any y’s in the
integral. So, before evaluating the integral we’ll need to substitute in for y as well.
The surface area is then,


92


Calculus II

2

⌠
S = ⎮ 2π 9 − x 2
⌡ −2

3
9 − x2

dx

2

= ∫ 6π dx
−2

= 24π

Previously we made the comment that we could use either ds in the surface area
formulas. Let’s work an example in which using either ds won’t create integrals that are
too difficult to evaluate and so we can check both ds’s.
Example 2 Determine the surface area of the solid obtained by rotating y = 3 x ,
1 ≤ y ≤ 2 about the y-axis. Use both ds’s to compute the surface area.
Solution
Note that we’ve been given the function set up for the first ds and limits that work for the
second ds.

Solution 1
This solution will use the first ds listed above. We’ll start with the derivative and root.
dy 1 − 2
= x 3
dx 3
4

4

1
9x 3 +1
9x 3 +1
⎛ dy ⎞
1+ ⎜ ⎟ = 1+ 4 =
=
4
2
⎝ dx ⎠
9x 3
9x 3
3x 3
2

We’ll also need to get new limits. That isn’t too bad however. All we need to do is plug
in the given y’s and get that the range of x’s is 1 ≤ x ≤ 8 . The integral for the surface area
is then,
8

4
⌠
9x 3 +1
S = ⎮ 2π x
dx
2
⎮
3
3x
⌡1
4
2π ⌠ 8 1
⎮ x 3 9 x 3 + 1 dx
3 ⌡1
Note that this time we didn’t need to substitute in for the x as we did in the previous
example. In this case we picked up a dx from the ds and so we don’t need to do a
substitution for the x. In fact if we had substituted for x we would have put y’s into
integral which would have caused problems.

=

Using the substitution
4

1

u = 9x 3 +1

du = 12 x 3 dx

the integral becomes,


93


Calculus II

S=
=
=

π

145

π

3 145
2

18 ∫10
27

u

u du

10

π ⎛

⎞
⎜ 145 − 10 ⎟ = 199.48
27 ⎝
⎠
3
2

3
2

Solution 2
This time we’ll use the second ds. So, we’ll first need to solve the equation for x. We’ll
also go ahead and get the derivative and root while we’re at it.
dx
x = y3
= 3y2
dy
2

⎛ dx ⎞
1+ ⎜ ⎟ = 1+ 9 y4
⎝ dy ⎠
The surface area is then,
2

S = ∫ 2π x 1 + 9 y 4 dy
1

We used the original y limits this time because we picked up a dy from the ds. Also note
that the presence of the dy means that this time, unlike the first solution, we’ll need to
substitute in for the x. Doing that gives,
2

S = ∫ 2π y 3 1 + 9 y 4 dy
1

=
=

π

18 ∫

145

10

u = 1+ 9 y4

u du

π ⎛

3
3
⎞
145 2 − 10 2 ⎟ = 199.48
⎜
27 ⎝
⎠

Note that after the substitution the integral was identical to the first solution and so the
work was skipped.
As this example has shown we can used either ds to get the surface area. It is important
to point out as well that with one ds we had to do a substitution for the x and with the
other we didn’t. This will always work out that way.
Note as well that in the case of the last example it was just as easy to use either ds. That
often won’t be the case. In many examples only one of the ds will be convenient to work
with so we’ll always need to determine which ds is liable to be the easiest to work with
before starting the problem.


94


Calculus II

Center of Mass
In this section we are going to find the center of mass or centroid of a thin plate with
uniform density ρ. The center of mass or centroid of a region is the point in which the
region will be perfectly balanced horizontally if suspended from that point.
So, let’s suppose that the plate is the region bounded by the two curves f(x) and g(x) on
the interval [a,b]. So, we want to find the center of mass of the region below.

We’ll first need the mass of this plate. The mass is,
M = ρ ( Area of plate )
= ρ ∫ f ( x ) − g ( x ) dx
b

a

Next we’ll need the moments of the region. There are two moments, denoted by Mx and
My. The moments measure the tendency of the region to rotate about the x and y-axis
respectively. The moments are given by,

)

(

b
2
2
1
M x = ρ⌠
⎡ f ( x ) ⎤ − ⎡ g ( x ) ⎤ dx
⎮
⎣
⎦ ⎣
⎦
⌡a 2

M y = ρ ∫ x ( f ( x ) − g ( x ) ) dx
b

a

The coordinates of the center of mass, ( x , y ) , are then,


95


Calculus II

∫ x ( f ( x ) − g ( x ) ) dx = 1 x f x − g x dx
( ( ) ( ))
A∫
f ( x ) − g ( x ) dx
∫
b

x=

My
M

=

b

a

b

a

a

y=

Mx
=
M

)

(

b
⌠ 1 ⎡ f ( x ) ⎤ 2 − ⎡ g ( x ) ⎤ 2 dx
⎮
⎦ ⎣
⎦
⌡a 2 ⎣

∫

b
a

f ( x ) − g ( x ) dx

)

(

2
2
1 ⌠b1
⎡ f ( x ) ⎤ − ⎡ g ( x ) ⎤ dx
⎮
⎣
⎦ ⎣
⎦
A ⌡a 2

=

where,
A = ∫ f ( x ) − g ( x ) dx
b

a

Note that the density, ρ, of the plate cancels out and so isn’t really needed.
Let’s work a couple of examples.
Example 1 Determine the center of mass for the region bounded by y = 2sin ( 2 x ) , y = 0

⎡ π⎤
on the interval ⎢ 0, ⎥ .
⎣ 2⎦
Solution
Here is a sketch of the region with the center of mass denoted with a small box.

Let’s first get the area of the region.
π

A = ∫ 2 2sin ( 2 x ) dx
0

π

= − cos ( 2 x ) 02
=2
Now, the moments (without density since it will just drop out) are,


96


Calculus II

π

M x = ∫ 2 2sin 2 ( 2 x ) dx
0

π

= ∫ 2 1 − cos ( 4 x ) dx
0

π

1
⎛
⎞4
= ⎜ x − sin ( 4 x ) ⎟
4
⎝
⎠0
=

π
2

π

M y = ∫ 2 2 x sin ( 2 x ) dx

integrating by parts...

0

π

π

2
0

0

= − x cos ( 2 x ) + ∫ 2 cos ( 2 x ) dx
π

= − x cos ( 2 x ) 02
=

π
2
1
+ sin ( 2 x )
2
0

π
2

The coordinates of the center of mass are then,
x=
y=

π

2 =π
2
4

π

2 =π
2
4

⎛π π ⎞
So, the center of mass for this region is ⎜ , ⎟ .
⎝4 4⎠
Example 2 Determine the center of mass for the region bounded by y = x 3 and y = x .
Solution
The two curves intersect at x=0 and x=1 and here is a sketch of the region with the center
of mass marked with a box.


97


Calculus II

We’ll first get the area of the region.
A=∫

1
0

x − x 3 dx
1

⎛2 3 1 4⎞
= ⎜ x2 − x ⎟
4 ⎠0
⎝3
5
=
12

Now the moments, again without density, are
1
1
M x = ⌠ ( x − x 6 ) dx
⎮
⌡0 2
1

1⎛1
1 ⎞
= ⎜ x2 − x7 ⎟
2⎝2
7 ⎠0
=

5
28
1

My = ∫ x
0

1

(

)

x − x3 dx

3
2

= ∫ x − x 4 dx
0

1

⎛2 5 1 5⎞
= ⎜ x2 − x ⎟
5 ⎠0
⎝5
1
=
5
The coordinates of the center of mass is then,


98


Calculus II

x=

1 5 12
=
5 12 25

y=

5 28 3
=
5 12 7

⎛ 12 3 ⎞
The coordinates of the center of mass are then, ⎜ , ⎟ .
⎝ 25 7 ⎠

Hydrostatic Pressure and Force
In this section we are going to submerge a vertical plate in water and we want to know
the force that is exerted on the plate due to the pressure of the water. This force is often
called the hydrostatic force.
There are two basic formulas that we’ll be using here. First, if we are d meters below the
surface then the hydrostatic pressure is given by,
P = ρ gd
where, ρ is the density of the fluid and g is the gravitational acceleration. We are going
to assume that the fluid in question is water and since we are going to be using the metric
system these quantities become,
ρ = 1000 kg/m3
g = 9.81 m/s 2
The second formula that we need is the following. Assume that a constant pressure P is
acting on a surface with area A. Then the hydrostatic force that acts on the area is,
F = PA
Note that we won’t be able to find the hydrostatic force on a vertical plate using this
formula since the pressure will vary with depth. We will however need this for our work.
The best way to see how these problems work is to do an example or two.
Example 1 Determine the hydrostatic force on the following triangular plate that is
submerged in water as shown.


99


Calculus II

Solution
The first thing to do here is set up an axis system. We’ll use the following axis system.

So, we are going to orient the x-axis so that positive x is downward, x=0 corresponds to
the waters surface and x=4 corresponds to the depth of the tip of the triangle.
Next we are break up the triangle into n strips each of equal width Δx and in each
interval [ xi −1 , xi ] choose any point xi* . In order to make the computations easier we are
going to make two assumptions about these strips. First, we will ignore the fact that the
ends are actually going to be slanted and assume the strips are rectangular. If Δx is
sufficiently small this will not affect our computations much. Second, we will assume
that Δx is small enough that the hydrostatic pressure on each strip is essentially constant.
Below is a representative strip.


100


Calculus II

The height of this strip is Δx and the width is 2a. We can use similar triangles to
determine a.
3
a
3
=
⇒
a = 3 − xi*
*
4 4 − xi
4
Now, since we are assuming the pressure on this strip is constant, the pressure is given
by,
Pi = ρ gd = 1000 ( 9.81) xi* = 9810 xi*
and the hydrostatic force is
3 ⎞
3 ⎞
⎛
⎛
Fi = Pi A = Pi ( 2aΔx ) = 9810 xi* ( 2 ) ⎜ 3 − xi* ⎟ Δx = 19620 xi* ⎜ 3 − xi* ⎟ Δx
4 ⎠
4 ⎠
⎝
⎝
The approximate hydrostatic force on the plate is then the sum of the forces on all the
strips or,
n
3 ⎞
⎛
F ≈ ∑19620 xi* ⎜ 3 − xi* ⎟ Δx
4 ⎠
⎝
i =1
Taking the limit will get the exact hydrostatic force,
n
3 ⎞
⎛
F = lim ∑19620 xi* ⎜ 3 − xi* ⎟ Δx
n →∞
4 ⎠
⎝
i =1
Using the definition of the definite integral this is nothing more than,


101


Calculus II

4

3 ⎞
⌠
⎛
F = ⎮ 19620 ⎜ 3x − x 2 ⎟ dx
4 ⎠
⌡0
⎝
The hydrostatic force is then,
4

3 ⎞
⌠
⎛
F = ⎮ 19620 ⎜ 3 x − x 2 ⎟ dx
4 ⎠
⌡0
⎝
4

1 ⎞
⎛3
= 19620 ⎜ x 2 − x 3 ⎟
4 ⎠0
⎝2
= 156960 N

Example 2 Find the hydrostatic force on a circular plate of radius 2 that is submerged 6
meters in the water.
Solution
First, we’re going to assume that the top of the circular plate is 6 meters under the water.
Next, we will set up the axis system so that the origin of the axis system is at the center of
the plate. Finally, we will again split up the plate into n strips each of width Δy and
we’ll choose a point yi* from each strip. We’ll also assume that the strips are rectangular
again to help with the computations. Here is a sketch of the setup.


102


Calculus II

The depth below the waters surface of each strip is,
di = 8 − yi*
and that in turn gives us the pressure on the strip,
Pi = ρ gdi = 9810 ( 8 − yi* )
The area of each strip is,
Ai = 2 4 − ( yi* ) Δy
2

The hydrostatic force on each strip is,
Fi = Pi Ai = 9810 ( 8 − yi* ) ( 2 ) 4 − ( yi* ) Δy
2

The total force on the plate is,
F = lim ∑19620 ( 8 − yi* ) 4 − ( yi* ) Δy
n

n →∞

2

i =1

= 19620 ∫

2
−2

(8 − y )

4 − y 2 dy

To do this integral we’ll need to split it up into two integrals.
2

2

−2

−2

F = 19620 ∫ 8 4 − y 2 dy − 19620 ∫ y 4 − y 2 dy
The first integral requires the trig substitution y = 2sin θ and the second integral needs
the substitution v = 4 − y 2 . After using these substitution we get,
π 2

F = 627840 ∫
= 313920 ∫

−π 2

π 2
−π 2

cos 2 θ dθ + 9810 ∫

0
0

v dv

1 + cos ( 2θ ) dθ + 0
π

1
⎛
⎞2
= 313920 ⎜ θ + sin ( 2θ ) ⎟
2
⎝
⎠ −π

2

= 313920π

Note that we know the second integral will be zero because the upper and lower limit is
the same.


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Calculus II

Probability
In this last application of integrals that we’ll be looking at we’re going to look at
probability. Before actually getting into the applications we need to get a couple of
definitions out of the way.
Suppose that we wanted to look at the age of a person, the height of a person, the amount
of time spent waiting in line, or maybe the lifetime of a battery. Each of these quantities
have values that will range over an interval of integers. Because of this these are called
continuous random variables. Continuous random variables are often represented by X.
Every continuous random variable, X, has a probability density function, f(x).
Probability density functions satisfy the following conditions.
1.

f ( x ) ≥ 0 for all x.

2.

∫

∞
−∞

f ( x ) dx = 1

Probability density functions can be used to determine the probability that a continuous
random variable lies between two values, say a and b. This probability is denoted by
P ( a ≤ X ≤ b ) and is given by,
P ( a ≤ X ≤ b ) = ∫ f ( x ) dx
b

a

Let’s take a look at an example of this.
x3
(10 − x ) for 0 ≤ x ≤ 10 and f ( x ) = 0 for all other values of
5000
x. Answer each of the following questions about this function.
(a) Show that f(x) is a probability density function.
(b) Find P (1 ≤ X ≤ 4 )
Example 1 Let f ( x ) =

(c) Find P ( x ≥ 6 )
Solution

(a) To show this is a probability density function we’ll need to show that

∫

∞
−∞

∫

∞
−∞

f ( x ) dx = 1 .

10

x3
f ( x ) dx = ⌠
(10 − x ) dx
⎮
⌡ 0 5000
10

⎛ x4
x5 ⎞
=⎜
−
⎟
⎝ 2000 25000 ⎠ 0
=1


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Calculus II

Note the change in limits on the integral. The function is only non-zero in these ranges
and so the integral can be reduced down to only the interval where the function is not
zero.

(b) In this case we need to evaluate the following integral.
4
x3
P (1 ≤ X ≤ 4 ) = ⌠
(10 − x ) dx
⎮
⌡1 5000
4

⎛ x4
x5 ⎞
=⎜
−
⎟
⎝ 2000 25000 ⎠ 1
= 0.08658
So the probability of X being between 1 and 4 is 8.658%.

(c) Note that in this case P ( x ≥ 6 ) is equivalent to P ( 6 ≤ X ≤ 10 ) since 10 is the largest
value that X can be. So the probability that X is greater than or equal to 6 is,
10
x3
P ( X ≥ 6) = ⌠
(10 − x ) dx
⎮
⌡ 6 5000
10

⎛ x4
x5 ⎞
=⎜
−
⎟
⎝ 2000 25000 ⎠ 6
= 0.66304
This probability is then 66.304%.

Probability density functions can also be used to determine the mean of a continuous
random variable. The mean is given by,
∞

μ = ∫ xf ( x ) dx
−∞

Let’s work one more example.
Example 2 It has been determined that the probability density function for the wait in
line at a counter is given by,
if t < 0
⎧0
⎪
f (t ) = ⎨
t
−
⎪0.1e 10 if t ≥ 0
⎩
where t is the number of minutes spent waiting in line. Answer each of the following
questions about this probability density function.
(a) Verify that this is in fact a probability density function.
(b) Determine the probability that a person will wait in line for at least 6 minutes.
(c) Determine the mean wait in line.
Solution
(a) Not much to do here other than compute the integral.

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Calculus II

∫

∞

∞

f ( t ) dt = ∫ 0.1e

−∞

−

t
10

0

u

dt

= lim ∫ 0.1e
u →∞

−

t
10

0

dt

u

⎛ −t ⎞
= lim ⎜ −e 10 ⎟
u →∞
⎝
⎠0
u
−
⎛
⎞
10
= lim ⎜1 − e ⎟
u →∞
⎝
⎠
=1

So it is a probability density function.
(b) The probability that we’re looking for here is P ( x ≥ 6 ) .
∞

P ( X ≥ 6 ) = ∫ 0.1e

−

t
10

6

u

dt

= lim ∫ 0.1e
u →∞

−

t
10

6

dt

u

⎛ −t ⎞
= lim ⎜ −e 10 ⎟
u →∞
⎝
⎠6
u
−
⎛ −6
⎞
= lim ⎜ e 10 − e 10 ⎟
u →∞
⎝
⎠
−

3
5

= e = 0.548812
So the probability that a person will wait in line for more than 6 minutes is 54.8811%.

(c) Here’s the mean wait time.
∞

μ = ∫ t f ( t ) dt
−∞
∞

= ∫ 0.1t e
0

u

−

t
10

dt

= lim ∫ 0.1t e
u →∞

0

−

t
10

integrating by parts....

dt
u

t
−
⎛
⎞
10
= lim ⎜ − ( t + 10 ) e ⎟
u →∞
⎝
⎠0
u
−
⎛
⎞
= lim ⎜10 − ( u + 10 ) e 10 ⎟
u →∞
⎝
⎠
= 10
So, it looks like the average wait time is 10 minutes.


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Calculus II

Parametric Equations and Polar Coordinates
Introduction
In this section we will be looking at parametric equations and polar coordinates. While
the two subjects don’t appear to have that much in common on the surface we will see
that several of the topics in polar coordinates can be done in terms of parametric
equations and so in that sense they make a good match in this chapter.
We will also be looking at how to do many of the standard calculus topics such as
tangents and area in terms of parametric equations and polar coordinates.
Here is a list of topics that we’ll be covering in this chapter.
Parametric Equations and Curves – An introduction to parametric equations and
parametric curves (i.e. graphs of parametric equations)
Tangents with Parametric Equations – Finding tangent lines to parametric curves.
Area with Parametric Equations – Finding the area under a parametric curve.
Arc Length with Parametric Equations – Determining the length of a parametric
curve.
Surface Area with Parametric Equations – Here we will determine the surface area of
a solid obtained by rotating a parametric curve about an axis.
Polar Coordinates – We’ll introduce polar coordinates in this section. We’ll look at
converting between polar coordinates and Cartesian coordinates. We will also look at
some basic graphs in polar coordinates.
Tangents with Polar Coordinates – Finding tangent lines of polar curves.
Area with Polar Coordinates – Finding the area enclosed by a polar curve.
Arc Length with Polar Coordinates – Determining the length of a polar curve.
Surface Area with Polar Coordinates – Here we will determine the surface area of a
solid obtained by rotating a polar curve about an axis.
Arc Length and Surface Area Revisited – In this section we will summarize all the arc
length and surface area formulas from the last two chapters.


107


Calculus II

Parametric Equations and Curves
To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at
functions in the form y = f ( x ) or x = h ( y ) and almost all of the formulas that we’ve
developed require that functions be in one of these two forms. The problem is that not all
curves or equations that we’d like to look at fall easily into this form.
Take, for example, a circle. It is easy enough to write down the equation of a circle
centered at the origin with radius r.
x2 + y 2 = r 2
However, we will never be able to write the equation of a circle down as a single
equation in either of the forms above. Sure we can solve for x or y as the following two
formulas show
y = ± r 2 − x2
x = ± r 2 − y2
but there are in fact two functions in each of these. Each formula gives a portion of the
circle.
y = r 2 − x2
y = − r 2 − x2

( top )
( bottom )

x = r 2 − y2
x = − r2 − y2

( right side )
( left side )

Unfortunately we usually are working on the whole circle, or simply can’t say that we’re
going to be working only on one portion of it. Even if we can narrow things down to
only one of these portions the function is still fairly unpleasant to work with.
There are also a great many curves out there that we can’t even write down as a single
equation in terms of only x and y.
To deal with some of these problems we introduce parametric equations. Instead of
defining y in terms of x ( y = f ( x ) ) or x in terms of y ( x = h ( y ) ) we define both x and y
in terms of a third variable called a parameter.
x = f (t )
y = g (t )
This third variable is usually denoted by t (as we did here) but doesn’t have to be of
course. Sometimes we will restrict the values of t that we’ll use and at other times we
won’t. This will often be dependent on the problem and just what we are attempting to
do.
Each value of t defines a point ( x, y ) = ( f ( t ) , g ( t ) ) that we can plot. The collection of
points that we get by letting t be all possible values is the graph of the parametric
equations and is called the parametric curve.
Sketching a parametric curve is not always an easy thing to do. Let’s take a look at an
example to see one way of sketching a parametric curve. This example will also illustrate
why this method is usually not the best.


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Calculus II

Example 1 Sketch the parametric curve for the following set of parametric equations.
x = t2 + t
y = 2t − 1
Solution
At this point our only option for sketching a parametric curve is to pick values of t, plug
them into the parametric equations and then plot the points. So, let’s plug in some t’s.

t
-2
-1
1
−
2
0
1

x
2
0
1
−
4
0
2

y
-5
-3
-2
-1
1

The first question that should be asked as this point is : How did we know to use the
values of t that we did, especially the third choice? Unfortunately there is no real answer
to this question. We simply pick t’s until we are fairly confident that we’ve got a good
idea of what the curve looks like. It is this problem with picking “good” values of t that
make this method of sketching parametric curves one of the poorer choices. Sometimes
we have no choice, but if we do have a choice we should avoid it. We’ll discuss an
alternate graphing method in later examples.
We have one more idea to discuss before we actually sketch the curve. Parametric curves
have a direction of motion. The direction of motion is given by increasing t. So, when
plotting parametric curves we also include arrows that show the direction of motion. We
will often give the value of t that gave specific points on the graph to make it clear the
value of t that have that particular point.
Here is the sketch of this parametric curve.

So, we had a parabola that opens to the right.

109


Calculus II

Before addressing a much easier way to sketch this graph let’s first address the issue of
limits on the parameter. In the previous example we didn’t have any limits on the
parameter. This will not always be the case. So, let’s look a slight variation of the
previous example.
x = t2 + t
y = 2t − 1
−1 ≤ t ≤ 2
Solution
Note that the only difference here is the presence of the limits on t. All these limits do is
tell us that we can’t take any value of t outside of this range. Therefore, the parametric
curve will only be a portion of the curve above. Here is the parametric curve for this
example.

It is now time to take a look at an easier method of sketching this parametric curve. This
method uses the fact that in many cases we can actually eliminate the parameter from the
parametric equations and get a function involving only x and y. There will be two small
problems with this method, but it will be easy to address those problems.
Just how we eliminate the parameter will depend upon the parametric equations that
we’ve got. Let’s see how to eliminate the parameter for the set of parametric equations
that we’ve been working with to this point.
Example 3 Eliminate the parameter from the following set of parametric equations.
x = t2 + t
y = 2t − 1
Solution
One of the easiest ways to eliminate the parameter is to simply solve one of the equations
for the parameter (t, in this case) and substitute that into the other equation. Note that
while this may be the easiest, it’s usually not the best as we’ll see soon enough.

In this case we can solve y for t.


110


Calculus II

t=

1
( y + 1)
2

Plugging this into the equation for x gives,
2

1
3
⎛1
⎞ 1
x = ⎜ ( y + 1) ⎟ + ( y + 1) = y 2 + y +
4
4
⎝2
⎠ 2
Sure enough this is a parabola that opens to the right.
Getting a sketch of the parametric curve once we’ve eliminated the parameter is fairly
simple. All we need to do is graph the equation that we found by eliminating the
parameter.
As noted already however, there are two small problems with this method. The first is
direction of motion. The equation involving only x and y will NOT give the direction of
motion of the parametric curve. This is an easy problem to fix however. All we need to
do is plug in some values of t into the parametric equations and we can determine
direction of motion from that. How many values of t we plug in will depend upon the
parametric equations. In some cases only two will be required and in others we might
need more points.
The second problem is best illustrated in an example as we’ll be seeing this in the
remaining examples.
Let’s sketch another parametric curve.
Clearly indicate direction of motion.
x = 5cos t
y = 2sin t
0 ≤ t ≤ 2π
Solution
In this case we could eliminate the parameter as we did in the previous section by solving
one of these for t and plugging this into the other. For example,
⎛
⎛ x⎞
⎛ x ⎞⎞
t = cos −1 ⎜ ⎟
⇒
y = 2sin ⎜ cos −1 ⎜ ⎟ ⎟
⎝5⎠
⎝ 5 ⎠⎠
⎝
Can you see the problem with doing this? We have a greater chance of correctly
graphing the original parametric equations that we do graphing this!

There are many ways to eliminate the parameter form the parametric equations and
solving for t is usually not the best way to do it. In this case let’s notice that we could do
the following.
x 2 y 2 25cos 2 t 4sin 2 t
+
=
+
= cos 2 t + sin 2 t = 1
25 4
25
4
Eliminating the middle steps gives us,


111


Calculus II

x2 y2
+
=1
25 4
and so it looks like we’ve got an ellipse.
Next, we need to determine the direction of motion. Note that in this case we’ll need
more than two points to do this. Given any two points on an ellipse we could get
between them by going either clockwise or counter-clockwise about the circle. So, we’ll
need at least three points to accurately determine the direction of motion.
While doing this we should also keep in mind that we’ve been given a range of t’s to
work with and as we saw in Example 2 this may mean that we will only get a portion of
the actual ellipse. So, let’s choose t’s that will cover the whole range. This will give us
the direction of motion and enough information to determine what portion of the ellipse is
in fact traced out.
Note that this is the second problem alluded to above in eliminating the parameter. Once
we have eliminated the parameter we’ve not only eliminated the direction of motion, but
we’ve also eliminated any information about what portion of the actual graph is traced
out by the parametric equations. We will always need to keep in mind that this a
potential problem when eliminating the parameter.
So, here is a table of values for this set of parametric equations.
t
0

π

2

π
3π
2
2π

x
5

y
0

0

2

-5

0

0

-2

5

0

It looks like we are moving in a counter-clockwise direction about the ellipse and it also
looks like we’ll make exactly one complete trace of the ellipse in the range given.
Here is a sketch of the parametric curve.


112


Calculus II

Clearly indicate direction of motion.
x = 5cos ( 3t )
y = 2sin ( 3t )
0 ≤ t ≤ 2π
Solution
Note that the only difference in these parametric equations is that we replaced the t with
3t. We can eliminate the parameter here in exactly the same way that we did in the
previous example.
2
2
x 2 y 2 25cos ( 3t ) 4sin ( 3t )
+
=
+
= cos 2 ( 3t ) + sin 2 ( 3t ) = 1
25 4
25
4

So, we get the same ellipse that we did in the previous example. However, we don’t get
the same parametric curve in some sense. We saw in the previous example that we make
one complete trace of the ellipse in the range 0 ≤ t ≤ 2π . In this set of parametric curves
we don’t have just a t however. In this set we’ve got a 3t. This means that we’ll
complete one complete trace when
2π
3t = 2π
t=
⇒
3
So, while we have the same ellipse that we got in the previous example it will be traced
out three times in the range of t’s given instead of only once as we got in that example.
By picking values of t we can see that the direction of motion isn’t changed in this case.
However, because we’re going around faster than before we should probably use a
different set this time to make sure we get an accurate idea of the direction of motion.
t
0

π

6


x
5

y
0

0

2

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Calculus II

π

-5 0

3
Here’s the sketch.

So, we saw in the last two examples two sets of parametric equations that in some way
gave the same graph. Yet, because they traced out the graph a different number of times
we really do need to think of them as different parametric curves. This may seem like a
difference that we don’t need to worry about, but as we will see in later sections this can
be a very important difference. In some of the later sections we are going to need a curve
that is traced out exactly once.
Clearly identify the direction of motion. If the curve is traced out more than once give a
range of the parameter for which the curve will trace out exactly once.
x = sin 2 t
y = 2 cos t
Solution
We can eliminate the parameter much as we did in the previous two examples. However,
we’ll need to note that the x already contains a sin 2 t and so we won’t need to square the
x. We will however, need to square the y.
y2
y2
x+
= sin 2 t + cos 2 t = 1
⇒
x = 1−
4
4
In this case we get a parabola that opens to the left.

We will need to be very, very careful however in sketching this parametric curve. We
will NOT get the whole parabola. To see this let’s note the following,
−1 ≤ sin t ≤ 1
−1 ≤ cos t ≤ 1


⇒
⇒

0 ≤ sin 2 t ≤ 1
− 2 ≤ 2 cos t ≤ 2

114

⇒
⇒

0 ≤ x ≤1
−2≤ y ≤ 2


Calculus II

We will only get the portion of the parabola that is traced out in this range of x and y.
Before sketching let’s also get the direction of motion. Here are some points for a range
of t’s.
t
0

π

2

π
3π
2
2π

x
0

y
2

1

0

0 -2
1

0

0

2

So, in the range 0 ≤ t ≤ 2π we start at (0,2) and end up back at that same point. Recalling
that we must travel along the parabola this means that we must retrace our path to get
back to the starting point.
So, it looks like we’ve got a parametric curve that is traced out over and over in both
directions and we will trace out once in the range 0 ≤ t ≤ π .
Here is a sketch of the curve with a few value of t noted on it.

To this point we’ve seen examples that would trace out the complete graph that we got by
eliminating the parameter if we took a large enough range of t’s. However, in the
previous example we’ve now seen that this will not always be the case. It is more than
possible to have a set of parametric equations which will continuously trace out just a
portion of the curve.


115


Calculus II

We can usually determine this by looking for limits on x and y that are imposed up us by
the parametric equation.
We will often use parametric equations to describe the path of an object or particle. Let’s
take a look at an example of that.
Example 7 The path of a particle is given by the following set of parametric equations.
x = 3cos ( 2t )
y = 1 + cos 2 ( 2t )
Completely describe the path of this particle. Do this by sketching the path, determining
limits on x and y and giving a range of t’s for which the path will be traced out exactly
once (provide it traces out more than once of course).
Solution
Eliminating the parameter this time will be a little different. We only have cosines this
time and we’ll use that to our advantage. We can solve the x equation for cosine and plug
that into the equation for y. This gives,
2

x2
⎛ x⎞
y = 1+ ⎜ ⎟ = 1+
9
⎝3⎠

x
cos ( 2t ) =
3

This time we’ve got a parabola that opens upward. We also have the following limits on
x and y.
−1 ≤ cos ( 2t ) ≤ 1
− 3 ≤ 3cos ( 2t ) ≤ 3
−3 ≤ x ≤ 3
0 ≤ cos 2 ( 2t ) ≤ 1

1 ≤ 1 + cos 2 ( 2t ) ≤ 2

1≤ y ≤ 2

So, again we only trace out a portion of the curve. Here’s a set of evaluations so we can
determine a range of t’s for one trace of the curve.
t
0

x
3
0

π

4

π

y
2
1

-3 2

2
3π
4

0
3

π

1
2

So, it looks like the particle will again, continuously trace out this portion of the curve
and will make one trace in the range 0 ≤ t ≤

π

2

. Here is a sketch of the particles path with

a few value of t on it.


116


Calculus II

We should give a small warning at this point. Because of the ideas involved in them we
concentrated on parametric curves that retraced portions of the curve more than once. Do
not however, get too locked into the idea that this will always happen. Many, if not most
parametric curves will only trace out once. The first one we looked at is a good example
of this. That parametric curve will never repeat any portion of itself.
There is one final topic to be discussed in this section before moving on. So far we’ve
started with parametric equations and eliminated the parameter to determine the
parametric curve.
However, there are times in which we want to go the other way. Given a function or
equation we might want to write down a set of parametric equations for it. In these cases
we say that we parameterize the function.
If we take Examples 4 and 5 as examples we can do this for ellipses (and hence circles).
Given the ellipse
x2 y 2
+
=1
a 2 b2
a set of parametric equations for it would be,

x = a cos t

y = b sin t

There is usually more than one set of parametric equations for the ellipse. Any of the
following will also parameterize the same ellipse.
x = a cos (ω t )

y = b sin (ω t )

x = a sin (ω t )

y = b cos (ω t )

x = − a cos (ω t )

y = b sin (ω t )

There are many more of course, but you get the idea. Also note that while all of these
have parametric curves in the shape of the ellipse they will all have different speeds as
they travel around the ellipse, they will have different starting places (if we think of t=0
as the starting place anyway) and they have potentially different directions of motion.

117


Calculus II

As noted we can use these to get the parametric equations for a circle centered at the
origin of radius r as well. On possible way to parameterize a circle is,
x = r cos t

y = r sin t

Finally, even though there may not seem to be any reason to, we can also parameterize
functions in the form y = f ( x ) or x = h ( y ) . In these cases we parameterize them in the
following way,
x=t

x = h (t )

y = f (t )

y=t

Tangents with Parametric Equations
In this section we want to find the tangent lines to the parametric equations given by,
x = f (t )
y = g (t )
To do this let’s first recall how to find the tangent line to y = F ( x ) at x=a. Here the
tangent line is given by,
dy
y = F ( a ) + m ( x − a ) , where m =
= F′(a)
dx x = a
dy
from the parametric
dx
equations we could simply reuse this formula since we will be able to use the parametric
equations to find the x and y coordinates of the point.

Now, notice that if we could figure out how to get the derivative

So, just for a second let’s suppose that we where able to eliminate the parameter from the
parametric form and write the parametric equations in the form y = F ( x ) . Now, plug
the parametric equations in for x and y. Yes, it seem silly to eliminate the parameter, then
immediately put it back in, but it’s what we need to do in order to get our hands on the
derivative. Doing this gives,
g (t ) = F ( f (t ))
Now, differentiate with respect to t and notice that we’ll need to use the Chain Rule on
the right hand side.
g′ (t ) = F ′ ( f (t )) f ′ (t )
Let’s do another change in notation. We need to be careful with our derivatives here.
Derivatives of the lower case function are with respect to t while derivatives of upper


118


Calculus II

case functions are with respect to x. So, to make sure that we keep this straight let’s
rewrite things as follows.
dy
dx
= F′( x)
dt
dt
At this point we should remind ourselves just what we here after. We needed a formula
dy
or F ′ ( x ) that we is in terms of the parametric formulas. Notice however that we
for
dx
can get that from the above equation.
dy
dy
dx
= dt , provided
≠0
dx
dx
dt
dt
Notice as well that this will be a function of t and not x.
As an aside, notice that we could also get the following formula with a similar derivation
if we needed to,
dx
dx
dy
= dt , provided
≠0
dy
dy
dt
dt
Why would we want to do this? Well, recall that in the arc length section of the
Applications of Integral section we actually needed this derivative on occasion.
So, let’s find a tangent line.
Example 1 Find the tangent line(s) to the parametric curve given by
x = t 5 − 4t 3
y = t2
at (0,4).
Solution
Note that there is apparently the potential for more than one tangent line here! We will
look into this more after we’re done with the example.

The first thing that we should do is find the derivative so we can get the slope of the
tangent line.
dy
dy
2t
2
= dt = 4
= 3
2
dx
5t − 12t
5t − 12t
dx
dt
At this point we’ve got a small problem. The derivative is in terms of t and all we’ve got


119


Calculus II

is an x-y coordinate pair. The next step then is to determine that value(s) of t which will
give this point. We find these by plugging the x and y values into the parametric
equations and solving for t.
0 = t 5 − 4t 3 = t 3 ( t 2 − 4 )

⇒

t = 0, ±2

4 = t2

⇒

t = ±2

Any value of t which appears in both lists will give the point. So, since there are two
values of t that give the point we will in fact get two tangent lines. That’s definitely not
something that happened back in Calculus I and we’re going to need to look into this a
little more. However, before we do that let’s actually get the tangent lines.
t = -2
Since we already know the x and y-coordinates of the point all that we need to do is find
the slope of the tangent line.
dy
1
m=
=−
dx t = − 2
8
The tangent line (at t = -2) is then,
1
y = 4− x
8
t=2
Again, all we need is the slope.
m=

dy
dx

=
t =2

1
8

The tangent line (at t = 2) is then,
1
y = 4+ x
8
Now, let’s take a look at just how we could possibly get two tangents lines at a point.
This was definitely not possible back in Calculus I where we first ran across tangent
lines.
A quick graph of the parametric curve will explain what is going on here.


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Calculus II

So, the parametric curve crosses itself! That explains how there can be more than one
tangent line. There is one tangent line for each instance that the curve goes through the
point.
The next topic that we need to discuss in this section is that of horizontal and vertical
tangents. We can easily identify where these will occur (or at least the t’s that will give
them) by looking at the derivative formula.
dy
dy
= dt
dx
dx
dt
Horizontal tangents will occur where the derivative is zero and that means that we’ll get
horizontal tangent at values of t for which we have,
dy
dx
= 0, provided
≠0
dt
dt
Vertical tangents will occur where the derivative is not defined and so we’ll get vertical
tangents at values of t for which we have,
dx
dy
= 0, provided
≠0
dt
dt
Example 2 Determine the x-y coordinates of the points where the following parametric
equations will have horizontal or vertical tangents.
x = t 3 − 3t
y = 3t 2 − 9
Solution
We’ll first need the derivatives of the parametric equations.


121


Calculus II

dx
= 3t 2 − 3 = 3 ( t 2 − 1)
dt
Horizontal Tangents
We’ll have horizontal tangents where,
6t = 0

⇒

dy
= 6t
dt

t=0

Now, this is the value of t which gives the horizontal tangents and we were asked to find
the x-y coordinates of the point. To get these we just need to plug t into the parametric
equations. Therefore, the only horizontal tangent will occur at the point (0,-9).
Vertical Tangents
In this case we need to solve,
3 ( t 2 − 1) = 0

⇒

t = ±1

The two vertical tangents will occur at the points (2,-6) and (-2,-6).
For the sake of completeness and at least partial verification here is the sketch of the
parametric curve.

The final topic that we need to discuss in this section really isn’t related to tangent lines,
but does fit in nicely with the derivation of the derivative that we needed to get the slope
of the tangent line.
Before moving into the new topic let’s first remind ourselves of the formula for the first
derivative and in the process rewrite it slightly.


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Calculus II

d
( y)
dy d
dt
= ( y) =
dx
dx dx
dt
Written in this way we can see that the formula actually tells us how to differentiate a
function y (as a function of t) with respect to x (when x is also a function of t) when we
are using parametric equations.
Now let’s move onto the final topic of this section. We would also like to know how to
get the second derivative of y with respect to x.
d2y
dx 2
Getting a formula for this is fairly simple if we remember the rewritten formula for the
first derivative above.
d ⎛ dy ⎞
d 2 y d ⎛ dy ⎞ dt ⎜ dx ⎟
= ⎜ ⎟= ⎝ ⎠
2
dx
dx
dx ⎝ dx ⎠
dt
Note that,
d2y
2
d y
≠ dt
d 2x
dx 2
dt 2
2

Let’s work a quick example.
Example 3 Find the second derivative for the following set of parametric equations.
x = t 5 − 4t 3
y = t2
Solution
This is the set of parametric equations that we used in the first example and so we already
have the following computations completed.
dy
= 2t
dt
dx
= 5t 4 − 12t 2
dt
dy
2
= 3
dx 5t − 12t

We will first need the following,


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Calculus II

2
d⎛
2
24 − 30t 2
⎞ −2 (15t − 12 )
=
⎜
⎟=
2
dt ⎝ 5t 3 − 12t ⎠ ( 5t 3 − 12t )2
( 5t 3 − 12t )

The second derivative is then,
d ⎛ dy ⎞
d y dt ⎜ dx ⎟
= ⎝ ⎠
2
dx
dx
dt
24 − 30t 2
2

( 5t
=
=
=

3

− 12t )

2

5t 4 − 12t 2
24 − 30t 2

( 5t

4

− 12t 2 )( 5t 3 − 12t )

2

24 − 30t 2

t ( 5t 3 − 12t )

3

So, why would we want the second derivative? Well, recall from your Calculus I class
that with the second derivative we can determine where a curve is concave up and
concave down. We could do the same thing with parametric equations if we wanted to.
Example 4 Determine the values of t for which the parametric curve given by the
following set of parametric equations is concave up and concave down.
x = 1− t2
y = t7 + t5
Solution
To compute the second derivative we’ll first need the following.
dy
= 7t 6 + 5t 4
dt
dx
= −2t
dt
dy 7t 6 + 5t 4
1
=
= − ( 7t 5 + 5t 3 )
dx
−2t
2

Note that we can also use the first derivative above to get some information about the
increasing/decreasing nature of the curve as well. In this case it looks like the parametric
curve will be increasing if t<0 and decreasing if t>0.
Now let’s move on to the second derivative.
1
− ( 35t 4 + 15t 2 )
d2y
1
= 2
= ( 35t 3 + 15t )
2
dx
−2t
4

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Calculus II

It’s clear, hopefully, that the second derivative will only be zero at t=0. Using this we
can see that the second derivative will be negative if t<0 and positive if t>0. So the
parametric curve will be concave down for t<0 and concave up for t>0.
Here is a sketch of the curve for completeness sake.

Area with Parametric Equations
In this section we will find a formula for determining the area under a parametric curve
given by the parametric equations,
x = f (t )
y = g (t )
We will also need to further add in the assumption that the curve is traced out exactly
once as t increases from α to β.
We will do this in much the same way that we found the first derivative in the previous
section. We will first recall how to find the area under y = F ( x ) on a ≤ x ≤ b .
A = ∫ F ( x ) dx
b

a

We will now think of the parametric equation x = f ( t ) as a substitution in the integral.
We will also assume that a = f (α ) and b = f ( β ) for the purposes of this formula.
There is actually no reason to assume that this will always be the case and so we’ll give a
corresponding formula later if it’s the opposite case ( b = f (α ) and a = f ( β ) ).
So, if this is going to be a substitution we’ll need,
dx = f ′ ( t ) dt


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Calculus II

Plugging this into the area formula above and making sure to change the limits to their
corresponding t values gives us,
A = ∫ F ( f ( t ) ) f ′ ( t ) dt
β

α

Since we don’t know what F(x) is we’ll use the fact that
y = F ( x ) = F ( f (t )) = g (t )
and we arrive at the formula that we want.
β

A = ∫ g ( t ) f ′ ( t ) dt
α

Now, if we should happen to have b = f (α ) and a = f ( β ) the formula would be,
α

A = ∫ g ( t ) f ′ ( t ) dt
β

Let’s work an example.
Example 1 Determine the area under the parametric curve given by the following
parametric equations.
x = 6 (θ − sin θ )
y = 6 (1 − cos θ )
0 ≤ θ ≤ 2π
Solution
First, notice that we’ve switched the parameter to θ for this problem. This is to make
sure that we don’t get too locked into always having t as the parameter.

Now, we could graph this to verify that the curve is traced out exactly once for the given
range if we wanted to. We are going to be looking at this curve in more detail after this
example so we won’t sketch its graph here.
There really isn’t too much to this example other than plugging the parametric equations
into the formula. We’ll first need the derivative of the parametric equation for x
however.
dx
= 6 (1 − cos θ )
dθ
The area is then,


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Calculus II

A=∫

2π
0

36 (1 − cos θ ) dθ
2

2π

= 36 ∫ 1 − 2 cos θ + cos 2 θ dθ
0

2π
3
1
= 36⌠
− 2 cos θ + cos ( 2θ ) dθ
⎮
⌡0 2
2
2π

1
⎛3
⎞
= 36 ⎜ θ − 2sin θ + sin ( 2θ ) ⎟
4
⎝2
⎠0
= 108π

The parametric curve (without the limits) we used in the previous example is called a
cycloid. In its general form the cycloid is,
x = r (θ − sin θ )

y = r (1 − cos θ )

The cycloid represents the following situation. Consider a wheel of radius r. Let the
point where the wheel touches the ground initially be called P. Then start rolling the
wheel to the right. As the wheel rolls to the right trace out the path of the point P. The
path that the point P traces out is called a cycloid and is given by the equations above. In
these equations we can think of θ as the angle through which the point P has rotated.
Here is a sketch of the cycloid used in the previous example.

We can see that one arch of the cycloid is traced out in the range 0 ≤ θ ≤ 2π . This makes
sense when you consider that the point P will be back on the ground after it has rotated
through and angle of 2π.

Arc Length with Parametric Equations
In the previous two sections we’ve looked at a couple of Calculus I topics in terms of
parametric equations. We now need to look at a couple of Calculus II topics in terms of
In this section we will look at the arc length of the parametric curve given by,
x = f (t )
y = g (t )
α ≤t≤β


127


Calculus II

We will also be assuming that the curve is traced out exactly once as t increases from α
to β. Also, for the purposes of the derivation that we’re going to use we will assume that
the curve is traced out from left to right as t increases. This is equivalent to saying,
dx
≥0
dt

for α ≤ t ≤ β

This is not actually required for the final formula, but as noted above we’ll need it for our
derivation. If the curve isn’t traced out from left to right we would need to go through a
slightly more complicated derivation.
So, let’s start out the derivation by recalling the arc length formula as we first derived it
in the arc length section of the Applications of Integrals chapter.
L = ∫ ds
where,
2

⎛ dy ⎞
ds = 1 + ⎜ ⎟ dx
⎝ dx ⎠
2

⎛ dx ⎞
ds = 1 + ⎜ ⎟ dy
⎝ dy ⎠

if y = f ( x ) , a ≤ x ≤ b
if y = h ( y ) , c ≤ y ≤ d

We will use the first ds above because we have a nice formula for the derivative in terms
of the parametric equations (see the Tangents with Parametric Equations section). To use
this we’ll also need to know that,
dx
dx = f ′ ( t ) dt = dt
dt
The arc length formula then becomes,
β

⌠
⎛ dy
⎮
⎜
L = ⎮ 1 + ⎜ dt
dx
⎮
⎜
⎮
⎝ dt
⌡α

⎞
⎟
⎟
⎟
⎠

2

β

2
⌠
⎛ dy ⎞
⎮
⎜ ⎟
dx
dt
dt = ⎮ 1 + ⎝ ⎠2
dt
⎮
⎛ dx ⎞
⎜ ⎟
⎮
⎝ dt ⎠
⌡α

dx
dt
dt

This is a particularly unpleasant formula. However, if we factor out the denominator
from the square root we arrive at,
β

⌠
1
L=⎮
⎮ dx
⎮
⌡ α dt


2

⎛ dx ⎞ ⎛ dy ⎞
⎜ ⎟ +⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠

128

2

dx
dt
dt


Calculus II

Now, making use of our assumption that the curve is being traced out from left to right
we can drop the absolute value bars on the derivative which will allow us to cancel the
two derivatives that are outside the square root this gives,
β

⌠
L=⎮
⌡α

2

2

⎛ dx ⎞ ⎛ dy ⎞
⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠

Notice that we could have used the second formula for ds above is we had assumed
instead that
dy
≥0
for α ≤ t ≤ β
dt
If we had gone this route in the derivation we would have gotten the same formula.
Example 1 Determine the length of the parametric curve given by the following
x = 3sin ( t )
y = 3cos ( t )
0 ≤ t ≤ 2π
Solution
We know that this is a circle of radius 3 centered at the origin from our prior discussion
about graphing parametric curves. We also know from this discussion that it will be
traced out exactly once in this range.

So, we can use the formula we derived above. We’ll first need the following,
dx
dy
= 3cos ( t )
= −3sin ( t )
dt
dt
The length is then,
L=∫

=∫

2π

9sin 2 ( t ) + 9 cos 2 ( t ) dt

0
2π
0

= 3∫

3 sin 2 ( t ) + cos 2 ( t ) dt

2π
0

dt

= 6π
Since this is a circle we could have just used the fact that the length of the circle is just
the circumference of the circle. This is a nice way, in this case, to verify our result.
Let’s take a look at one possible consequence of a curve is traced out more than once and
we try to find the length of the curve without taking this into account.
Example 2 Use the arc length formula for the following parametric equations.
x = 3sin ( 3t )
y = 3cos ( 3t )
0 ≤ t ≤ 2π


129


Calculus II

Solution
Notice that this is the identical circle that we had in the previous example and so the
length is still 6π. However, for the range given we know trace out the curve three times
instead once as required for the formula. Despite that restriction let’s use the formula
anyway and see what happens.

In this case the derivatives are,
dx
= 9 cos ( 3t )
dt
and the length formula gives,
L=∫
=∫

2π
0
2π
0

dy
= −9sin ( 3t )
dt
81sin 2 ( t ) + 81cos 2 ( t ) dt

9 dt

= 18π

The answer we got form the arc length formula in this example was 3 times the actual
length. Recalling that we also determined that this circle would trace out three times in
the range given, the answer should make some sense.
If we had wanted to determine the length of the circle for this set of parametric equations
we would need to determine a range of t for which this circle is traced out exactly once.
2π
. Using this range of t we get the following for the length.
This is, 0 ≤ t ≤
3
L=∫

2π
3
0

=∫

2π
3
0

81sin 2 ( t ) + 81cos 2 ( t ) dt
9 dt

= 6π

which is the correct answer.
Be careful to not make the assumption that this is always what will happen if the curve is
traced out more than once. Just because the curve traces out n times does not mean that
the arc length formula will give us n times the actual length of the curve!
Before moving on to the next section let’s notice that we can put the arc length formula
derived in this section into the same form that we had when we first looked at arc length.
The only difference is that we will add in a definition for ds when we have parametric
equations.
The arc length formula can be summarized as,
L = ∫ ds
where,


130


Calculus II

2

⎛ dy ⎞
ds = 1 + ⎜ ⎟ dx
⎝ dx ⎠

if y = f ( x ) , a ≤ x ≤ b

2

⎛ dx ⎞
ds = 1 + ⎜ ⎟ dy
⎝ dy ⎠
2

if y = h ( y ) , c ≤ y ≤ d
2

⎛ dx ⎞ ⎛ dy ⎞
ds = ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠

if x = f ( t ) , y = g ( t ) , α ≤ t ≤ β

Surface Area with Parametric Equations
In this final section dealing with calculus using parametric equations we will take a look
at determining the surface area of a region obtained by rotating a parametric curve about
the x or y-axis.
We will rotate the parametric curve given by,
x = f (t )
y = g (t )

α ≤t≤β

about the x or y-axis. We are going to assume that the curve is traced out exactly once as
t increases from α to β. At this point there actually isn’t all that much to do. We know
that the surface area can be found by using one of the following two formulas depending
on the axis of rotation (recall the Surface Area section of the Applications of Integrals
chapter).
S = ∫ 2π y ds
S = ∫ 2π x ds


All that we need is a formula for ds to use and from the previous section we have,
2

2

⎛ dx ⎞ ⎛ dy ⎞
ds = ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠

if x = f ( t ) , y = g ( t ) , α ≤ t ≤ β

which is exactly what we need.
We will need to be careful with the x or y that is in the original surface area formula.
Back when we first looked at surface area we saw that sometimes we had to substitute for
the variable in the integral and at other times we didn’t. This was dependent upon the ds
that we used. In this case however, we will always have to substitute for the variable.
The ds that we use for parametric equations introduces a dt into the integral and that
means that everything needs to be in terms of t. Therefore, we will need to substitute the
appropriate parametric equation for x or y depending on the axis of rotation.
Let’s take a quick look at an example.


131


Calculus II

Example 1 Determine the surface area of the solid obtained by rotating the following
parametric curve about the x-axis.

x = cos3 θ

y = sin 3 θ

0 ≤θ ≤

π

2

Solution
We’ll first need the derivatives of the parametric equations.
dx
dy
= −3cos 2 θ sin θ
= 3sin 2 θ cos θ
dt
dt

Before plugging into the surface area formula let’s get the ds out of the way.
ds = 9 cos 4 θ sin 2 θ + 9sin 4 θ cos 2 θ dt

= 3 cos θ sin θ

cos 2 θ + sin 2 θ

= 3cos θ sin θ
Notice that we could drop the absolute value bars since both sine and cosine are positive
in this range of θ given.
Now let’s get the surface area and don’t forget to also plug in for the y.
S = ∫ 2π y ds
π

= 2π ∫ 2 sin 3 θ ( 3cos θ sin θ ) dθ
0

π

= 6π ∫ 2 sin 4 θ cos θ dθ

u = sin θ

0

1

= 6π ∫ u 4 du
0

=

6π
5

Polar Coordinates
Up to this point we’ve dealt exclusively with the Cartesian (or Rectangular, or x-y)
coordinate system. However, as we will see, this is not always the easiest coordinate
system to work in. So, in this section we will start looking at the polar coordinate system.
Coordinate systems are really nothing more than a way to define a point. For instance in
the Cartesian coordinate system at point is given the coordinates (x,y) and we use this to
define the point by starting at the origin and then moving x units horizontally followed by
y units vertically. This is shown in the sketch below.


132


Calculus II

This is not, however, the only way to define a point in two dimensional space. Instead of
moving vertically and horizontally from the origin to get to the point we could instead go
straight out of the origin until we hit the point and then determine the angle this line
makes with the positive x-axis. We could then use the distance of the point from the
origin and the amount we needed to rotate from the positive x-axis as the coordinates of
the point. This is shown in the sketch below.

Coordinates in this form are called polar coordinates.
The above discussion may lead one to think that r must be a positive number. However,
π⎞
⎛ π⎞
⎛
we also allow r to be negative. Below is a sketch of the two points ⎜ 2, ⎟ and ⎜ −2, ⎟ .
6⎠
⎝
⎝ 6⎠


133


Calculus II

From this sketch we can see that if r is positive the point will be in the same quadrant as
θ. On the other hand if r is negative the point will end up in the quadrant exactly
π⎞
⎛
opposite θ. Notice as well that the coordinates ⎜ −2, ⎟ describe the same point as the
6⎠
⎝
7π
⎛ 7π ⎞
⎛ 7π ⎞
from
coordinates ⎜ 2,
⎟ do. The coordinates ⎜ 2,
⎟ tells us to rotate an angle of
6
⎝ 6 ⎠
⎝ 6 ⎠
the positive x-axis, this would put us on the dashed line in the sketch above, and then
move out a distance of 2.
This leads to an important difference between Cartesian coordinates and polar
coordinates. In Cartesian coordinates there is exactly one set of coordinates for any given
point. With polar coordinates this isn’t true.
In polar coordinates there is literally an infinite number of coordinates for a given point.
For instance, the following four points are all coordinates for the same point.
5π
⎛ π⎞ ⎛
⎜ 5, ⎟ = ⎜ 5, −
3
⎝ 3⎠ ⎝

4π
⎞ ⎛
⎟ = ⎜ −5,
3
⎠ ⎝

2π ⎞
⎞ ⎛
⎟ = ⎜ −5, −
⎟
3 ⎠
⎠ ⎝

Here is a sketch of the angles used in these four sets of coordinates.


134


Calculus II

In the second coordinate pair we rotated in a clock-wise direction to get to the point. We
shouldn’t forget about rotating in the clock-wise direction. Sometimes it’s what we have
to do.
The last two coordinate pairs use the fact that if we end up in the opposite quadrant from
the point we can use a negative r to get back to the point and of course there is both a
counter clock-wise and a clock-wise rotation to get to the angle.
These four points only represent the coordinates of the point without rotating around the
system more than once. If we allow the angle to make as many complete rotation about
the axis system as we want then there are an infinite number of coordinates for the same
point. In fact the point ( r , θ ) can be represented by any of the following coordinate
pairs.

( r ,θ + 2π n )

( −r ,θ + ( 2n + 1) π ) ,

where n is any integer.

Next we should talk about the origin of the coordinate system. In polar coordinates the
origin is often called the pole. Because aren’t actually moving away from the origin/pole
we know that r=0. However, we can still rotate around the system by any angle we want
and so the coordinates of the origin/pole are ( 0, θ ) .
Now that we’ve got a grasp on polar coordinates we need to think about converting
between the two coordinate systems. Well start out with the following sketch reminding
us how both coordinate systems work.


135


Calculus II

Note that we’ve got a right triangle above and with that we can get the following
equations that will convert polar coordinates into Cartesian coordinates.
x = r cos θ
y = r sin θ
Converting from Cartesian is almost as easy. Let’s first notice the following.
2
2
x 2 + y 2 = ( r cos θ ) + ( r sin θ )
= r 2 cos 2 θ + r 2 sin 2 θ
= r 2 ( cos 2 θ + sin 2 θ )
= r2
This is a very useful formula that we should remember, however we are after an equation
for r so let’s take the square root of both sides. This gives,

r = x2 + y2
Note that technically we should have a plus or minus in front of the root since we know
that r can be either positive or negative. We will run with the convention of positive r
here.
Getting an equation for θ is almost as simple. We’ll start with,
y r sin θ
=
= tan θ
x r cos θ
Taking the inverse tangent of both sides gives,
⎛ y⎞
θ = tan −1 ⎜ ⎟
⎝x⎠


136


Calculus II

We will need to be careful with this because inverse tangents only return values in the
range −

π

<θ <

π

2
2
is given by θ + π .

. Recall that there is a second possible angle and that the second angle

Summarizing then gives the following formulas for converting from Cartesian
coordinates to polar coordinates.
r 2 = x2 + y2
r = x2 + y 2

⎛ y⎞
⎝ ⎠

θ = tan −1 ⎜ ⎟
x
Let’s work a quick example.
Example 1 Convert each of the following points into the given coordinate system.
2π ⎞
⎛
(a) ⎜ −4,
⎟ into Cartesian coordinates.
3 ⎠
⎝
(b) (-1,-1) into polar coordinates.
Solution
(a) This conversion is easy enough. All we need to do is plug the points into the
formulas.
⎛ 2π ⎞
⎛ 1⎞
x = −4 cos ⎜
⎟ = −4 ⎜ − ⎟ = 2
⎝ 3 ⎠
⎝ 2⎠
⎛ 3⎞
⎛ 2π ⎞
y = −4sin ⎜
⎟ = −4 ⎜
⎜ 2 ⎟ = −2 3
⎟
⎝ 3 ⎠
⎝
⎠
So, in Cartesian coordinates this point is 2, −2 3 .

(

)

(b) Let’s first get r.
r=

( −1) + ( −1)
2

2

= 2

Now, let’s get θ.

π
⎛ −1 ⎞
−1
⎟ = tan (1) =
4
⎝ −1 ⎠
This is not the correct angle however. This value of θ is in the first quadrant and the
point we’ve been given is in the third quadrant. As noted above we can get the correct
angle by adding π onto this. Therefore, the actual angle is,
π
5π
θ = +π =
4
4
θ = tan −1 ⎜


137


Calculus II

5π ⎞
⎛
So, in polar coordinates the point is ⎜ 2, ⎟ . Note as well that we could have used the
4 ⎠
⎝
first θ that we got by using a negative r. In this case the point could also be written in
π⎞
⎛
polar coordinates as ⎜ − 2, ⎟ .
4⎠
⎝
We can also use the above formulas to convert equations from one coordinate system to
the other.
Example 2 Convert each of the following into an equation in the given coordinate
system.
(a) Convert 2 x − 5 x3 = 1 + xy into polar coordinates.
(b) Convert r = −8cos θ into Cartesian coordinates.
Solution
(a) In this case there really isn’t much to do other than plugging in the formulas for x and
y (i.e. the Cartesian coordinates) in terms of r and θ (i.e. the polar coordinates).
3
2 ( r cos θ ) − 5 ( r cos θ ) = 1 + ( r cos θ )( r sin θ )
2r cos θ − 5r 3 cos3 θ = 1 + r 2 cos θ sin θ

(b) This one is a little trickier, but not by much. First notice that we could substitute
straight for the r. However, there is no straight substitution for the cosine that will give
us only Cartesian coordinates. If we had an r on the right along with the cosine then we
could do a direct substitution. So, if an r on the right side would be convenient let’s put
one there, just don’t forget to put on the right side as well.
r 2 = −8r cos θ
We can now make some substitutions that will convert this into Cartesian coordinates.
x 2 + y 2 = −8 x
Before moving on to the next subject let’s do a little more work on the second part of the
previous example.
The equation given in the second part is actually a fairly well known graph; it just isn’t in
a form that most people will quickly recognize. To identify it let’s take the Cartesian
coordinate equation and do a little rearranging.
x2 + 8x + y 2 = 0
Now, complete the square on the x portion of the equation.
x 2 + 8 x + 16 + y 2 = 16

( x + 4)


2

+ y 2 = 16

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Calculus II

So, this was a circle of radius 4 and center (-4,0).
This leads us into the final topic of this section.

Common Polar Coordinate Graphs
Let’s identify a few of the more common graphs in polar coordinates. We’ll also take a
look at a couple of special polar graphs.
Lines
Some lines have fairly simple equations in polar coordinates.
1. θ = β .
We can see that this is a line by converting to Cartesian coordinates as follows
θ =β
⎛ y⎞
tan −1 ⎜ ⎟ = β
⎝x⎠
y
= tan β
x
y = ( tan β ) x

This is a line that goes through the origin and makes an angle of β with the
positive x-axis.
2. r cos θ = a
This is easy enough to convert to Cartesian coordinates to x=a. So, this is a
vertical line.
3. r sin θ = b
Likewise, this converts to y=b and so is a horizontal line.
Example 3 Graph θ =

3π
, r cos θ = 4 and r sin θ = −3 on the same axis system.
4

Solution
There really isn’t too much to this one other than doing the graph so here it is.


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Calculus II

Circles
Let’s take a look at the equations of circles in polar coordinates.
1. r = a .
This equation is saying that no matter what angle we’ve got the distance from the
origin must be a. If you think about it that is exactly the definition of a circle of
radius a centered at the origin.
So, this is a circle of radius a centered at the origin. This is also one of the
reasons why we might want to work in polar coordinates. The equation of a circle
centered at the origin has a very nice equation, unlike the corresponding equation
in Cartesian coordinates.
2. r = 2a cos θ .
We looked at a specific example of one of these when we were converting
equations to Cartesian coordinates.
This is a circle of radius a and center ( a, 0 ) . Note that a might be negative (as it
was in our example above) and so the absolute value bars are required on the
radius. They should not be used however on the center.
3. r = 2b sin θ .
This is similar to the previous one. It is a circle of radius b and center ( 0,b ) .
4. r = 2a cos θ + 2b sin θ .
This is a combination of the previous two and by completing the square twice it
can be shown that this is a circle of radius

a 2 + b 2 and center ( a, b ) . In other

words, this is the general equation of a circle that isn’t centered at the origin.
Example 4 Graph r = 7 , r = 4 cos θ , and r = −7 sin θ on the same axis system.


140


Calculus II

Solution
The first one is a circle of radius 7 centered at the origin. The second is a circle of radius
7⎞
7
⎛
2 centered at (2,0). The third is a circle of radius centered at ⎜ 0, − ⎟ . Here is the
2⎠
2
⎝
graph of the three equations.

Note that it takes a range of 0 ≤ θ ≤ 2π for a complete graph of r=a and it only takes a
range of 0 ≤ θ ≤ π to graph the other circles given here.
Cardioids and Limacons
These can be broken up into the following three cases.
1. Cardioids : r = a ± a cos θ and r = a ± a sin θ .
These have a graph that is vaguely heart shaped and always contain the origin.
2. Limacons with an inner loop : r = a ± b cos θ and r = a ± b sin θ with a < b .
These will have an inner loop and will always contain the origin.
3. Limacons without an inner loop : r = a ± b cos θ and r = a ± b sin θ with a > b .
These do not have an inner loop and do not contain the origin.
Example 5 Graph r = 5 − 5sin θ , r = 7 − 6 cos θ , and r = 2 + 4 cos θ .
Solution
These will all graph out once in the range 0 ≤ θ ≤ 2π . Here is a table of values for each
followed by graphs of each.

θ
0

r = 5 − 5sin θ
5

r = 7 − 6 cos θ
1

141

r = 2 + 4 cos θ
6


Calculus II

π
2

π
3π
2
2π


0

7

2

5

13

-2

10

7

2

5

1

6

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Calculus II

There is one final thing that we need to do in this section. In the third graph in the
previous example we had an inner loop. We will, on occasion, need to know the value of
θ for which the graph will pass through the origin. To find these all we need to do is set
the equation equal to zero and solve.
0 = 2 + 4 cos θ
1
cos θ = −
2

⇒

θ=

2π 4π
,
3 3

Tangents with Polar Coordinates
We now need to discuss some calculus topics in terms of polar coordinates.
We will start with finding tangent lines to polar curves. In this case we are going to
assume that the equation is in the form r = f (θ ) . With the equation in this form we can
dy
we derived when we looked at tangent
dx
lines with parametric equations. To do this however requires us to come up with a set of
parametric equations to represent the curve. This is actually pretty easy to do.
actually use the equation for the derivative

From our work in the previous section we have the following set of conversion equations
for going from polar coordinates to Cartesian coordinates.
x = r cos θ
y = r sin θ
Now, we’ll use the fact that we’re assuming that the equation is in the form r = f (θ ) .
Substituting this into these equations gives the following set of parametric equations
(with θ as the parameter) for the curve.


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Calculus II

x = f (θ ) cos θ

y = f (θ ) sin θ

Now, we will need the following derivatives.
dx
= f ′ (θ ) cos θ − f (θ ) sin θ
dθ
dr
cos θ − r sin θ
=
dθ
The derivative

dy
= f ′ (θ ) sin θ + f (θ ) cos θ
dθ
dr
sin θ + r cos θ
=
dθ

dy
is then,
dx
dr
sin θ + r cos θ
dy dθ
=
dx dr cos θ − r sin θ
dθ

Note that rather than trying to remember this formula it would probably be easier to
remember how we derived it and just remember the formula for parametric equations.
Let’s work a quick example with this.
Example 1 Determine the equation of the tangent line to r = 3 + 8sin θ at θ =

π
6

.

Solution
We’ll first need the following derivative.
dr
= 8cos θ
dθ
dy
becomes,
The formula for the derivative
dx
dy 8cos θ sin θ + ( 3 + 8sin θ ) cos θ
16 cos θ sin θ + 3cos θ
=
=
2
dx
8cos θ − ( 3 + 8sin θ ) sin θ
8cos 2 θ − 3sin θ − 8sin 2 θ

The slope of the tangent line is,
dy
=
m=
dx θ = π
6

Now, at θ =

3 3
2 = 11 3
3
5
4−
2

4 3+

π

we have r=7. We’ll need to get the corresponding x-y coordinates so we
6
can get the tangent line.


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Calculus II

⎛π ⎞ 7 3
x = 7 cos ⎜ ⎟ =
2
⎝6⎠
The tangent line is then,
y=

⎛π ⎞ 7
y = 7 sin ⎜ ⎟ =
⎝6⎠ 2

7 11 3 ⎛
7 3⎞
+
⎜x−
⎟
⎜
2
5 ⎝
2 ⎟
⎠

For the sake of completeness here is a graph of the curve and the tangent line.

Area with Polar Coordinates
In this section we are going to look at areas enclosed by polar curves. Note as well that
we said “enclosed by” instead of “under” as we typically have in these problems. These
problems work a little differently in polar coordinates. Here is a sketch of what the area
that we’ll be finding in this section looks like.


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Calculus II

We’ll be looking for the shaded area in the sketch above. The formula for finding this
area is,
β
1 2
A=⌠
r dθ
⎮
⌡α 2
Notice that we use r in the integral instead of f (θ ) so make sure and substitute
accordingly when doing the integral.
Example 1 Determine the area of the inner loop of r = 2 + 4 cos θ .
Solution
We graphed this function back when we first started looking at polar coordinates. Here is
the sketch again for the sake of completeness.


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Calculus II

So, we will need the limits that will enclose this area as we trace out the curve. This will
be the angles for which the curve passes through the origin. We can get these by setting
the equation equal to zero and solving.
0 = 2 + 4 cos θ
1
2π 4π
cos θ = −
,
⇒
θ=
2
3 3
So, the area is then,
4π

3 1
2
A=⌠
⎮ 2π ( 2 + 4 cos θ ) dθ
⌡ 2
3

4π
3

1
2
=⌠
⎮ 2π ( 4 + 16 cos θ + 16 cos θ ) dθ
⌡ 2
3

=∫

4π
3
2π
3

2 + 8cos θ + 4 (1 + cos ( 2θ ) ) dθ

=∫

4π
3
2π
3

6 + 8cos θ + 4 cos ( 2θ ) dθ
4π

= ( 6θ + 8sin θ + 2sin ( 2θ ) ) 23π
3

= 4π − 6 3 = 2.174
So, that’s how we determine areas that are enclosed by a single curve, but what about
situations like the following sketch were we want to find the area between two curves.


147


Calculus II

In this case we can use the above formula to find the area enclosed by both and then the
actual area is the difference between the two. The formula for this is,
β
1 2 2
A=⌠
⎮
( r − r ) dθ
⌡α 2 o i
Example 2 Determine the area that lies inside r = 3 + 2sin θ and outside r = 2 .
Solution
Here is a sketch of the region that we are after.

To determine this area we’ll need to know that value of θ for which the two curves
intersect. We can determine these points by setting the two equations and solving.


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Calculus II

3 + 2sin θ = 2
1
⇒
2
Here is a sketch of the figure with these angles added.
sin θ = −

θ=

7π 11π
,
6 6

Note as well here that we also acknowledged that another representation for the angle
11π
π
is − . This is important for this problem. In order to use the formula above the
6
6
7π
area must be enclosed as we increase from the smaller to larger angle. So, if we use
6
11π
we will not enclose the shaded area, instead we will enclose the bottom most of
to
6
π
7π
we will enclose the area that
the three regions. However if we use the angles − to
6
6
we’re after.
So, the area is then,


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Calculus II

7π

(

)

6 1
2
2
A=⌠
( 3 + 2sin θ ) − ( 2 ) dθ
⎮ π
⌡− 2
6

7π

6 1
2
=⌠
⎮ π ( 5 + 12sin θ + 4sin θ ) dθ
⌡− 2
6

7π
6

1
=⌠
⎮ π ( 7 + 12sin θ − 2 cos ( 2θ ) ) dθ
⌡− 2
6

7π
6
1
= ( 7θ − 12 cos θ − sin ( 2θ ) )
π
2
−

6

=

11 3 14π
+
= 24.187
2
3

Let’s work a slight modification of the previous example.
Example 3 Determine the area of the region outside r = 3 + 2sin θ and inside r = 2 .
Solution
This time we’re looking for the following region.

So, this is the region that we get by using the limits


150

7π
11π
to
. The area is then,
6
6


Calculus II

11π

(

)

6 1
2
2
A=⌠
( 2 ) − ( 3 + 2sin θ ) dθ
⎮ 7π
⌡ 2
6

11π

6 1
2
=⌠
⎮ 7π ( −5 − 12sin θ − 4sin θ ) dθ
⌡ 2
6

11π
6

1
=⌠
⎮ 7π ( −7 − 12sin θ + 2 cos ( 2θ ) ) dθ
⌡ 2
6

11π
6
1
= ( −7θ + 12 cos θ + sin ( 2θ ) )
7π
2
6

11 3 7π
−
= 2.196
2
3
Notice that for this area the “outer” and “inner” function where opposite!

=

Let’s do one final modification of this example.
Example 4 Determine the area that is inside both r = 3 + 2sin θ and r = 2 .
Solution
Here is the sketch for this example.

We are not going to be able to do this problem in the same fashion that we did the
previous two. There is no set of limits that will allow us to enclose this area as we
increase from one to the other.
In this case however, that is not a major problem. There are two ways to do get the area
in this problem.


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Calculus II

Solution 1
Notice that the circle is divided up into two portions and we’re after the upper portion.
Also notice that we found the area of the lower portion in Example 3. Therefore, the area
is,
Area = Area of Circle − Area from Example 3

= π ( 2 ) − 2.196
2

= 10.370
Solution 2
In this case we do pretty much the same thing except this time we’ll think of the area as
the other portion of the limacon than the portion that we were dealing with in Example 2.
So, in this case the area is,
Area = Area of Limacon − Area from Example 2
2π
1
2
=⌠
( 3 + 2sin θ ) dθ − 24.187
⎮
⌡0 2
2π
1
=⌠
⎮
( 9 + 12sin θ + 4sin 2 θ ) dθ − 24.187
⌡0 2
2π
1
=⌠
⎮
(11 + 12sin θ + 2 cos ( 2θ ) ) dθ − 24.187
⌡0 2
2π

1
= (11θ − 12 cos (θ ) + sin ( 2θ ) ) − 24.187
2
0
= 11π − 24.187
= 10.370

A slightly longer approach, but sometimes we are forced to take this longer approach.
As this last example has shown we will not be able to get all areas in polar coordinates
straight from an integral.

Arc Length with Polar Coordinates
We now need to move into the Calculus II applications of integrals and how we do them
in terms of polar coordinates. In this section we’ll look at the arc length of the curve
given by,
r = f (θ )
α ≤θ ≤ β


152


Calculus II

where we also assume that the curve is traced out exactly once. Just as we did with the
tangent lines in polar coordinates we’ll first write the curve in terms of a set of parametric
equations.
x = r cos θ
y = r sin θ

= f (θ ) cos θ
= f (θ ) sin θ
and we can now use the parametric formula for finding the arc length.
We’ll need the following derivatives for these computations.
dx
dy
= f ′ (θ ) cos θ − f (θ ) sin θ
= f ′ (θ ) sin θ + f (θ ) cos θ
dθ
dθ
dr
dr
cos θ − r sin θ
sin θ + r cos θ
=
=
dθ
dθ
We’ll need the following for our ds.
2
2
2
2
⎛ dx ⎞ ⎛ dy ⎞ ⎛ dr
⎞ ⎛ dr
⎞
cos θ − r sin θ ⎟ + ⎜
sin θ + r cos θ ⎟
⎜
⎟ +⎜
⎟ =⎜
⎝ dθ ⎠ ⎝ dθ ⎠ ⎝ dθ
⎠ ⎝ dθ
⎠
2

dr
⎛ dr ⎞
2
cos θ sin θ + r 2 sin 2 θ
=⎜
⎟ cos θ − 2r
dθ
⎝ dθ ⎠
2

dr
⎛ dr ⎞
2
cos θ sin θ + r 2 cos 2 θ
+⎜
⎟ sin θ + 2r
dθ
⎝ dθ ⎠
2

⎛ dr ⎞
2
2
2
2
2
=⎜
⎟ ( cos θ + sin θ ) + r ( cos θ + sin θ )
dθ ⎠
⎝
⎛ dr ⎞
= r2 + ⎜
⎟
⎝ dθ ⎠

2

The arc length formula for polar coordinates is then,
L = ∫ ds
where,
2

⎛ dr ⎞
ds = r 2 + ⎜
⎟ dθ
⎝ dθ ⎠
Let’s work a quick example of this.
Example 1 Determine the length of r = θ 0 ≤ θ ≤ 1 .
Solution
Okay, let’s just jump straight into the formula since this is a fairly simple function.

L=∫


1
0

θ 2 + 1 dθ

153


Calculus II

We’ll need to use a trig substitution here.
θ = tan x
dθ = sec 2 x dx
θ =0
0 = tan x
x=0

θ =1

1 = tan x

x=

π
4

θ + 1 = tan x + 1 = sec x = sec x = sec x
2

2

2

The arc length is then,
L=∫

1
0

θ 2 + 1 dθ

π

= ∫ 4 sec3 x dx
0

π
4
1
= ( sec x tan x + ln sec x + tan x )
2
0

=

1
2

(

(

2 + ln 1 + 2

))

Surface Area with Polar Coordinates
We will be looking at surface area in polar coordinates in this section. Note however that
all we’re going to do is give the equations since most of these integrals tend to be fairly
difficult.
We want to find the surface area of the region found by rotating,
r = f (θ )
α ≤θ ≤ β
about the x or y-axis.
As we did in the tangent and arc length sections we’ll write the curve in terms of a set of
x = r cos θ
y = r sin θ

= f (θ ) cos θ

= f (θ ) sin θ

If we now use the parametric formula for finding the surface area we’ll get,
S = ∫ 2π y ds
S = ∫ 2π x ds


where,
2

⎛ dr ⎞
ds = r 2 + ⎜
⎟ dθ
⎝ dθ ⎠


r = f (θ ) , α ≤ θ ≤ β

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Calculus II

Note that because we will pick up a dθ from the ds we’ll need to substitute one of the
parametric equation in for x or y depending on the axis of rotation. This will often mean
that the integrals will be somewhat unpleasant.

Arc Length and Surface Area Revisited
We won’t be working any examples in this section. This section is here solely for the
purpose of summarizing up all the arc length and surface area problems.
Over the course of the last two chapters the topic of arc length and surface area has arisen
many times and each time we got a new formula out of the mix. Students often get a
little overwhelmed with all the formulas.
However, there really aren’t as many formulas as it might seem at first glance. There is
exactly one arc length formula and exactly two surface area formulas. These are,
L = ∫ ds
S = ∫ 2π y ds


S = ∫ 2π x ds


The problems arise because we have quite a few ds’s that we can use. Again students
often have trouble deciding which one to use. The examples/problems usually suggest
the correct one to use however. Here is a complete listing of all the ds’s that we’ve seen
and when they are used.
2

⎛ dy ⎞
ds = 1 + ⎜ ⎟ dx
⎝ dx ⎠

if y = f ( x ) , a ≤ x ≤ b

2

⎛ dx ⎞
ds = 1 + ⎜ ⎟ dy
⎝ dy ⎠
2

if x = h ( y ) , c ≤ y ≤ d
2

⎛ dx ⎞ ⎛ dy ⎞
ds = ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
2

⎛ dr ⎞
ds = r + ⎜
⎟ dθ
⎝ dθ ⎠
2

if x = f ( t ) , y = g ( t ) , α ≤ t ≤ β
if r = f (θ ) , α ≤ θ ≤ β

Depending on the form of the function we can quickly tell which ds to use.
There is only one other thing to worry about in terms of the surface area formula. The ds
will introduce a new differential to the integral. Before integrating make sure all the


155


Calculus II

variables are in terms of this new differential. For example if we have parametric
equations well use the third ds and then we’ll need to make sure and substitute for the x
or y depending on which axis we rotate about to get everything in terms of t.
Likewise, if we have a function in the form x = h ( y ) then we’ll use the second ds and if
the rotation is about the y-axis we’ll need to substitute for the x in the integral. On the
other hand if we rotate about the x-axis we won’t need to do a substitution for the y.
Keep these rules in mind and you’ll always be able to determine which formula to use
and how to correctly do the integral.

Sequences and Series
Introduction
In this chapter we’ll be taking a look at sequences and (infinite) series. Actually this
chapter will deal almost exclusively with series, however, we also need to understand
some of the basics of sequences in order to properly deal with series. We will therefore,
spend a little time on sequences as well.
Series is one of those topics that many students simply don’t see the use in studying and
to be honest many students will never see series outside of their calculus class. However,
series do play an important role in the field of ordinary differential equations and without
series large portions of the field of partial differential equations would not be possible.
In other words, series is an important topic even if you won’t ever see any of the
applications. Most of the applications are beyond the scope of most Calculus courses and
tend to occur in classes that many students don’t take. So, as you go through this
material keep in mind that these do have applications even if we won’t really be covering
many of them in this class.
Here is a list of topics in this chapter.
Sequences – We will start the chapter off with a brief discussion of sequences.
Series – The Basics – In this section we will discuss some of the basics of infinite series.
Series – Convergence/Divergence – Most of this chapter will be about the
convergence/divergence of a series so we will give the basic ideas and definitions in this
section.
Series – Special Series – We will look at the Geometric Series, Telescoping Series, and
Harmonic Series in this section.


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Calculus II

Integral Test – Using the Integral Test to determine if a series converges or diverges.
Comparison Test/Limit Comparison Test – Using the Comparison Test and Limit
Comparison Tests to determine if a series converges or diverges.
Alternating Series Test – Using the Alternating Series Test to determine if a series
converges or diverges.
Absolute Convergence – A brief discussion on absolute convergence and how it differs
from convergence.
Ratio Test – Using the Ratio Test to determine if a series converges or diverges.
Root Test – Using the Root Test to determine if a series converges or diverges.
Strategy for Series – A set of general guidelines to use when deciding which test to use.
Estimating the Value of a Series – Here we will look at estimating the value of an
infinite series.
Power Series – An introduction to power series and some of the basic concepts.
Power Series and Functions – In this section we will start looking at how to find a
power series representation of a function.
Taylor Series – Here we will discuss how to find the Taylor/Maclaurin Series for a
function.
Applications of Series – In this section we will take a quick look at a couple of
applications of series.
Binomial Series – A brief look at binomial series.

Sequences
Let’s start off this section with a discussion of just what a sequence is. A sequence is
nothing more than a list of numbers written in a specific order. The list may or may not
have an infinite number of terms in them although we will be dealing exclusively with
infinite sequences in this class. General sequence terms are denoted as follows,


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Calculus II

a1 − first term
a2 − second term
an − nth term
an +1 − ( n + 1) term
st

Because we will be dealing with infinite sequences each term in the sequence will be
followed by another term as noted above. In the notation above we need to be very
careful with the subscripts. The subscript of n+1 denotes the next term in the sequence
and NOT one plus the nth term! In other words,
an +1 ≠ an + 1
so be very careful when writing subscripts to make sure that the “+1” doesn’t migrate out
of the subscript!
There is a variety of ways of denoting a sequence. Each of the following are equivalent
ways of denoting a sequence.
∞
{a1 , a2 ,… , an , an+1 ,…}
{an }
{an }n=1
In the second and third notations above an is usually given by a formula.
A couple of notes are now in order about these notations. First, note the difference
between the second and third notations above. If the starting point is not important or is
implied in some way by the problem it is often not written down as we did in the third
notation. Next, we used a starting point of n=1 in the third notation only so we could
write one down. There is absolutely no reason to believe that a sequence will start at
n=1. A sequence will start where ever it needs to start.
Let’s take a look at a couple of sequences.
Example 1 Write down the first few terms of each of the following sequences.
∞

⎧ n + 1⎫
(a) ⎨ 2 ⎬
⎩ n ⎭n =1
∞

⎧ ( −1)n +1 ⎫
⎪
⎪
(b) ⎨
⎬
n
⎪ 2 ⎪n = 0
⎩
⎭
(c) {bn }n =1 , where bn = nth digit of π
∞

Solution
(a) To get the first few sequence terms here all we need to do is plug in values of n into
the formula given and we’ll get the sequence terms.


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Calculus II

⎧
⎫
∞
3 4 5 6
⎪
⎪
⎧ n + 1⎫
⎨ 2 ⎬ = ⎨ 2 , , , , ,…⎬
⎩ n ⎭n =1 ⎪ n =1 4 9 16 25 ⎪
⎩ n = 2 n =3 n = 4 n =5 ⎭
Note the inclusion of the “…” at then end! This is an important piece of notation as it is
the only thing that tells us that the sequence continues on and doesn’t terminate at the last
term.
(b) This one is similar to the first one. The main difference is that this sequence doesn’t
start at n=1.
∞

⎧ ( −1)n +1 ⎫
⎪
⎪
⎧ 1 1 1 1
⎫
⎨
⎬ = ⎨−1, , − , , − ,…⎬
n
⎪ 2 ⎪n =0 ⎩ 2 4 8 16 ⎭
⎩
⎭
Note that the terms in this sequence alternate in signs. Sequences of this kind are
sometimes called alternating sequences.
(c) This sequence is different from the first two in the sense that it doesn’t have a specific
formula for each term. However, it does tell us what each term should be. Each term
should be the nth digit of π. So we know that π = 3.14159265359…

The sequence is then,

{3,1, 4,1,5,9, 2, 6,5,3,5,…}

In the first two parts of the previous example note that we were really treating the
formulas as functions that can only have integers plugged into them. Or,
n +1
f ( n) = 2
n

g (n)

( −1)
=

n +1

2n

This is an important idea in the study of sequences (and series). Treating the sequence
terms as function evaluations will allow us to do many things with sequences that
couldn’t do otherwise. Before delving further into this idea however we need to get a
couple more ideas out of the way.
First we want to think about “graphing” a sequence. To graph the sequence {an } we plot
the points ( n, an ) as n ranges over all possible values on a graph.
∞

⎧ n + 1⎫
For instance, let’s graph the sequence ⎨ 2 ⎬ . The first few points on the graph are
⎩ n ⎭n =1
then,
3
4
5
6
(1, 2 ) , ⎛ 2, ⎞ , ⎛ 3, ⎞ , ⎛ 4, ⎞ , ⎛ 5, ⎞ , …
⎜
⎟ ⎜
⎟ ⎜
⎟ ⎜
⎟
⎝ 4 ⎠ ⎝ 9 ⎠ ⎝ 16 ⎠ ⎝ 25 ⎠

The graph is then,


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Calculus II

This graph leads us to an important idea about sequences. Notice that as n increases the
sequence terms from our sequence terms, in this case, get closer and closer to zero. We
then say that zero is the limit (or sometimes the limiting value) of the sequence and
write,
n +1
lim an = lim 2 = 0
n →∞
n →∞ n
This notation should look familiar to you. It is the same notation we used when we
talked about the limit of a function. In fact, if you recall, we said earlier that we could
think of sequences as functions in some way and so this notation shouldn’t be too
surprising.
Using the ideas that we developed for limits of functions we can write down the
following working definition for limits of sequences.
Definitions
1. We say that

lim an = L
n →∞

if we can make an as close to L as we want for all sufficiently large n. In other
words, the value of the an’s approach L as n approaches infinity.
2. We say that
lim an = ∞
n →∞

if we can make an as large as we want for all sufficiently large n. Again, in other
words, the value of the an’s get larger and larger without bound as n approaches
infinity.
3. We say that


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Calculus II

lim an = −∞
n →∞

if we can make an as large and negative as we want for all sufficiently large n.
Again, in other words, the value of the an’s are negative and get larger and larger
without bound as n approaches infinity.
We now have a bit of terminology to get out of the way. If lim an exists and is finite we
n →∞

say that the sequence is convergent. If lim an doesn’t exist or is infinite we say the
n →∞

sequence diverges. Note that sometimes we will say the sequence diverges to ∞ if
lim an = ∞ and if lim an = −∞ we will sometimes say that the sequence diverges to −∞ .
n →∞

n →∞

So just how do we find the limits of sequences? We will use the following theorems and
facts.
Theorem 1
Given the sequence {an } if we have a function f(x) such that f ( n ) = an and
lim f ( x ) = L then lim an = L
x →∞

n →∞

This theorem is basically telling us that we take the limits of sequences much like we take
the limit of functions. In fact, in most cases we’ll not even really use this theorem by
explicitly writing down a function. We will more often just treat the limit as if it were a
limit of a function.
Theorem 2
If lim an = 0 then lim an = 0 .
n →∞

n →∞

This theorem is convenient for sequences that alternate in signs and note that it will only
work if the sequence has a limit of zero.
Theorem 3

The sequence {r n }

∞

n=0

converges if −1 < r ≤ 1 and diverges for all other value of r. Also,
⎧0 if − 1 < r < 1
lim r n = ⎨
n →∞
⎩1 if r = 1

Note that the sequence in this theorem will converge for r = 1 and diverge for r = −1
Facts


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Calculus II

lim ( an ± bn ) = lim an ± lim bn
n →∞

n →∞

n →∞

lim can = c lim an
n →∞

(

n →∞

lim ( an bn ) = lim an
n →∞

n →∞

)( lim b )
n →∞

n

an lim an
= n→∞ , provided lim bn ≠ 0
n →∞ b
n →∞
lim bn
n

lim

n →∞

Example 2 Determine if the following sequences converge or diverge. If the sequence
converges determine its limit.
∞

⎧ 3n 2 − 1 ⎫
(a) ⎨
2⎬
⎩10n + 5n ⎭n = 2
∞

⎧ e2 n ⎫
(b) ⎨ ⎬
⎩ n ⎭n =1
∞

⎧ ( −1)n ⎫
⎪
⎪
(c) ⎨
⎬
⎪ n ⎪n =1
⎩
⎭

(d)

{( −1) }n=0
n ∞

Solution
(a) In this case all we need to do is recall the method that was developed in Calculus I to
deal with the limits of rational functions. See the Limits Involving Infinity section of my
Calculus I notes for a review of this if you need to.

To do a limit in this form all we need to do is factor from the numerator and denominator
the largest power of n, cancel and then take the limit.
1 ⎞
⎛
1
n2 ⎜ 3 − 2 ⎟
3− 2
2
3n − 1
n ⎠
n =3
= lim ⎝
= lim
lim
2
n →∞ 10n + 5n
n →∞
⎛ 10
⎞ n→∞ 10 + 5 5
n2 ⎜ + 5 ⎟
n
⎝ n
⎠
3
.
5
(b) We will need to be careful with this one. We will need to use L’Hospital’s Rule on
this sequence. The problem is that L’Hospital’s Rule only works on functions and not on
sequences. Normally this would be a problem, but we’ve got Theorem 1 from above to
help us out. Let’s define
e2 x
f ( x) =
x
So the sequence converges and its limit is


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Calculus II

and note that,

e2n
n

f (n) =

Theorem 1 says that all we need to do is take the limit of the function.
e2 n
e2 x
2e 2 x
lim
= lim
= lim
=∞
n →∞ n
x →∞ x
x →∞ 1
So, the sequence in this part diverges (to ∞ ).
More often than not we just do L’Hospital’s Rule on the sequence terms without first
converting to x’s since the work will be identical regardless of whether we use x or n.
However, we really should remember that technically we can’t do the derivatives while
dealing with sequence terms.

(c) We will also need to be careful with this sequence. We might be tempted to just say
that the limit of the sequence terms is zero (and we’d be correct). However, technically
we can’t take the limit of sequences whose terms alternate, because we don’t know how
to do limits of functions that exhibit that same behavior. Also, we want to be very careful
to not rely too much on intuition with these problems. As we will see in this section, and
in later sections, our intuition can lead us astray in these problem if we aren’t careful.
So, let’s work this one by the book. We will need to use Theorem 2 on this problem.

( −1)
lim
n →∞

n

n

1
=0
n →∞ n

= lim

Therefore, since the limit of the sequence terms with absolute value bars on them goes to
zero we know by Theorem 2 that,

( −1)
lim

n

=0
n
which also means that the sequence converges to a value of zero.
n →∞

(d) For this sequence all that we need to do is acknowledge that lim ( −1) doesn't exist to
n

n→∞

get that the sequence is divergent. If you’re not convinced that this limit doesn’t exist
write down the first few terms of the sequence.

{( −1) }

n ∞
n=0

= {1, −1,1, −1,1, −1,1, −1,1, −1,…}

In order for a limit to exist the terms must be settling down towards a specific value and
these clearly will never do that.
Note that we could also use Theorem 3 on this sequence if we wanted to.
Before moving on we need to give a warning about misusing Theorem 2. Theorem 2
only works if the limit is zero. If the limit of the absolute value of the sequence terms is


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Calculus II

not zero then the theorem will not hold. The last part of the previous example is a good
example of this. Notice that
lim ( −1) = lim1 = 1
n

n →∞

n →∞

and yet, lim ( −1) doesn’t even exist let alone equal 1. So, be careful using this theorem.
n

n →∞

We now need to take a look at some more terminology and definitions for sequences.
Given any sequence {an } we have the following.
1. We call the sequence increasing if an ≤ an +1 for every n.
2. We call the sequence decreasing if an ≥ an +1 for every n.

3. If {an } is an increasing sequence or {an } is a decreasing sequence we call it

monotonic.
4. If there exists a number m such that m ≤ an for every n we say the sequence is
bounded below. The number m is sometimes called a lower bound for the
sequence.
5. If there exists a number M such that an ≤ M for every n we say the sequence is
bounded above. The number M is sometimes called an upper bound for the
sequence.
6. If the sequence is both bounded below and bounded above we call the sequence
bounded.
Note that in order for a sequence to be increasing or decreasing it must be
increasing/decreasing for every n. In other words, a sequence that increases for three
terms and then decreases for the rest of the terms is NOT a decreasing sequence!
Before moving on we should make a quick point about the bounds for a sequence that is
bounded above and/or below. We’ll make the point about lower bounds, but we could
just as easily make it about upper bounds.
A sequence is bounded below if we can find any number m such that m ≤ an for every n.
Note however that if we find one number m to use for a lower bound then any number
smaller than m will also be a lower bound. Also, just because we find one lower bound
that doesn’t mean there won’t be a better lower bound for the sequence than the one we
found. In other words, there are an infinite number of lower bounds for a sequence that is
bounded below, some will be better than others. In my class all that I’m after will be a
lower bound. I don’t necessarily need the best lower bound, just a number that will be a
lower bound for the sequence.
We also have the following theorem about bounded and monotonic sequences.

Theorem 4
If {an } is bounded and monotonic then {an } is convergent.


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Calculus II

Example 3 Determine if the following sequences are monotonic and/or bounded.

(a) {− n 2 }

∞

(b)

n=0
n +1 ∞

{( −1) }

n =1

∞

⎧2⎫
(c) ⎨ 2 ⎬
⎩ n ⎭ n =5
Solution
(a) This sequence is a decreasing sequence (and hence monotonic) because,
2
− n 2 > − ( n + 1)

for every n. Also, since the sequence terms will be either zero or negative this sequence
is bounded above. We can use any positive number or zero as the bound, M, however,
it’s standard to choose the smallest possible bound if we can and it’s a nice number. So,
we’ll choose M=0 since,
−n2 ≤ 0
for every n
This sequence is not bounded below however since we can always get below any
potential bound by taking n large enough. Therefore, while the sequence is bounded
above it is not bounded.
As a side note we can also note that this sequence diverges (to −∞ if we want to be
specific).

(b) The sequence terms in this sequence alternate between 1 and -1 and so the sequence is
not monotonic. It is bounded however since it is bounded above by 1 and bounded below
by -1.
This sequence is also divergent.

(c) This sequence is a decreasing sequence (and hence monotonic) since,
2
2
>
2
2
n
( n + 1)
The terms in this sequence are all positive and so it is bounded below by zero. Also,
since the sequence is a decreasing sequence the first sequence term will be the largest and
2
so we can see that the sequence will also be bounded above by
. Therefore, this
25
sequence is bounded.
Theorem 4 says that this sequence is then convergent since it is both bounded and
monotonic. A quick limit can verify that this sequence is convergent and its value is
zero.


165


Calculus II

Now, let’s work a couple more examples that are designed to make sure that we don’t get
too used to relying on our intuition with these problems.
Example 4 Determine if the following sequences are monotonic and/or bounded.
∞

⎧ n ⎫
(a) ⎨
⎬
⎩ n + 1 ⎭n =1
∞

⎧
⎫
n3
(b) ⎨ 4
⎬
⎩ n + 10000 ⎭n =0
Solution
(a) We’ll start with the bounded part of this example first and then come back and deal
with the increasing/decreasing question since that is where students often make mistakes
with problems of this type.
First, n is positive and so the sequence terms are all positive. The sequence is therefore
bounded below by zero. Likewise each sequence term is the quotient of a number
divided by a larger number and so is guaranteed to be less that one. The sequence is then
bounded above by one. So, this sequence is bounded.
Now let’s think about the monotonic question. First, students will often make the
mistake of assuming that because the denominator is larger the quotient must be
decreasing. This will not always be the case and in this case we would be wrong. This
sequence is increasing.
To see this we will need to resort to Calculus I techniques. First consider the following
function and its derivative.
x
1
f ( x) =
f ′( x) =
2
x +1
( x + 1)
We can see that the first derivative is always positive and so from Calculus I we know
that the function must then be an increasing function. So, how does this help us? Notice
that,
n
f ( n) =
= an
n +1
Therefore,
n
n +1
an =
an < an +1
= f ( n ) < f ( n + 1) =
= an +1
⇒
n +1
n +1+1
In other words, the sequence must be increasing. Note that now that we know the
sequence is an increasing sequence we can get a better lower bound for the sequence.
Since the sequence is increasing the first term in the sequence must be the smallest term
1
and so since we are starting at n = 1 we could also use a lower bound of for this
2


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Calculus II

sequence. It is important to remember that any number that is always less than or equal
to all the sequence terms can be a lower bound. Some are better than others however.
Also, the sequence converges to 1, which again is a verification of Theorem 4.
Before moving on to the next part there is a natural question that many students will have
at this point. That is this. Why did we use Calculus to determine the
increasing/decreasing nature of the sequence when we could have just plugged in a
couple of n’s and quickly determined the say thing?
The answer to this question is the next part!

(b) This is a messy looking sequence, but it needs to be in order to make the point of this
part.
First, notice that, as with the previous part, the sequence terms are all positive and will all
be less that one (since the numerator is guaranteed to be less than the denominator) and
so the sequence is bounded.
Now, let’s move on to the increasing/decreasing question. First, as with the last problem,
many students will look at the exponents in the numerator and denominator and
determine based on that that sequence terms must decrease.
This however, isn’t a decreasing sequence. Let’s take a look at the first few terms to see
this.
1
1
a1 =
≈ 0.00009999
a2 =
≈ 0.0007987
10001
1252
27
4
a3 =
a4 =
≈ 0.005678
≈ 0.006240
10081
641
1
27
a5 =
a6 =
≈ 0.011756
≈ 0.019122
85
1412
343
32
a7 =
a8 =
≈ 0.02766
≈ 0.03632
12401
881
729
1
a9 =
a10 =
≈ 0.04402
= 0.05
16561
20
The first 10 (at least) terms of this sequence are all increasing and so clearly the sequence
can’t be a decreasing sequence. Recall that a sequence can only be decreasing if ALL the
terms are decreasing.
Now, we can’t make another mistake and assume that because the first few terms
increase then whole sequence must also increase. If we did that we would also be
mistaken.
This sequence is neither decreasing or increasing. The only sure way to see this is to do


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Calculus II

the Calculus I approach to increasing/decreasing functions.
In this case we’ll need the following function and its derivative.
− x 2 ( x 4 − 30000 )
x3
f ( x) = 4
f ′( x) =
2
x + 10000
( x4 + 10000 )
This function will have the following three critical points,
x = 0, x = 4 30000 ≈ 13.1607, x = − 4 30000 ≈ −13.1607
Why critical points? Remember these are the only places where the function may change
sign! Our sequence starts at n=0 and so we can ignore the third one since it lies outside
the values of n that we’re considering. By plugging in some test values of x we can
quickly determine that the derivative is positive for 0 < x < 4 30000 and so the function
is increasing in this range. Likewise, we can see that the derivative is negative for
x > 4 30000 and so the function will be decreasing in this range.
So, our sequence will be increasing for 0 ≤ n ≤ 13 and decreasing for n ≥ 13 . Therefore
the function is not monotonic.
Notice however Theorem 4 does NOT say that this sequence will then diverge. In fact,
this sequence converges to a value of zero. Be careful not to misuse Theorem 4.
So, as the last example, has shown we need to be careful in making assumptions about
sequences. Our intuition will often not be sufficient to get the correct answer and we can
NEVER make assumptions about a sequence based on the value of the first few terms.
As the last part showed there are sequences which will increase or decrease for a few
terms and then change direction after that.
Note as well that we said “first few terms” here, but it is completely possible for a
sequence to decrease for the first 10,000 terms and then start increasing for the remaining
terms. In other words, there is no “magical” value of n for which all we have to do is
check up to that point and then we’ll now that the whole sequence will do.
The only time that we’ll be able to avoid using Calculus I techniques to determine the
increasing/decreasing nature of a sequence is in sequences like part (c) of Example 3. In
this case increasing n only changed (in fact increased) the denominator and so we were
able to determine the behavior of the sequence based on that.
In Example 4 however, increasing n increased both the denominator and the numerator.
In cases like this there is no way to determine which increase will “win out” and cause
the sequence terms to increase or decrease and so we need to resort to Calculus I
techniques to answer the question.


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Calculus II

Series – The Basics
In this section we will introduce the topic that we will be discussing for the rest of this
chapter. That topic is series. So just what is a series? Well, let’s start with a sequence
∞
{an }n=1 (note the n=1 is for convenience, it can be anything) and then add up all the terms
of the sequence.
a1 + a2 + a3 + + an +
This is not an easy thing to write down on a regular basis and so we introduce the
following notation.
a1 + a2 + a3 +

+ an +

∞

= ∑ an
n =1

We have just defined what a series is. A series is nothing more than the summation of a
list of numbers or sequence. Since we started out with an infinite sequence we will be
summing up an infinite list of numbers. Because of this the series above is sometimes
called an infinite series. The n is often called an index of summation or just index for
short.
Note as well that we will use

∑a

n

to represent an infinite series in which the starting

point for the index is not important. We will do this in quite a few facts and theorems
that we’ll be seeing throughout this chapter. In these facts the starting point of the series
will not affect the result and so to simplify the notation and to avoid giving the
impression that the starting point is important we will drop the index from the notation.
Do not forget however, that there is a starting point and that this will be an infinite series.
The notation used for series is called a number of things. The most common names are :
series notation, summation notation, and sigma notation.
The series notation (or summation notation or sigma notation, which ever you prefer)
tells us to add all the items from the sequence starting at the value of the index that is
below the sigma. Also note that the letter that we use for the index is not important. The
following two series are identical.
∞
∞
3
3
=∑ 2
∑ n 2 + 1 i =0 i + 1
n=0
As long as we’re starting at the same spot (n=0 or i=0 respectively) and we’re adding the
same terms so the series will be identical.
The point of this section is to cover the basic ideas, concepts and manipulations involved
in series. We need to be somewhat familiar with these before we get into the later
sections.
So, let’s start off the discussion with some basic facts about arithmetic with series.


169


Calculus II

∑ ca

n

∞

∞

= c ∑ an
∞

∑ a ± ∑b = ∑(a
n=k

n

n=k

n

n=k

n

c is any number
± bn )

⎛ ∞ ⎞⎛ ∞ ⎞ ∞
⎜ ∑ an ⎟⎜ ∑ bn ⎟ = ∑ cn
⎝ n =0 ⎠⎝ n =0 ⎠ n =0

n

where cn = ∑ ai bn −i
i =0

Before discussing these in detail let’s again talk a little bit about the notation. In the first
formula the initial value of the index is not important to the formula and so was dropped
completely. Note as well that we dropped the infinity from the top of the sigma in this
case as well. We will always drop the infinity when we drop the initial value of the
index. When there is no initial value given for the series the assumption is that we can
use any value there and we neglected putting one in the formula to avoid giving the
impression that the starting point was important.
In the second series we had a starting point, but it wasn’t a specific starting point. In this
case we needed to acknowledge that in order to add or subtract two series we need the
two series to start at the same point. Just what this value happens to be isn’t important,
but it must be the same for both series and to denote this we used the general starting
point n=k.
In the third series we gave a specific starting point (n=0). From this point on it can be
assumed that if a fact/theorem gives a specific starting point for a series then that starting
point is required in order of the fact/theorem to be valid. That is the case with this fact
about multiplying series. There are similar facts for series that don’t start at zero,
however we won’t be doing much in the way of multiplying series in this chapter and so
we don’t really need them.
Okay, now that the notational stuff is out of the way, let’s think about these facts a little.
The first fact should make some sense. We can always factor a constant out of a series.
We know that we can do this with a finite sum,
2 x 2 + 4 x + 6 = 2 ( x 2 + 2 x + 3)
and so it should make some sense that we can also do it for infinite sums (i.e. infinite
series).
Likewise, the second fact isn’t too bad to understand. Adding/Subtracting two (or more)
series involves nothing more than adding/subtracting similar terms
The formula that always causes difficulty is the multiplication of series formula. First,
let’s note the following,
⎛ ∞ ⎞⎛ ∞ ⎞ ∞
⎜ ∑ an ⎟ ⎜ ∑ bn ⎟ ≠ ∑ ( an bn )
⎝ n=0 ⎠ ⎝ n=0 ⎠ n=0
To convince yourself that this isn’t true consider the following product of two finite
sums.

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Calculus II

( 2 + x ) ( 3 − 5 x + x 2 ) = 6 − 7 x − 3x 2 + x3
Yeah, it was just the multiplication of two polynomials, but it makes the point. In doing
the multiplication we didn’t just multiply the constant terms, them the x terms, etc.
Instead we had to distribute the 2 through the second polynomial, then distribute the x
through the second polynomial and finally combine like terms.
Multiplying infinite series needs to be done in the same manner. Remember that a series
is really a giant summation and so here is what we’re really asking for multiplication,
⎛ ∞ ⎞⎛ ∞ ⎞
⎜ ∑ an ⎟ ⎜ ∑ bn ⎟ = ( a0 + a1 + a2 + a3 + )( b0 + b1 + b2 + b3 + )
⎝ n=0 ⎠ ⎝ n=0 ⎠
To do this multiplication we would have to distribute the a0 through the second term,
distribute the a1 through, etc then combine like terms. This is pretty much impossible
since both series have an infinite set of terms in them. The formula given above is the
general formula for this, but again, is usually impossible to use in practice.
The reality is that multiplication of series is a somewhat difficult process and in general is
avoided at all costs if possible. We will take a brief look at it towards the end of the
chapter when we’ve got more work under our belt and we run across a situation where it
might actually be what we want to do. Until then, don’t worry about multiplying series.
The next topic that we need to discuss in this section is that of index shift. To be honest
this is not a topic that we’ll see all that often in this course. In fact, we’ll use it once in
the next section and then not use it again in all likelihood. Despite the fact that we won’t
use it much in this course doesn’t mean however that it isn’t used often in other classes
where you might run across series. So, we will cover it briefly here so that you can say
you’ve seen it.
The basic idea behind index shifts is to start a series at a different value for whatever the
reason (and yes, there are legitimate reasons for doing that).
Consider the following series,
n+5
n
n=2 2
Suppose that for some reason we wanted to start this series at n=0, but we didn’t want to
change the value of the series. This means that we can’t just change the n=2 to n=0 as
this would add in two new terms to the series changing its value.
∞

∑

Performing an index shift is a fairly simple process to do. We’ll start by defining a new
index, say i, as follows,
i = n−2
Now, when n=2, we will get i=0. Notice as well that if n = ∞ then i = ∞ − 2 = ∞ , so
only the lower limit will change here. Next, notice that we can solve this for n to get,
n=i+2


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Calculus II

We can now completely rewrite the series in terms of the index i instead of the index n.
∞
n + 5 ∞ (i + 2) + 5 ∞ i + 7
∑ 2 n = ∑ 2i + 2 = ∑ 2i + 2
n=2
i =0
i =0
To finish the problem out we’ll recall that the letter we used for the index doesn’t matter
and so we’ll change the final i back into an n to get,
∞
n+5 ∞ n+7
∑ 2n = ∑ 2n + 2
n=2
n=0
To convince yourselves that these really are the same summation let’s write out the first
couple of terms for each of them,
∞
n + 5 7 8 9 10
∑ 2 n = 2 2 + 23 + 2 4 + 25 +
n=2
∞
n + 7 7 8 9 10
∑ 2 n + 2 = 2 2 + 23 + 2 4 + 25 +
n=0
So, sure enough they are the same series.
There is actually an easier way to do an index shift. The method given above is the
technically correct way of doing an index shift. However, notice in the above example
we decreased the initial value of the index by 2 and all the n’s in the series terms
increased by 2 as well.
This will always work in this manner. If we decrease the initial value of the index by a
set amount then all the other n’s in the series term will increase by the same amount.
Likewise, if we increase the initial value of the index by a set amount, then all the n’s in
the series term will decrease by the same amount.
Let’s do a couple of examples using this shorthand method for doing index shifts.
Example 1 Perform the following index shifts.
∞

(a) Write

∑ ar

n −1

as a series that starts at n=0.

n =1
∞

(b) Write

n2
∑ 1 − 3n+1 as a series that starts at n=3.
n =1

Solution
(a) In this case we need to decrease the initial value by 1 and so the n’s (okay the single
n) in the term must increase by 1 as well.
∞

∑ ar
n =1

n −1

∞

∞

n=0

n=0

= ∑ ar ( n +1)−1 = ∑ ar n

(b) For this problem we want to increase the initial value by 2 and so all the n’s in the
series term must decrease by 2.


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Calculus II

∞
∞
( n − 2)
( n − 2)
n2
∑ 1 − 3n+1 = ∑ 1 − 3( n−2)+1 = ∑ 1 − 3n−1
n =1
n =3
n =3
2

∞

2

The final topic in this section is again a topic that we’ll not be seeing all that often in this
class, although we will be seeing it more often than the index shifts. This final topic is
really more about alternate ways to write series when the situation requires it.
Let’s start with the following series (note that the n=1 starting point is only for
convenience, we can start anywhere).
∞

∑a
n =1

n

= a1 + a2 + a3 + a4 + a5 +

Notice that if we ignore the first term the remaining terms will also be a series that will
start at n=2 instead of n=1 So, we can rewrite the original series as follows,
∞

∞

n =1

n=2

∑ an = a1 + ∑ an
In the example we say that we’ve stripped out the first term.
We could have stripped out more terms if we wanted to. In the following series we’ve
stripped out the first two terms and the first four terms respectively.
∞

∞

n =1

n =3

∑ an = a1 + a2 + ∑ an
∞

∞

n =1

n =5

∑ an = a1 + a2 + a3 + a4 + ∑ an
Being able to strip out terms will, on occasion, simplify our work or allow us to reuse a
prior result so it’s an important idea to remember.
Notice that in the second example above we could have also denoted the four terms that
we stripped out as a finite series as follows,
∞

∞

4

∞

n =1

n =5

n =1

n =5

∑ an = a1 + a2 + a3 + a4 + ∑ an = ∑ an + ∑ an
This is a convenient notation when we are stripping out a large number of terms or if we
need to strip out an undetermined number of terms. In general, we can write a series as
follows,
∞

N

∑a = ∑a
n =1

n

n =1

n

+

∞

∑

n = N +1

an

We’ll leave this section with an important warning about terminology. Don’t get
sequences and series confused! A sequence is a list of numbers written in a specific order
while a series is a summation of the list of numbers. So, provided it even makes sense to
do the addition (see the next section), a series will be a single value.


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Calculus II

So, once again, a sequence is a list of numbers while a series is a single number, provided
it makes sense to even compute the series. Students will often confuse the two and try to
use facts pertaining to one on the other. However, since they are different beasts this just
won’t work. There will be problems where we are using both sequences and series, but
we’ll always have to remember that they are different.

Series – Convergence/Divergence
In the previous section we spent some time getting familiar with series, but to be honest,
most of what we did in that section won’t be used on a regular basis in this chapter. We
covered that material because we need to be aware of how series work and can be
manipulated, but we also covered it so that we could start getting our feet wet in the
subject of series.
It is now time to start talking about an idea involved in series that we will deal with to
one extent of another in almost all of the remaining sections of this chapter.
There was a very important question about series that was only mentioned in passing
towards the very end of the previous section. The question is simply this : Does it even
make sense to add up an infinite sequence of numbers? Technically we can always write
down an infinite series or summation, but that doesn’t mean that it makes sense to do it.
The real question that we’ll be asking here is does the (infinite) series/summation yield a
finite value or infinite value? Of course that also assumes that the series yields a value at
all! As we will see it will be possible for a series to not even have a value.
To answer this question we’ll need some more terminology out of the way. Let’s start
with the following series.
∞

∑a
n =1

n

Note that we’re starting at n=1 only for convenience. We could start the series
anywhere, but the following notation and terminology demands that we start somewhere
and so for the sake of the work we choose to start at n=1.
Now, instead of adding all the terms out to infinity, let’s look at the following finite
summations/series.


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Calculus II

s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
s4 = a1 + a2 + a3 + a4
sn = a1 + a2 + a3 + a4 +

n

+ an = ∑ ai
i =1

These are called partial sums. Notice that the partial sums will form an infinite
∞
sequence, {sn }n =1 , and that while it might not make sense to perform a summation of an
infinite list of numbers these are all summations of a finite list of numbers and so are
guaranteed to be finite numbers. Well they will be finite numbers provided we don’t end
up with a division by zero error somewhere in the list. In all of the work that we’ll be
doing in this chapter we will assume that all the sequence terms exist and are finite
numbers.
From the section on sequences we know how to determine if the sequence of partial sums
converges or diverges. Also notice that as n → ∞ the sequence terms, sn, should start
looking more and more like the infinite series. In fact it can be shown that if the
sequence of partial sums is convergent and if we define,
s = lim sn
n →∞

then,
∞

∑a

n

n =1

=s

In these cases we call the series convergent and we call s the sum or value of the series.
If the sequence of partial sums is divergent (i.e. either the limit doesn’t exist or is infinite)
then we call the series divergent.
In other words, the series is convergent if the sequence of partial sums is convergent and
hence has a finite value. Likewise the series will be divergent if the sequence of partial
sums is divergent. In the case of a divergent series, either the series will have an infinite
value or won’t have a value at all depending on whether or not the limit of the sequence
of partial sums exists or is infinite.
Let’s take a look at some series and see if we can determine if they are convergent or
divergent.
Example 1 Determine if the following series is convergent or divergent. If it converges
determine its sum.
∞

∑n
n =1

Solution


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Calculus II

To determine if the series is convergent we first need to get our hands on a formula for
the general term in the sequence of partial sums.
n

sn = ∑ i
i =1

This is a known series and its value can be shown to be,
n
n ( n + 1)
sn = ∑ i =
2
i =1
Don’t worry if you didn’t know this formula (I’d be surprised if anyone knew it…) as
you won’t be required to know it in my course.
So, to determine if the series is convergent we will first need to see if the sequence of
partial sums,
∞

⎧ n ( n + 1) ⎫
⎨
⎬
⎩ 2 ⎭n =1
is convergent or divergent. That’s not terribly difficult in this case.
n ( n + 1)
=∞
lim
n →∞
2

Therefore, the sequence of partial sums diverges to ∞ and so the series also diverges.
Note that we can say that the series has the value of ∞ in these cases, although the series
is still called divergent.
So, as we saw in this example we had to know a fairly obscure formula in order to
determine the convergence of this series. In general finding a formula for the general
term in the sequence of partial sums is a very difficult process. In fact after the next
section we’ll not be doing much with the partial sums of series due to the extreme
difficulty faced in find the general formula.
We will continue with a few more examples however, since this is technically how we
determine convergence of a series. Also, the remaining examples we’ll be looking at in
this section will lead us to a very important fact about the convergence of series.
So, let’s take a look at a couple more examples.
Example 2 Determine if the following series converges or diverges. If it converges
determine its sum.
∞
1
∑ n2 − 1
n=2
Solution
This is actually one of the few series in which we are able to determine a formula for the
general term in the sequence of partial fractions. However, in this section we are more
interested in the general idea of convergence and divergence and so we’ll put off
discussing the process for finding the formula until the next section.


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Calculus II

The general formula for the partial sums is,
n
1
3 1
1
sn = ∑ 2
= −
−
4 2n 2 ( n + 1)
i =2 i − 1
and in this case we have,
⎛3 1
1 ⎞ 3
lim sn = lim ⎜ −
−
⎟=
n →∞
n →∞ ⎜ 4
2n 2 ( n + 1) ⎟ 4
⎝
⎠
The sequence of partial sums converges and so the series converges as well and its value
is,
∞
1
3
∑ n2 − 1 = 4
n=2
determine its sum.
∞

∑ ( −1)

n

n=0

Solution
In this case we really don’t need a general formula for the partial sums to determine the
convergence of this series. Let’s just write down the first few partial sums.

s0 = 1
s1 = 1 − 1 = 0
s2 = 1 − 1 + 1 = 1
s3 = 1 − 1 + 1 − 1 = 0
etc.
So, it looks like the sequence of partial sums is,
∞
{sn }n=0 = {1, 0,1, 0,1, 0,1, 0,1,…}
and this sequence diverges since lim sn doesn’t exist. Therefore, the series also diverges.
n →∞

Note as well that unlike the first example, which was also divergent, this series doesn’t
even have a value.
determine its sum.
∞
1
∑ 3n−1
n =1
Solution
Here is the general formula for the partial sums for this series.


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Calculus II

n

sn = ∑
i =1

1
3⎛
1⎞
= ⎜1 − n ⎟
i −1
3
2⎝ 3 ⎠

Again, do not worry about knowing this formula. This is not something that you’ll ever
be asked to know in my class.
In this case the limit of the sequence of partial sums is,
3⎛
1
lim sn = lim ⎜1 − n
n →∞
n →∞ 2
⎝ 3

⎞ 3
⎟=
⎠ 2

The sequence of partial sums is convergent and so the series will also be convergent. The
value of the series is,
∞
1
3
∑ 3n−1 = 2
n =1
As we already noted, do not get excited about determining the general formula for the
sequence of partial sums. There is only going to be one type of series where you will
need to determine this formula and the process in that case isn’t too bad. In fact, you
already know how to do most of the work in the process as you’ll see in the next section.
So, we’ve determined the convergence of four series now. Two of the series converged
and two diverged. Let’s go back and examine the series terms for each of these. For
each of the series let’s take the limit as n goes to infinity of the series term (not the partial
sums!!).
lim n = ∞

this series diverged

n →∞

lim

n →∞

1
=0
n −1

this series converged

2

lim ( −1) doesn't exist

this series diverged

1
=0
n →∞ 3n −1

this series converged

n

n →∞

lim

Notice that for the two series that converged the series term itself was zero in the limit.
This will always be true for convergent series.

Theorem
If ∑ an converges then lim an = 0 .
n →∞

Be careful to not misuse this theorem however! This theorem gives us a requirement for
convergence but not a guarantee of convergence. In other words, the converse is NOT
true. If lim an = 0 the series may actually diverge! Consider the following two series.
n →∞


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Calculus II

∞

∞
1
1
∑n
∑ n2
n =1
n =1
In both cases the series terms are zero in the limit as n goes to infinity, yet only the
second series converges. The first series diverges. It will be a couple of sections before
we can prove this, so at this point please believe this and know that you’ll be able to
prove the convergence of these two series in a couple of sections.

Again, as noted above, all this theorem does is give us a requirement for a series to
converge. In order for a series to converge the series terms must go to zero in the limit.
If the series terms do not go to zero in the limit then there is no way the series can
converge since this would violate the theorem.
This leads us to the first of many tests for the convergence/divergence of a series that
we’ll be seeing in this chapter.

Divergence Test
If lim an ≠ 0 then
n →∞

∑a

n

will diverge.

Again, do NOT misuse this test. This test only says that a series is guaranteed to diverge
if the series terms don’t go to zero in the limit. If the series terms do happen to go to zero
the series may or may not converge! Again, recall the following two series,
∞
1
diverges
∑n
n =1
∞
1
converges
∑ n2
n =1
One of the more common mistakes that students make when the first get into series is to
assume that if lim an = 0 then ∑ an will converge. There is just no way to guarantee
n →∞

this so be careful!
Let’s take a quick look at an example of how this test can be used.
Example 5 Determine if the following series is convergent or divergent.
∞
4n 2 − n 3
∑ 10 + 2n3
n=0
Solution
With almost every series the first thing that we should do is take a look at the series terms
and see if they go to zero of not. If it’s clear that the terms don’t go to zero use the
Divergence Test and be done with the problem.

That’s what we’ll do here.


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Calculus II

4n 2 − n 3
1
=− ≠0
3
n →∞ 10 + 2n
2
The limit of the series terms isn’t zero and so by the Divergence Test the series diverges.
lim

The divergence test is the first test of many tests that we will be looking at over the
course of the next several sections. You will need to keep track of all these tests, the
conditions under which they can be used and their conclusions all in one place so you can
quickly refer back to them as you need to.
Next we should talk briefly about arithmetic of series and convergence/divergence. If

∑a

n

and

∑b

n

are both convergent series then so are

∞

∑ ca

n

and

∑(a
n=k

n

± bn ) . Further

more, these series will have the following sums or values.

∑ ca

n

= c ∑ an

∞

∞

∞

n=k

n=k

n=k

∑ ( an ± bn ) = ∑ an ± ∑ bn
We’ll see an example of this in the next section after we get a few more examples under
our belt. At this point just remember that a sum of convergent sequences is convergent
and multiplying a convergent sequence by a number will not change its convergence.
We need to be a little careful with these facts when it comes to divergent series. In the
first case if ∑ an is divergent then ∑ can will also be divergent (provided c isn’t zero of
course) since multiplying a series that is infinite in value or doesn’t have a value by a
finite value (i.e. c) won’t change that fact. However, it is possible to have ∑ an and

∑ bn are both be divergent series and yet have

∞

∑(a
n=k

n

± bn ) be a convergent series.

Before leaving this section there is one more topic that we need to briefly discuss. Since
the main topic of this section is the convergence of a series we should mention a stronger
type of convergence.
A series

∑a

n

is said to converge absolutely if

∑a

n

also converges. Absolute

convergence is stronger than convergence in the sense that a series that is absolutely
convergent will also be convergent, but a series that is convergent may or may not be
absolutely convergent.
In fact if

∑a

n

converges and

∑a

n

diverges the series

∑a

n

is sometimes called

conditionally convergent.
At this point we don’t really have the tools at hand to properly investigate this topic in
detail and so we’ll not say anything more about this subject for a while. When we finally


180


Calculus II

have the tools in hand to discuss this topic in more detail we will revisit it. Until then
don’t worry about it. The idea is mentioned here only because we were already
discussing convergence in this section.

Series – Special Series
In this section we are going to take a brief look at three special series. Actually, special
may not be the correct term. All three have been named which makes them special in
some way, however the main reason that we’re going to look at two of them in this
section is that they are the only types of series that we’ll be looking at for which we will
be able to get actual values for the series. The third type is divergent and so won’t have a
value to worry about.
In general, determining the value of a series is very difficult and outside of these two
kinds of series that we’ll look at in this section we will not be determining the value of
series in this chapter.
So, let’s get started.

Geometric Series
A geometric series is any series that can be wrote in the form,
∞

∑ ar

n −1

n =1

or, with an index shift the geometric series will often be written as,
∞

∑ ar

n

n=0

These are identical series and will have identical values, provided they converge of
course.
If we start with the first form it can be shown that the partial sums are,
a (1 − r n )
a
ar n
sn =
=
−
1− r
1− r 1− r
The series will converge provided the partial sums form a convergent sequence, so let’s
take the limit of the partial sums.
⎛ a
ar n ⎞
lim sn = lim ⎜
−
⎟
n →∞
n →∞ 1 − r
1− r ⎠
⎝
a
ar n
= lim
− lim
n →∞ 1 − r
n →∞ 1 − r
a
a
lim r n
=
−
1 − r 1 − r n→∞


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Calculus II

Now, from Theorem 3 from the Sequence section we know that the limit above will exist
and be finite provided −1 < r ≤ 1 . However, note that we can’t let r=1 since this will give
division by zero. Therefore, this will exist and be finite provided −1 < r < 1 and in this
case the limit is zero and so we get,
a
lim sn =
n →∞
1− r
Therefore, a geometric series will converge if −1 < r < 1 , which is usually written r < 1 ,
its value is,
∞

∞

n =1

n =0

∑ ar n−1 = ∑ ar n =

a
1− r

Note that in using this formula we’ll need to make sure that we are in the correct form. In
other words, if the series starts at n=0 then the exponent on the r must be n. Likewise if
the series starts at n=1 then the exponent on the r must be n-1.
Example 1 Determine if the following series converge or diverge. If they converge give
the value of the series.
∞

(a)

∑9

− n + 2 n +1

4

n =1
∞

(b)

∑
n=0

( −4 )

3n

5n −1

Solution
(a) This series doesn’t really look like a geometric series. However, notice that both
parts of the series term are numbers raised to a power. This means that it can be put into
the form of a geometric series. We will just need to decide which form is the correct
form. Since the series starts at n=1 we will want the exponents on the numbers to be n-1.

It will be fairly easy to get this into the correct form. Let’s first rewrite things slightly.
One of the n’s in the exponent has a negative in front of it and that can’t be there in the
geometric form. So, let’s first get rid of that.
∞
∞
∞
4n +1
− n−2
9− n + 2 4n +1 = ∑ 9 ( ) 4n +1 = ∑ n − 2
∑
n =1
n =1
n =1 9
Now let’s get the correct exponent on each of the numbers. This can be done using
simple exponent properties.
∞
∞
4n +1 ∞ 4n −142
− n + 2 n +1
∑ 9 4 = ∑ 9n−2 = ∑ 9n−19−1
n =1
n =1
n =1
Now, rewrite the term a little.


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Calculus II

∞

∞

n =1

n =1

∑ 9− n+2 4n+1 = ∑16 ( 9 )

4n −1 ∞
⎛4⎞
= ∑144 ⎜ ⎟
n −1
9
⎝9⎠
n =1

So, this is a geometric series with a=144 and r =

n −1

4
< 1 . Therefore, since r < 1 we know
9

the series will converge and its value will be,
∞
144 9
1296
∑ 9− n+2 4n+1 = 4 = 5 (144 ) = 5
n =1
1−
9
(b) Again, this doesn’t look like a geometric series, but it can be put into the correct form.
In this case the series starts at n=0 so we’ll need the exponents to be n on the terms. Note
that this means we’re going to need to rewrite the exponent on the numerator a little
∞

∑

( −4 )

n=0

3n

5n −1

∞

=∑
n=0

( ( −4 ) )

3 n

5n 5−1

∞

=∑
n=0

( −64 )
5
5n

n

⎛ −64 ⎞
= ∑ 5⎜
⎟
n=0 ⎝ 5 ⎠
∞

n

So, we’ve got it into the correct form and we can see that a=5 and r = −

64
. Also note
5

that r ≥ 1 and so this series diverges.
Back in the Series – Basics section we talked about stripping out terms from a series, but
didn’t really provide any examples of how this idea could be used in practice. We can
now do some examples.
Example 2 Use the results from the previous example to determine the value of the
following series.
∞

(a)

∑9

− n + 2 n +1

4

n=0
∞

(b)

∑9

− n + 2 n +1

4

n =3

Solution
(a) In this case we could just acknowledge that this is a geometric series that starts at n=0
and so we could put it into the correct form and be done with it. However, this does
provide us with a nice example of how to use the idea of stripping out terms to our
advantage.

Let’s notice that if we strip out the first term from this series we arrive at,
∞

∞

∞

n=0

n =1

n =1

∑ 9− n+2 4n+1 = 92 41 + ∑ 9− n+2 4n+1 = 324 + ∑ 9− n+ 2 4n+1
From the previous example we know the value of the new series that arises here and so
the value of the series in this example is,


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Calculus II

∞

∑9

− n + 2 n +1

4

= 324 +

n=0

1296 2916
=
5
5

(b) In this case we can’t strip out terms from the given series to arrive at the series used
in the previous example. However, we can start with the series used in the previous
example and strip terms out of it to get the series in this example. So, let’s do that. We
will strip out the first two terms from the series we looked at in the previous example.
∞

∞

∞

n =1

n =3

n =3

∑ 9− n+2 4n+1 = 9142 + 90 43 + ∑ 9− n+ 2 4n+1 = 208 + ∑ 9− n+ 2 4n+1
We can now use the value of the series from the previous example to get the value of this
series.
∞
∞
1296
256
9− n + 2 4n +1 = ∑ 9− n + 2 4n +1 − 208 =
− 208 =
∑
5
5
n =3
n =1
Notice that we didn’t discuss the convergence of either of the series in the above
example. Here’s why. Consider the following series written in two separate ways (i.e.
we stripped out a couple of terms from it).
∞

∑a
n=0

n

∞

= a0 + a1 + a2 + ∑ an
n =3

∞

∑a

Let’s suppose that we know

n =3

n

is a convergent series. This means that it’s got a finite

value and adding three finite terms onto this will not change that fact. So the value of
∞

∑a
n=0

n

is also finite and so is convergent.
∞

Likewise, suppose that

∑a
n=0

n

is convergent. In this case if we subtract three finite values
∞

from this value we will remain finite and arrive at the value of

∑a
n =3

n

. This is now a

finite value and so this series will also be convergent.
In other words, if we have two series and they differ only by the presence, or absence, of
a finite number of terms they will either both be convergent or they will both be
divergent. The difference of a few terms one way or the other will not change the
convergence of a series. This is an important idea and we will use it several times in the
following sections to simplify some of the tests that we’ll be looking at.

Telescoping Series
It’s now time to look at the second of the three series in this section. In this portion we
are going to look at a series that is called a telescoping series. The name in this case
comes from what happens with the partial sums and is best shown in an example.

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Calculus II

Example 3 Determine if the following series converges or diverges. If it converges find
its value.
∞
1
∑ n2 + 3n + 2
n=0
Solution
We first need the partial sums for this series.
n
1
sn = ∑ 2
i = 0 i + 3i + 2

Now, let’s notice that we can use partial fractions on the series term to get,
1
1
1
1
=
=
−
2
i + 3i + 2 ( i + 2 )( i + 1) i + 1 i + 2
I’ll leave the details of the partial fractions to you. By now you should be fairly adept at
this since we spent a fair amount of time doing partial fractions back in the Integration
Techniques chapter. If you need a refresher you should go back and review that section.
So, what does this do for us? Well, let’s start writing out the terms of the general partial
sum for this series using the partial fraction form.
n
1 ⎞
⎛ 1
sn = ∑ ⎜
−
⎟
i+2⎠
i =0 ⎝ i + 1
⎛1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎛1
1 ⎞ ⎛ 1
1 ⎞
= ⎜ − ⎟+⎜ − ⎟+⎜ − ⎟+ +⎜ −
−
⎟+⎜
⎟
⎝1 2 ⎠ ⎝ 2 3 ⎠ ⎝ 3 4 ⎠
⎝ n n +1 ⎠ ⎝ n +1 n + 2 ⎠
= 1−

1
n+2

Notice that every term except the first and last term canceled out. This is the origin of the
name telescoping series.
This also means that we can determine the convergence of this series by taking the limit
of the partial sums.
1 ⎞
⎛
lim sn = lim ⎜1 −
⎟ =1
n →∞
n →∞
⎝ n+2⎠
The sequence of partial sums is convergent and so the series is convergent and has a
value of
∞
1
∑ n2 + 3n + 2 = 1
n=0
In telescoping series be careful to not assume that successive terms will be the ones that
cancel. Consider the following example.


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Calculus II

Example 4 Determine if the following series converges or diverges. If it converges find
its value.
∞
1
∑ n 2 + 4n + 3
n =1
Solution
As with the last example we’ll leave the partial fractions details to you to verify. The
partial sums are,
n
1
1 ⎞
⎛ 1
⎞ 1 n ⎛ 1
−
sn = ∑ ⎜ 2 − 2 ⎟ = ∑ ⎜
⎟
i + 3 ⎠ 2 i =1 ⎝ i + 1 i + 3 ⎠
i =1 ⎝ i + 1

1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎢⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
2 ⎢⎝ 2 4 ⎠ ⎝ 3 5 ⎠ ⎝ 4 6 ⎠
⎣
1 ⎡1 1
1
1 ⎤
= ⎢ + −
−
2 ⎣ 2 3 n + 2 n + 3⎥
⎦

=

⎛1
1 ⎞ ⎛ 1
1 ⎞⎤
+⎜ −
−
⎟+⎜
⎟⎥
⎝ n n + 2 ⎠ ⎝ n + 1 n + 3 ⎠⎥
⎦

In this case instead of successive terms canceling a term will cancel with a term that is
farther down the list. The end result this time is two initial and two final terms are left.
1
Notice as well that in order to help with the work a little we factored the out of the
2
series.
The limit of the partial sums is,
1⎛5
1
1 ⎞ 5
lim sn = lim ⎜ −
−
⎟=
n →∞
n →∞ 2 6
n + 2 n + 3 ⎠ 12
⎝
So, this series is convergent (because the partial sums form a convergent sequence) and
its value is,
∞
1
5
∑ n2 + 4n + 3 = 12
n =1
Note that it’s not always obvious if a series is telescoping or not until you try to get the
partial sums and then see if they are in fact telescoping. There is no test that will tell us
that we’ve got a telescoping series right off the bat. Also note that just because you can
do partial fractions on a series term does not mean that the series will be a telescoping
series. The following series, for example, is not a telescoping series despite the fact that
we can partial fraction the series terms.
∞
3 + 2n
1 ⎞
⎛ 1
= ∑⎜
∑ n2 + 3n + 2 n=1 ⎝ n + 1 + n + 2 ⎟
⎠
n =1
∞

In order for a series to be a telescoping we must get terms to cancel and all of these terms
are positive and so none will cancel.


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Calculus II

Next, we need to go back and address an issue that was first raised in the previous
section. In that section we stated that the sum or difference of convergent series was also
convergent and that the presence of a multiplicative constant would not affect the
convergence of a series. Now that we have a few more series in hand let’s work a quick
example showing that.
Example 5 Determine the value of the following series.
∞
4
⎛
⎞
∑ ⎜ n2 + 4n + 3 − 9− n+2 4n+1 ⎟
⎠
n =1 ⎝
Solution
To get the value of this series all we need to do is rewrite it and then use the previous
results.
∞
∞
∞
4
4
⎛
− n + 2 n +1 ⎞
∑ ⎜ n2 + 4n + 3 − 9 4 ⎟ = ∑ n2 + 4n + 3 − ∑ 9− n+ 2 4n+1
⎠ n =1
n =1 ⎝
n =1
∞

∞
1
− ∑ 9− n + 2 4n +1
2
n =1 n + 4n + 3
n =1

= 4∑

⎛ 5 ⎞ 1296
= 4⎜ ⎟ −
5
⎝ 12 ⎠
3863
=−
15
We didn’t discuss the convergence of this series because it was the sum of two
convergent series and that guaranteed that the original series would also be convergent.

Harmonic Series
This is the third and final series that we’re going to look at in this chapter. Here is the
harmonic series.
∞
1
∑n
n =1
The harmonic series is divergent and we’ll need to wait until the next section to show
that. This series is here because it’s got a name and so I wanted to put it here with the
other two named series that we looked at in this section. We’re also going to use the
harmonic series to illustrate a couple of ideas about divergent series that we’ve already
discussed for convergent series. We’ll do that with the following example.
Example 6 Show that each of the following series are divergent.
∞
5
(a) ∑
n =1 n
∞
1
(b) ∑
n=4 n
Solution


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Calculus II

(a) To see that this series is divergent all we need to do is use the fact that we can factor a
constant out of a series as follows,
∞
∞
5
1
∑ n = 5∑ n
n =1
n =1
∞
1
Now, ∑ is divergent and so five times this will still not be a finite number and so the
n =1 n
series has to be divergent. In other words, if we multiply a divergent series by a constant
it will still be divergent.
(b) In this case we’ll start with the harmonic series and strip out the first three terms.
∞
∞
1
1 1 ∞ 1
1 ⎛ ∞ 1 ⎞ 11
⇒
∑ n = 1+ 2 + 3 + ∑ n
∑ n = ⎜∑ n ⎟− 6
n =1
n=4
n=4
⎝ n =1 ⎠
In this case we are subtracting a finite number from a divergent series. This subtraction
will not change the divergence of the series. We will either have infinity minus a finite
number, which is still infinity, or a series with no value minus a finite number, which will
still have no value.
Therefore, this series is divergent.
Just like with convergent series, adding/subtracting a finite number from a divergent
series is not going to change the fact the convergence of the series.
So, some general rules about the convergence/divergence of a series are now in order.
Multiplying a series by a constant will not change the convergence/divergence of the
series and adding or subtracting a constant from a series will not change the
convergence/divergence of the series. These are nice ideas to keep in mind.

Integral Test
The last topic that we discussed in the previous section was the harmonic series. In that
discussion we stated that the harmonic series was a divergent series. It is now time to
prove that statement. This proof will also get us started on the way to our next test for
convergence that we’ll be looking at.
So, we will be trying to prove that the harmonic series,
∞
1
∑n
n =1
diverges.


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Calculus II

We’ll start this off by looking at an apparently unrelated problem. Let’s start off by
1
asking what the area under f ( x ) = is on the interval [1, ∞ ) . From the section on
x
Improper Integrals we know that this is,
∞
⌠ 1 dx = ∞
⎮
⌡1 x
and so we called this integral divergent (yes, that’s the same term we’re using here with
series….).
So, just how does that help us to prove that the harmonic series diverges? Well, recall
that we can always estimate the area by breaking up the interval into segments and then
sketching in rectangles and using the sum of the area all of the rectangles as an estimate
of the actual area. Let’s do that for this problem as well and see what we get.
We will break up the interval into subintervals of width 1 and we’ll take the function
value at the left endpoint as the height of the rectangle. The image below shows the first
few rectangles for this area.

So, the area under the curve is approximately,
⎛1⎞
⎛1⎞
⎛1⎞
⎛1⎞
⎛1⎞
A ≈ ⎜ ⎟ (1) + ⎜ ⎟ (1) + ⎜ ⎟ (1) + ⎜ ⎟ (1) + ⎜ ⎟ (1) +
⎝1⎠
⎝2⎠
⎝3⎠
⎝ 4⎠
⎝5⎠
1 1 1 1 1
= + + + + +
1 2 3 4 5
Now note a couple of things about this approximation. First, each of the rectangles
overestimates the actual area and secondly the formula for the area is exactly the
harmonic series!
Putting these two facts together gives the following,
∞
1 ⌠∞1
A≈∑ >⎮
dx = ∞
⌡1 x
n =1 n


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Calculus II

Notice that this tells us that we must have,
∞
∞
1
1
>∞
⇒
∑n
∑n =∞
n =1
n =1
Since we can’t really be larger than infinity the harmonic series must also be infinite in
value. In other words, the harmonic series is in fact divergent.
So, we’ve managed to relate a series to an improper integral that we could compute and it
turns out that the improper integral and the series have exactly the same convergence.
Let’s see if this will also be true for a series that converges. When discussing the
Divergence Test we made the claim that
∞
1
∑ n2
n =1
converges. Let’s see if we can do something similar to the above process to prove this.
We will try to relate this to the area under f ( x ) =

1
is on the interval [1, ∞ ) . Again,
x2

from the Improper Integral section we know that,
∞
⌠ 1 dx = 1
⎮ 2
⌡1 x
and so this integral converges.
We will once again try to estimate the area under this curve. We will do this in an almost
identical manner as the previous part with the exception that we will instead of using the
left end points for the height of our rectangles we will use the right end points. Here is a
sketch of this case,

In this case the area estimation is,


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Calculus II

⎛ 1 ⎞
⎛1⎞
⎛ 1⎞
⎛1⎞
A ≈ ⎜ 2 ⎟ (1) + ⎜ 2 ⎟ (1) + ⎜ 2 ⎟ (1) + ⎜ 2 ⎟ (1) +
⎝2 ⎠
⎝3 ⎠
⎝4 ⎠
⎝5 ⎠
1 1 1 1
= 2+ 2+ 2+ 2+
2 3 4 5
This time, unlike the first case, the area will be an underestimation of the actual area and
the estimation is not quite the series that we are working with. Notice however that the
only difference is that we’re missing the first term. This means we can do the following,
∞
∞
1 1 1 1 1 1
1
= 2 + 2 + 2 + 2 + 2 + < 1 + ⌠ 2 dx = 1 + 1 = 2
⎮
∑ n2 1 2 3 4 5
⌡1 x
n =1
Area Estimation

Or, putting all this together we see that,
∞

1

∑n
n =1

2

<2

With the harmonic series this was all that we needed to say that the series was divergent.
With this series however, this isn’t quite enough. For instance −∞ < 2 and if the series
did have a value of −∞ then it would be divergent (when we want convergent). So, let’s
do a little more work.
First, let’s notice that all the series terms are positive (that’s important) and that the
partial sums are,
n
1
sn = ∑ 2
i =1 i
Because the terms are all positive we know that the partial sums must be an increasing
sequence. In other words,
n
1 n +1 1
sn = ∑ 2 < ∑ 2 = sn +1
i =1 i
i =1 i
In sn+1 we are adding a single positive term onto sn and so must get larger. Therefore, the
partial sums form an increasing (and hence monotonic) sequence.
Also note that, since the terms are all positive, we can say,
n
1 ∞ 1
sn = ∑ 2 < ∑ 2 < 2
⇒
i =1 i
n =1 n
and so the sequence of partial sums is a bounded sequence.

sn < 2

In the section on Sequences we gave a theorem that stated that a bounded and monotonic
sequence was guaranteed to be convergent. This means that the sequence of partial sums
is a convergent sequence. So, who cares right? Well recall that this means that the series
must then also be convergent!


191


Calculus II

So, once again we were able to relate a series to an improper integral (that we could
compute) and the series and the integral had the same convergence.
We went through a fair amount of work in both of these examples to determine the
convergence of the two series. Luckily for us we don’t need to do all this work every
time. The ideas in these two examples can be summarized in the following test.

Integral Test

Suppose that f(x) is a positive, decreasing function on the interval [ k , ∞ ) and that
f ( n ) = an then,

1. If
2. If

∫
∫

∞
k
∞
k

f ( x ) dx is convergent so is

∞

∑a

n

n=k

f ( x ) dx is divergent so is

.

∞

∑a
n=k

n

.

There are a couple of things to note about the integral test. First, the lower limit on the
improper integral must be the same value that starts the series. Second, the function does
not actually need to be decreasing everywhere in the interval. All that’s really required is
that eventually the function is decreasing. In other words, it is okay if the function
increases for a while, but eventually the function must start decreasing and then continue
to decrease from that point on.
There is one more very important point that must be made about this test. This test does
NOT give the value of a series. It will only give the convergence/divergence of the
series. That’s it. No value. We can use the above series as a perfect example of this.
All that the test gave us was that,
∞
1
∑ n2 < 2
n =1
So, we got an upper bound on the value of the series, but not an actual value for the
series. In fact, from this point on we will not be asking for the value of a series we will
only be asking whether a series converges or diverges. In a later section we look at
estimating values of series, but even in that section still won’t actually be getting values
of series.
Just for the sake of completeness the value of this series is known.
∞
1 π2
∑ n2 = 6 = 1.644934... < 2
n =1


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Calculus II

∞

1

∑ n ln n
n=2

Solution
In this case the function we’ll use is,

1
x ln x
This function is clearly positive and if we make x larger the denominator will get larger
and so the function is also decreasing. Therefore, all we need to do is determine the
convergence of the following integral.
∞
t
1
1
⌠
dx = lim ⌠
dx
u = ln x
⎮
⎮
t →∞ ⌡ x ln x
⌡ 2 x ln x
2
f ( x) =

= lim ( ln ( ln x ) )
t →∞

t
2

= lim ( ln ( ln t ) − ln 2 )
t →∞

=∞
The integral is divergent and so the series is also divergent by the Integral Test.
∞

∑ ne n

− 2

n=0

Solution
The function that we’ll use in this example is,
f ( x ) = xe − x

2

This function is always positive on the interval that we’re looking at. Now we need to
check that the function is decreasing. It is not clear that this function will always be
decreasing on the interval given. We can use our Calculus I knowledge to help us
however. The derivative of this function is,
f ′ ( x ) = e − x (1 − 2 x 2 )
2

This function has two critical points (which will tell us where the derivative changes
1
sign) at x = ±
. Since we are starting at n=0 we can ignore the negative critical point.
2
Picking a couple of test points we can see that the function in increasing on the interval
⎡ 1 ⎤
⎡ 1
⎞
⎢ 0, 2 ⎥ and it is decreasing on ⎢ 2 , ∞ ⎟ . Therefore, eventually the function will be
⎣
⎠
⎣
⎦
decreasing and that’s all that’s required for us to use the Integral Test.


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Calculus II

∫

∞
0

xe − x dx = lim ∫ xe− x dx
t

2

t →∞

2

u = − x2

0

t

⎛ 1 2⎞
= lim ⎜ − e− x ⎟
t →∞
⎝ 2
⎠0
⎛1 1 2 ⎞
= lim ⎜ − e− t ⎟
t →∞ 2
2
⎝
⎠
1
=
2
The integral is convergent and so the series must also be convergent by the Integral Test.
We can use the Integral test to get the following fact/test for some series.

Fact (p –series Test)
∞
1
If k>0 then ∑ p converges if p > 1 and diverges if p ≤ 1 .
n=k n
Sometimes the series in this fact are called p-series and so this fact is sometimes called
the p-series test.
Example 3 Determine if the following series are convergent or divergent.
∞
1
(a) ∑ 7
n=4 n
∞
1
(b) ∑
n
n =1
Solution
(a) In this case p=7>1 and so by this fact the series is convergent.

(b) For this series p =

1
≤ 1 and so the series is divergent by the fact.
2

It is important to note before leaving this section that in order to use the Integral Test the
series terms MUST be positive. If they are negative then the test doesn’t work. Also
remember that the test only determines the convergence of a series and does NOT give
the value of the series.

Comparison Test / Limit Comparison Test
In the previous section we saw how to relate a series to an improper integral to determine
the convergence of a series. While the integral test is a nice test, it does force us to do


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Calculus II

improper integrals which aren’t always easy and in some cases may be impossible to
evaluate.
For instance consider the following series.
∞

∑3

n

n=0

1
+n

In order to use the Integral Test we would have to integrate
∞
1
⌠
dx
⎮ x
⌡0 3 + x
and I’m not even sure if it’s possible to do this integral. Nicely enough for us there is
another test that we can use on this series that will be much easier to use.
First, let’s note that the series terms are positive. As with the Integral Test that will be
important in this section. Next let’s note that we must have x > 0 since we are
integrating on the interval 0 ≤ x < ∞ . Likewise, regardless of the value of x we will
always have 3x > 0 . So, if we drop the x from the denominator the denominator will get
smaller and hence the whole fraction will get larger. So,
1
1
< n
n
3 +n 3
Now,
∞

1

∑3
n=0

n

is a geometric series and we know that since r =

1
< 1 the series will converge and it’s
3

value will be,
∞

1

∑3
n=0

n

=

1
3
=
1
1− 3 2

Now, if we go back to our original series and write down the partial sums we get,
n
1
sn = ∑ i
i =1 3 + i
Since all the terms are positive adding a new term will only make the number larger and
so the sequence of partial sums must be an increasing sequence.
n
n +1
1
1
sn = ∑ i
<∑ i
= sn +1
i =1 3 + i
i =1 3 + i
Then since,
1
1
< n
3 +n 3
n


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Calculus II

and because the terms in these two sequences are positive we can also say that,
n
n
∞
1
1
1 3
3
sn = ∑ i
<∑ i <∑ n =
⇒
sn <
2
2
i =1 3 + i
i =1 3
n =1 3
Therefore, the sequence of partial sums is also a bounded sequence. Then from the
section on Sequences we know that a monotonic and bounded sequence is also
convergent.
So, the sequence of partial sums of our series is a convergent sequence. This means that
the series itself,
∞
1
∑ 3n + n
n=0
is also convergent.
So, what did we do here? We found a series whose terms were always larger than the
original series terms and this new series was also convergent. Then since the original
series terms were positive (very important) this meant that the original series was also
convergent.
To show that a series (with only positive terms) was divergent we could go through a
similar argument and find a new divergent series whose terms are always smaller than the
original series. In this case the original series would have to take a value larger than the
new series. However, since the new series is divergent its value will be infinite. This
means that the original series must also be infinite and hence divergent.
We can summarize all this in the following test.

Comparison Test
Suppose that we have two series

∑a

n

and

all n. Then,
1. If ∑ bn is convergent then so is
2.

∑b

n

with an , bn ≥ 0 for all n and an ≤ bn for

∑a .
If ∑ a is divergent then so is ∑ b .
n

n

n

In other words, we have two series of positive terms and the terms of one of the series is
always larger than the terms of the other series. Then if the larger series is convergent
the smaller series must also be convergent. Likewise, if the smaller series is divergent
then the larger series must also be divergent.
Do not misuse this test. Just because the smaller of the two series converges does not say
anything about the larger series. The larger series may still diverge. Likewise, just
because we know that the larger of two series diverges we can’t say that the smaller
series will also diverge! Be very careful in using this test


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Calculus II

Recall that we had a similar test for improper integrals back when we were looking at
integration techniques. So, if you could use the comparison test for improper integrals
you can use the comparison test for series as they are pretty much the same idea.
Let’s take a look at some examples.
∞
n
∑ n2 − cos2 ( n )
n =1
Solution
Since the cosine term in the denominator doesn’t get too large we can assume that the
series terms will behave like,
n 1
=
n2 n
which, as a series, will diverge. So, from this we can guess that the series will probably
diverge and so we’ll need to find a smaller series that will also diverge.

Recall that from the comparison test with improper integrals that we determined that we
can make a fraction smaller by either making the numerator smaller or the denominator
larger. In this case the two terms in the denominator are both positive. So, if we drop the
cosine term we will in fact be making the denominator larger since we will no longer be
subtracting off a positive quantity. Therefore,
n
n 1
> 2 =
2
2
n − cos ( n ) n
n
Then, since
∞

1

∑n
n =1

diverges (it’s harmonic and the p-series test) by the Comparison Test our original series
must also diverge.
Example 2 Determine if the following series converges or diverges.
∞
n2 + 2
∑ n4 + 5
n =1
Solution
In this case the “+2” and the “+5” don’t really add anything to the series and so the series
terms should behave pretty much like
n2 1
=
n4 n2
which will converge as a series. Therefore, we will need to find a larger series which
also converges.

This means that we’ll either have to make the numerator larger or the denominator
smaller. We can make the denominator smaller by dropping the “+5”. Doing this gives,

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Calculus II

n2 + 2 n2 + 2
<
n4 + 5
n4
At this point, notice that we can’t drop the “+2” from the numerator since this would
make the term smaller and that’s not what we want. However, this is actually all the
further that we need to go. Let’s take a look at the following series.
∞
n2 + 2 ∞ n2 ∞ 2
∑ n4 = ∑ n4 + ∑ n4
n =1
n =1
n =1
∞
∞
1
2
=∑ 2 +∑ 4
n =1 n
n =1 n
As shown, we can write the series as a sum of two series and both of these series are
convergent by the p-series test. Therefore, since each of these series are convergent we
know that the sum,
∞
n2 + 2
∑ n4
n =1
is also a convergent series. Recall that the sum of two convergent series will also be
convergent.
Now, since the terms of this series are larger than the terms of the original series we
know that the original series must also be convergent by the Comparison Test.
The comparison test is a nice test that allows us to do problems that either we couldn’t
have done with the integral test or at the best would have been very difficult to do with
the integral test. That doesn’t mean that it doesn’t have problems of its own.
Consider the following series.
∞

1
−n
n=0
This is not much different from the first series that we looked at. The original series
converged because the 3n gets very large very fast and will be significantly larger than the
n. Therefore, the n doesn’t really affect the convergence of the series in that case. The
fact that we are now subtracting the n off now instead of adding the n on really shouldn’t
change the convergence. We can say this because the 3n gets very large very fast and the
fact that we’re subtracting n off won’t really change the size of this term for all
sufficiently large value of n.

∑3

n

So, we would expect this series to converge. However, the comparison test won’t work
with this series. To use the comparison test on this series we would need to find a larger
series that we could easily determine the convergence of. In this case we can’t do what
we did with the original series. If we drop the n we will make the denominator larger
(since the n was subtracted off) and so the fraction will get smaller and just like when we
looked at the comparison test for improper integrals knowing that the smaller of two
series converges does not mean that the larger of the two will also converge.


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Calculus II

So, we will need something else to do help us determine the convergence of this series.
The following variant of the comparison test will allow us to determine the convergence
of this series.

Limit Comparison Test

∑a

n

and

∑b

n

with an , bn ≥ 0 for all n. Define,

an
bn
If c is positive (i.e. c > 0 ) and is finite (i.e. c < ∞ ) then either both series converge or
both series diverge.
c = lim

n →∞

Note that it doesn’t really matter which series term is in the numerator for this test, we
could just have easily defined c as,
b
c = lim n
n →∞ a
n
and we would get the same results. To see why this is, consider the following two
definitions.
a
b
c = lim n
c = lim n
n →∞ b
n →∞ a
n
n
Start with the first definition and rewrite it as follows, then take the limit.
a
1
1
1
c = lim n = lim
=
=
n →∞ b
n →∞ b
bn c
n
n
lim
n →∞ a
an
n
In other words, if c is positive and finite then so is c and if c is positive and finite then
so is c. Likewise if c = 0 then c = ∞ and if c = ∞ then c = 0 . Both definitions will give
the same results from the test so don’t worry about which series terms should be in the
numerator and which should be in the denominator. Choose this to make the limit easy to
compute.
Also, this really is a comparison test is some ways. If c is positive and finite this is
saying that both of the series terms will behave in generally the same fashion and so we
can expect the series themselves to also behave in a similar fashion. If c = 0 or c = ∞
we can’t say this and so the test fails to give any information.
The limit in this test will often be written as,
1
n →∞
bn
since often both terms will be fractions and this will make the limit easier to deal with.
c = lim an ⋅

Let’s see how this test works.


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Calculus II

∞
1
∑ 3n − n
n=0
Solution
To use the limit comparison test we need to find a second series that we can determine
the convergence of easily and has what we assume is the same convergence as the given
series. On top of that we will need to choose the new series in such a way as to give us
an easy limit to compute for c.

We’ve already guessed that this series converges and since it’s vaguely geometric let’s
use
∞
1
∑ 3n
n=0
as the second series. We know that this series converges and there is a chance that since
both series have the 3n in it the limit won’t be too bad.
Here’s the limit.
1 3n − n
n →∞ 3n
1
n
= lim1 − n
n →∞
3

c = lim

Now, we’ll need to use L’Hospital’s Rule on the second term in order to actually evaluate
this limit.
1
c = 1 − lim n
n →∞ 3 ln ( 3 )
=1

So, c is positive and finite so by the Comparison Test both series must converge since
∞
1
∑ 3n
n=0
converges.
∞
4n 2 + n

∑
n=2

3

n 7 + n3

Solution
Fractions involving only polynomials or polynomials under radicals will behave in the
same way as the largest power of n will behave in the limit. So, the terms in this series
should behave as,


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Calculus II

n2
3

n

7

n2

=

7
3

=

1
1

n
n3
and as a series this will diverge by the p-series test. In fact, this would make a nice
choice for our second series in the limit comparison test so let’s use it.
1

4n 2 + n n 3
lim
= lim
n →∞ 3 7
n + n3 1 n→∞

7

4

4n 3 + n 3
3

1 ⎞
⎛
n 7 ⎜1 + 4 ⎟
⎝ n ⎠

1⎞
⎛
n3 ⎜ 4 + ⎟
n⎠
= lim 7 ⎝
n →∞
1
n3 3 1+ 4
n
4
= 3 =4=c
1
7

So, c is positive and finite and so both limits will diverge since
∞
1

∑
n=2

1

n3

diverges.
Finally, to see why we need to c must be positive and finite (i.e. c ≠ 0 and c ≠ ∞ )
consider the following two series.
∞
∞
1
1
∑n
∑ n2
n =1
n =1
The first diverges and the second converges.
Now compute each of the following limits.
1 n2
lim i = lim n = ∞
n →∞ n 1
n →∞

1 n
1
i = lim = 0
2
n →∞ n
1 n→∞ n

lim

In the first case the limit from the limit comparison test yields c = ∞ and in the second
case the limit yields c = 0 . Clearly, both series do not have the same convergence.
Note however, that just because we get c = 0 or c = ∞ doesn’t mean that the series will
have the opposite convergence. To see this consider the series,
∞
∞
1
1
∑ n3
∑ n2
n =1
n =1
Both of these series converge and here are the two possible limits that the limit
comparison test uses.
1 n2
1
1 n3
lim 3 i = lim = 0
lim 2 i = lim n = ∞
n →∞ n
n →∞ n
1 n →∞ n
1 n→∞


201


Calculus II

So, even though both series had the same convergence we got both c = 0 and c = ∞ .
The point of all of this is to remind us that if we get c = 0 or c = ∞ from the limit
comparison test we will know that we have chosen the second series incorrectly and we’ll
need to find a different choice in order to get any information about the convergence of
the series.

Alternating Series Test
The last two tests that we looked at for series convergence have required that all the terms
in the series be positive. Of course there are many series out there that have negative
terms in them and so we now need to start looking at tests for these kinds of series.
The test that we are going to look into in this section will be a test for alternating series.
An alternating series is any series, ∑ an , for which the series terms can be written in
one of the following two forms.
n
an = ( −1) bn
an = ( −1)

n +1

bn ≥ 0
bn ≥ 0

bn

There are many other ways to deal with the alternating sign, but they can all be written as
on of the two forms above. For instance,
n+2
n
2
n
( −1) = ( −1) ( −1) = ( −1)

( −1)

= ( −1) ( −1) = ( −1)
There are of course many others, but they all follow the same basic pattern of reducing to
one of the first two forms given. If you should happen to run into a different form than
the first two, don’t worry about converting it to one of those forms, just be aware that it
can be and so the test from this section can be used.
n −1

n +1

−2

n +1

Suppose that we have a series

∑a

and either an = ( −1) bn or an = ( −1)
n

n

n +1

bn where

bn ≥ 0 for all n. Then if,
1. lim bn = 0 and,
n →∞

2.

{bn } is eventually a decreasing sequence

the series

∑a

n

is convergent.

There are a couple of things to note about this test. First, unlike the Integral Test and the
Comparison/Limit Comparison Test, this test will only tell us when a series converges
and not if a series will diverge.

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Calculus II

Secondly, in the second condition all that we need to require is that the series terms, bn
will eventually decreasing. It is possible for the first few terms of a series to increase and
still have the test be valid. All that is required is that eventually we will have bn ≥ bn +1 for
all n after some point.
To see why this is consider the following series,
∞

∑ ( −1)

n

n =1

{bn } is not a decreasing sequence and that for

Let’s suppose that for 1 ≤ n ≤ N

{bn } is a decreasing sequence.

bn
n ≥ N +1

The series can then be written as,

∞

∑ ( −1)

n

n =1

N

bn = ∑ ( −1) bn +
n

n =1

∞

∑ ( −1)

n = N +1

n

bn

The first series is a finite series (no matter how large N is, it will still be a finite series)
and so we can compute its value and it will be finite. Therefore, the convergence of the
series will depend solely on the convergence of the second (infinite) series. If the second
series has a finite value then the sum of two finite values is also finite and so the original
series will converge to a finite value. On the other hand if the second series is divergent
either because its value is infinite or it doesn’t have a value then adding a finite number
onto this will not change that fact and so the original series will be divergent.
The point of all this is that we don’t need to require that the series terms be decreasing for
all n. We only need to require that the series terms will eventually be decreasing since
we can always strip out the first few terms that aren’t actually decreasing and look only at
the terms that are actually decreasing.
Note that, in practice, we don’t actually strip out the terms that aren’t decreasing. All we
do is check that eventually the series terms are decreasing and then apply the test.
∞

∑

( −1)

n +1

n

n =1

Solution
First, identify the bn for the test.
∞

∑
n =1

( −1)
n

n +1

∞

= ∑ ( −1)
n =1

n +1

1
n

bn =

1
n

Now, all that we need to do is run through the two conditions in the test.


203


Calculus II

1
=0
n →∞
n →∞ n
1
1
bn = >
= bn +1
n n +1
lim bn = lim

Both conditions are met and so by the Alternating Series Test the series must converge.
The series from the previous example is sometimes called the Alternating Harmonic
Series.
In the previous example it was easy to see that the series terms decreased since increasing
n only increased the denominator for the term and hence made the term smaller. In
general however we will need to resort to Calculus I techniques to prove the series terms
decrease. We’ll see an example of this in a bit.
∞

∑
n =1

( −1)

n

n2
n2 + 5

Solution
First, identify the bn for the test.
∞

( −1)

n

∞
n2
n2
n
= ∑ ( −1) 2
∑ n2 + 5 n=1
n +5
n =1
Let’s check the conditions.

⇒

bn =

n2
n2 + 5

n2
=1≠ 0
n →∞
n →∞ n 2 + 5
So, the first condition isn’t met and so there is no reason to check the second. Since this
condition isn’t met we’ll need to use another test to check convergence. In these cases
where the first condition isn’t met it is usually best to use the divergence test.
n
( −1) n 2 = lim −1 n ⎛ lim n 2 ⎞
lim 2
( ) ⎜ n→∞ 2 ⎟
n →∞ n + 5
n →∞
n +5⎠
⎝
lim bn = lim

(
)
= ( lim ( −1) ) (1)
n

n →∞

= lim ( −1)

n

doesn't exist

n →∞

This limit doesn’t exist and so by the Divergence Test this series diverges.
∞

∑
n=0

( −1)

n −3

n

n+4

Solution
Notice that in this case the exponent on the “-1” isn’t n or n+1. That won’t change how


204


Calculus II

the test works however so we won’t worry about that. In this case we have,
n
bn =
n+4
so let’s check the conditions.
The first is easy enough to check.
lim bn = lim
n →∞

n →∞

n
=0
n+4

The second condition requires some work however. It is not immediately clear that these
terms will decrease. Increasing n to n+1 will increase both the numerator and the
denominator. Increasing the numerator says the term should also increase while
increasing the denominator says that the term should decrease. Since its not clear which
of these will win out we will need to resort to Calculus I techniques to show that the
terms decrease.
Let’s start with the following function and its derivative.
x
4− x
f ( x) =
f ′( x) =
2
x+4
2 x ( x + 4)
Now, there are three critical points for this function, x=-4, x=0, and x=4. The first is
outside the bound of our series so we won’t need to worry about that one. Using the test
points,
3
5
f ′ (1) =
f ′ ( 5) = −
50
810
and so we can see that the function in increasing on 0 ≤ x ≤ 4 and decreasing on x ≥ 4 .
Therefore, since f ( n ) = bn we know as well that the bn are also increasing on 0 ≤ n ≤ 4
and decreasing on n ≥ 4 .
The bn are then eventually decreasing and so the second condition is met.
Both conditions are met and so by the Alternating Series Test the series must be
converging.
As the previous example has shown, we sometimes need to do a fair amount of work to
show that the terms are decreasing. Do not just make the assumption that the terms will
be decreasing and let it go at that.
Let’s do one more example just to make a point.
∞
cos ( nπ )

∑
n=2

n

Solution


205


Calculus II

The point of this problem is really just to acknowledge that it is in fact an alternating
series. To see this we need to acknowledge that,
n
cos ( nπ ) = ( −1)
and so the series is really,
∞

∑
n=2

cos ( nπ )
n

∞

=∑

( −1)

n=2

n

⇒

n

bn =

1
n

Checking the two condition gives,
lim bn = lim
n →∞

bn =

n →∞

1
=0
n

1
1
>
= bn +1
n
n +1

The two conditions of the test are met and so by the Alternating Series Test the series is
convergent.
It should be pointed out that the rewrite we did in previous example only works because n
is an integer and because of the presence of the π. Without the π we couldn’t do this and
if n wasn’t guaranteed to be an integer we couldn’t do this.

Absolute Convergence
When we first talked about series convergence we briefly mentioned a stronger type of
convergence but didn’t do anything with it because we didn’t have any tools at our
disposal that we could use to work problems involving it. We now have some of those
tools so it’s now time to talk about absolute convergence in detail.
First, let’s go back over the definition of absolute convergence.
A series

∑a

is called absolutely convergent if

n

convergent and

∑a

n

∑a

n

is convergent. If

∑a

n

is

is divergent we call the series conditionally convergent.

We also have the following fact about absolute convergence.

Fact
If

∑a

n

is absolutely convergent then it is also convergent.

It is this fact that makes absolute convergence a “stronger” type of convergence. Series
that are absolutely convergent are guaranteed to be convergent. However, series that are
convergent may or may not be absolutely convergent.


206


Calculus II

Let’s take a quick look at a couple of examples of absolute convergence.
Example 1 Determine if each of the following series are absolute convergent,
conditionally convergent or divergent.
∞

(a)

∑

( −1)

n

n =1
∞

(b)

n

( −1)

∑

n+2

n2
∞
sin n
(c) ∑ 3
n =1 n
Solution
(a) This is the alternating harmonic series and we saw in the last section that it is a
convergent series so we don’t need to check that here. So, let’s see if it is an absolutely
convergent series. To do this we’ll need to check the convergence of.
n =1

∞

∑
n =1

( −1)

n

n

∞

1
n =1 n

=∑

This is the harmonic series and we know form the integral test section that it is divergent.
Therefore, this series is not absolutely convergent. It is however conditionally
convergent since the series itself does converge.

(b) In this case let’s just check absolute convergence first since if it’s absolutely
convergent we won’t need to bother checking convergence as we will get that for free.
∞

∑
n =1

( −1)
n

n+2

2

∞

1
2
n =1 n

=∑

This series is convergent by the p-series test and so the series is absolute convergent.
Note that this does say as well that it’s a convergent series.

(c) In this part we need to be a little careful. First, this is NOT an alternating series and
so we can’t use any tools from that section.
What we’ll do here is check for absolute convergence first again since that will also give
convergence. This means that we need to check the convergence of the following series.
∞
∞
sin n
sin n
∑ n3 = ∑ n3
n =1
n =1
To do this we’ll need to note that
−1 ≤ sin n ≤ 1

⇒

sin n ≤ 1

and so we have,
sin n 1
≤ 3
n3
n


207


Calculus II

Now we know that
∞

1

∑n
n =1

3

converges by the p-series test and so by the Comparison Test we also know that
∞
sin n
∑ n3
n =1
converges.
Therefore the original series is absolutely convergent (and hence convergent).

Ratio Test
In this section we are going to take a look at a test that we can use to see if a series is
absolutely convergent or not. Recall that if a series is absolutely convergent then we will
also know that it’s convergent and so we will more often than not use it to simply
determine the convergence of a series. We may as well acknowledge that it will give us
absolute convergence as well however.
Before proceeding with the test let’s do a quick reminder of factorials. This test will be
particularly useful for series that contain factorials (and we will see some in the
applications) so let’s make sure we can deal with them before we run into them in an
example.
If n is an integer such that n ≥ 0 then n factorial is defined as,
n ! = n ( n − 1)( n − 2 )

( 3)( 2 )(1)

0! = 1

if n ≥ 1
by definition

Let’s compute a couple real quick.
1! = 1
2! = 2 (1) = 2
3! = 3 ( 2 )(1) = 6
4! = 4 ( 3)( 2 )(1) = 24
5! = 5 ( 4 )( 3)( 2 )(1) = 120

In the last computation above, notice that we could rewrite the factorial in a couple of
different ways. For instance,


208


Calculus II

5! = 5 ( 4 )( 3)( 2 )(1) = 5 ⋅ 4!
4!

5! = 5 ( 4 )( 3)( 2 )(1) = 5 ( 4 ) ⋅ 3!
3!

In general we can always “strip out” terms from a factorial as follows.
n ! = n ( n − 1)( n − 2 ) ( n − k ) ( n − ( k + 1) ) ( 3)( 2 )(1)

( n − k ) ⋅ ( n − ( k + 1) )!
= n ( n − 1)( n − 2 ) ( n − k ) ⋅ ( n − k − 1) !
= n ( n − 1)( n − 2 )

We will need to do this on occasion so don’t forget about it.
Also, when dealing with factorials we need to be very careful with parenthesis. For
instance, ( 2n ) ! ≠ 2 n ! as we can see if we write each of the following factorials out.

( 2n )! = ( 2n )( 2n − 1)( 2n − 2 ) ( 3)( 2 )(1)
2 n ! = 2 ⎡( n )( n − 1)( n − 2 ) ( 3)( 2 )(1) ⎤
⎣
⎦
Again, we will run across factorials with parenthesis so don’t drop them. This is often
one of the more common mistakes that students make when the first run across factorials.
Okay, we are now ready for the test.

Ratio Test
Suppose we have the series

∑a

n

. Define,
L = lim

n →∞

an +1
an

Then,
1. if L < 1 the series is absolutely convergent (and hence convergent).
2. if L > 1 the series is divergent.
3. if L = 1 the series may be divergent, convergent, or absolutely convergent.
Notice that in the case of L = 1 the ratio test is pretty much worthless and we would need
to resort to a different test to determine the convergence of the series.
Also, the absolute value bars in the definition of L are absolutely required. If they are not
there it will be possible for us to get the incorrect answer.
Let’s take a look at some examples.


209


Calculus II

( −10 )
∑ 42n+1 ( n + 1)
n =1
n

∞

Solution
With this first example let’s be a little careful and make sure that we have everything
down correctly. Here are the series terms an.

( −10 )
an = 2 n +1
4 ( n + 1)
n

Recall that to compute an+1 all that we need to do is substitute n+1 for all the n’s in an.

( −10 )
( −10 )
= 2 n +3
an +1 = 2( n +1)+1
4
( ( n + 1) + 1) 4 ( n + 2 )
n +1

n +1

Now, to define L we will use,
1
n →∞
an
since this will be a little easier when dealing with fractions as we’ve got here. So,
L = lim an +1 ⋅

42 n +1 ( n + 1)
( −10 )
42 n +3 ( n + 2 ) ( −10 )n
−10 ( n + 1)
42 ( n + 2 )
n +1

L = lim

n →∞

= lim

n →∞

10
n +1
lim
16 n →∞ n + 2
10
= <1
16

=

So, L < 1 and so by the Ratio Test the series converges absolutely and hence will
converge.
As seen in the previous example there is usually a lot of canceling that will happen in
these. Make sure that you do this canceling. If you don’t do this kind of canceling it can
make the limit fairly difficult.
∞
n!
∑ 5n
n =0
Solution
Now that we’ve worked one in detail we won’t go into quite the detail with the rest of
these. Here is the limit.
( n + 1)! 5n = lim ( n + 1)!
L = lim
n →∞
5n +1 n ! n→∞ 5 n !


210


Calculus II

In order to do this limit we will need to eliminate the factorials. We simply can’t do the
limit with the factorials in it. To eliminate the factorials we will recall from our
discussion on factorials above that we can always “strip out” terms from a factorial. If
we do that with the numerator (in this case because it’s the larger of the two) we get,
( n + 1) n !
L = lim
n →∞
5 n!
at which point we can cancel the n! for the numerator an denominator to get,
( n + 1) = ∞ > 1
L = lim
n →∞
5
So, by the Ratio Test this series diverges.
∞
n2
∑ ( 2n − 1)!
n=2
Solution
In this case be careful in dealing with the factorials.

( n + 1) ( 2n − 1)!
L = lim
n →∞ 2 n + 1 − 1 !
( ( ) ) n2
2

( n + 1) ( 2n − 1)!
= lim
n →∞ ( 2n + 1) !
n2
2

( n + 1)
( 2n − 1)!
= lim
n →∞ ( 2n + 1)( 2n )( 2n − 1) !
n2
2
( n + 1)
= lim
n →∞ 2n + 1 2n
(
)( ) ( n 2 )
2

= 0 <1
So, by the Ratio Test this series converges.
∞
9n

∑
n =1

( −2 )

n +1

n

Solution
Do not mistake this for a geometric series. The n in the denominator means that this isn’t
a geometric series. So, let’s compute the limit.


211


Calculus II

L = lim

n →∞

= lim

n →∞

( −2 )

9n +1

( −2 ) ( n + 1)
n+2

n +1

9

n

n

9n
( −2 )( n + 1)

9
n
lim
n →∞ n + 1
2
9
= >1
2

=

Therefore, by the Ratio Test this series is divergent.
In the previous example the absolute value bars were required to get the correct answer.
9
If we hadn’t used them we would have gotten L = − < 1 which would have implied a
2
convergent series!
Now, let’s take a look at a couple of examples to see what happens when we get L = 1 .
Recall that the ratio test will not tell us anything about the convergence of these series. In
both of these examples we will first verify that we get L = 1 and then use other tests to
determine the convergence.
∞

( −1)

∑n
n=0

2

n

+1

Solution
Let’s first get L.

( −1)
n2 + 1
n2 + 1
= lim
=1
L = lim
n
2
2
n →∞
( n + 1) + 1 ( −1) n→∞ ( n + 1) + 1
n +1

So, as implied earlier we get L = 1 which means the ratio test is no good for determining
the convergence of this series. We will need to resort to another test for this series. This
series is an alternating series and so let’s check the two conditions from that test.
1
lim bn = lim 2
=0
n →∞
n →∞ n + 1
1
1
bn = 2
>
= bn +1
n + 1 ( n + 1)2 + 1
The two conditions are met and so by the Alternating Series Test this series is
convergent.


212


Calculus II

∞

n+2

∑ 2n + 7
n=0

Solution
Here’s the limit.

L = lim

n →∞

( n + 3)( 2n + 7 ) = 1
n+3
2n + 7
= lim
n →∞ ( 2n + 9 )( n + 2 )
2 ( n + 1) + 7 n + 2

Again, the ratio test tells us nothing here. We can however, quickly use the divergence
test on this. In fact that probably should have been our first choice on this one anyway.
n+2 1
lim
= ≠0
n →∞ 2n + 7
2
By the Divergence Test this series is divergent.
So, as we saw in the previous two examples if we get L = 1 from the ratio test the series
can be either convergent or divergent.
There is one more thing that we should note about the ratio test before we move onto the
next section. The last series was a polynomial divided by a polynomial and we saw that
we got L = 1 from the ratio test. This will always happen with rational expression
involving only polynomials or polynomials under radicals. So, in the future it isn’t even
worth it to try the ratio test on these kinds of problems since we now know that we will
get L = 1 .
Also, in the second to last example we saw an example of a rational expression times
n
( −1) . In this example we also saw that we got L = 1 . Again, as long as it is minus one
to a power (any power) this will happen as well and so again there really isn’t any use in
even trying the ratio test on these kinds of problems. Note however, if it had been
n +1
( −3) or 8n (i.e. anything other than “-1”) we wouldn’t have gotten L = 1 and so the
ratio test would work.

Root Test
This is the last test for series convergence that we’re going to be looking at. As with the
Ratio Test this test will also tell whether a series is absolutely convergent or not rather
than simple convergence.

Root Test
Suppose that we have the series

∑a

n

. Define,

L = lim n an = lim an
n →∞

1
n

n →∞

Then,

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Calculus II

As with the ratio test, if we get L = 1 the root test will tell us nothing and we’ll need to
use another test to determine the convergence of the series.
We will also need the following fact in some of these problems.

Fact
1

lim n n = 1
n →∞

∞
nn
∑ 31+2n
n =1
Solution
There really isn’t much to these problems other than computing the limit and then using
the root test. Here is the limit for this problem.
1
n

nn
L = lim 1+ 2 n
n →∞ 3

n

= lim

n →∞

3

1
+2
n

=

∞
= ∞ >1
32

So, by the Root Test this series is divergent.

⎛ 5n − 3n3 ⎞
∑ ⎜ 7 n3 + 2 ⎟
n=0 ⎝
⎠
∞

n

Solution
Again, there isn’t too much to this series.
1
n n

⎛ 5n − 3n ⎞
5n − 3n3
−3 3
L = lim ⎜ 3
=
= <1
⎟ = lim
n →∞
n →∞ 7 n 3 + 2
7
7
⎝ 7n + 2 ⎠
Therefore, by the Root Test this series converges absolutely and hence converges.
3

Note that we had to keep the absolute value bars on the fraction until we’d taken the limit
to get the sign correct.
∞

∑
n =3


( −12 )

n

n

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Calculus II

Solution
Here’s the limit for this series.

L = lim

( −12 )

n →∞

n

1
n n

= lim

12

n →∞

n

1
n

=

12
= 12 > 1
1

After using the fact from above we can see that the Root Test tells us that this series is
divergent.
Note that if we get L = 1 from the ratio test then the root test will also give also L = 1 and
so there isn’t any reason to try the root test on anything that gives L = 1 on the ratio test.

Strategy for Series
Now that we’ve got all of our tests out of the way it’s time to think about organizing all
of them into a general set of guidelines to help us determine the convergence of a series.
Note that these are a general set of guidelines and because some series can have more
than one test applied to them we will get a different result depending on the path that we
take through this set of guidelines. In fact, because more than one test may apply, you
should always go completely through the guidelines and identify all possible tests that
can be used on a given series. Once this has been done you can identity the test that you
feel will be the easiest for you to use.
With that said here is the set of guidelines for determining the convergence of a series.

1. With a quick glance does it look like the series terms don’t converge to zero in the
limit, i.e. does lim an ≠ 0 ? If so, use the Divergence Test. Note that you should
n →∞

only do the divergence test if a quick glance suggests that the series terms may
not converge to zero in the limit.
∞
∞
1
) or a geometric series ( ∑ ar n or ∑ ar n −1 )? If so
np
n=0
n =1
use the fact that p-series will only converge if p > 1 and a geometric series will
only converge if r < 1 . Remember as well that often some algebraic

2. Is the series a p-series ( ∑

manipulation is required to get a geometric series into the correct form.
3. Is the series similar to a p-series or a geometric series? If so, try the Comparison
Test.
4. Is the series a rational expression involving only polynomials or polynomials
under radicals (i.e. a fraction involving only polynomials or polynomials under


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Calculus II

radicals)? If so, try the Comparison Test and/or the Limit Comparison Test.
Remember however, that in order to use the Comparison Test and the Limit
Comparison Test the series terms all need to be positive.
5. Does the series contain factorials or constants raised to powers involving n? If so,
then the Ratio Test may work. Note that if the series term contains a factorial
then the only test that we’ve got that will work is the Ratio Test.
6. Can the series terms be written in the form an = ( −1) bn or an = ( −1)
n

n +1

bn ? If so,

then the Alternating Series Test may work.
7. Can the series terms be written in the form an = ( bn ) ? If so, then the Root Test
n

may work.
8. If a n = f ( n ) for some positive, decreasing function and

∫

∞

a

f ( x ) dx is easy to

evaluate then the Integral Test may work.

Again, remember that these are only a set of guidelines and not a set of hard and fast
rules to use when trying to determine the best test to use on a series. If more than one test
can be used try to use the test that will be the easiest for you to use and remember that
what is easy for someone else may not be easy for you!
Also just so we can put all the tests into one place here is a quick listing of all the test that
we’ve got.

Divergence Test
If lim an ≠ 0 then
n →∞

∑a

n

will diverge

Integral Test

Suppose that f(x) is a positive, decreasing function on the interval [ k , ∞ ) and that
f ( n ) = an then,

1. If
2. If

∫
∫

∞

k
∞

k

f ( x ) dx is convergent so is

∞

∑a

n

n=k

f ( x ) dx is divergent so is

Comparison Test

∞

∑a

.

and

∑b

n

n=k

∑a

n

.

n

with an , bn ≥ 0 for all n and an ≤ bn for

all n. Then,


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Calculus II

∑b
If ∑ a

1. If

n

2.

∑a .
is divergent then so is ∑ b .

is convergent then so is

n

n

n

Limit Comparison Test

∑a


n

and

∑b

n

with an , bn ≥ 0 for all n. Define,

an
bn
If c is positive (i.e. c > 0 ) and is finite (i.e. c < ∞ ) then either both series converge or
both series diverge.
c = lim

n →∞

Suppose that we have a series

∑a

and either an = ( −1) bn or an = ( −1)
n

n

n +1

bn where

bn ≥ 0 for all n. Then if,
1. lim bn = 0 and,
n →∞

2.

{bn } is eventually a decreasing sequence

the series

∑a

n

is convergent

Ratio Test
Suppose we have the series

∑a

n

. Define,
L = lim

n →∞

an +1
an

Then,

Root Test
Suppose that we have the series

∑a

n

. Define,

L = lim n an = lim an
n →∞

1
n

n →∞

Then,


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Calculus II

Estimating the Value of a Series
We have now spent quite a few sections determining the convergence of a series,
however, with the exception of geometric and telescoping series, we have not talked
about finding the value of a series. This is usually a very difficult thing to do and we still
aren’t going to talk about how to find the value of a series. What we will do is talk about
how to estimate the value of a series. Often that is all that you need to know.
Before we get into how to estimate the value of a series let’s remind ourselves how series
convergence works. It doesn’t make any sense to talk about the value of a series that
doesn’t converge and so we will be assuming that the series we’re working with
converges. Also, as well see the main method of estimating the value of series will come
out of this discussion.
∞

So, let’s start with the series ∑ an (the starting point is not important, but we need one to
n =1

do the work) and let’s suppose that it converges to s. Recall that this means that if we get
the partial sums,
n

sn = ∑ ai
i =1

then they will form a convergent sequence and its limit is s. In other words,
lim sn = s
n →∞

Now, just what does this mean for us? Well, since this limit converges it means that we
can make the partial sums, sn, as close to s as we want simply by taking n large enough.
In other words, if we take n large enough then we can say that,
sn ≈ s
This is one method of estimating the value of a series. We can just take a partial sum and
use that as an estimation of the value of the series. There are now two questions that we
should ask about this.
First, how good is the estimation? If we don’t have an idea of how good the estimation is
then it really doesn’t do all that much for us as an estimation.
Secondly, is there any way to make the estimate better? Sometimes we can use this as a
starting point and make the estimation better. We won’t always be able to do this, but if
we can that will be nice.
So, let’s start with a general discussion about the determining how good the estimation is.
Let’s first start with the full series and strip out the first n terms.


218


Calculus II

∞

∞

n

∑a = ∑a + ∑ a
i =1

i

i =1

i

i = n +1

(1)

i

Note that we converted over to an index of i in order to make the notation consistent with
prior notation. Recall that we can use any letter for the index and it won’t change the
value.
Now, notice that the first series (the n terms that we’ve stripped out) is nothing more than
the partial sum sn. The second series on the right (the one starting at i = n + 1 ) is called
the remainder and denoted by Rn. Finally let’s acknowledge that we also know the value
of the series since we are assuming it’s convergent. Taking this notation into account we
can rewrite (1) as,
s = sn + Rn
We can solve this for the remainder to get,
Rn = s − sn
So, the remainder tells us the difference, or error, between the exact value of the series
and the value of the partial sum that we are using as the estimation of the value of the
series.
Of course we can’t get our hands on the actual value of the remainder because we don’t
have the actual value of the series. However, we can use some of the tests that we’ve got
for convergence to get a pretty good estimate of the remainder provided we make some
assumptions about the series. Once we’ve got an estimate on the value of the remainder
we’ll also have an idea on just how good a job the partial sum does of estimating the
actual value of the series.
There are several tests that will allow us to get estimates of the remainder. We’ll go
through each one separately.

Integral Test
Recall that in this case we will need to assume that the series terms are all positive and
will eventually be decreasing. We originally derived the integral test by using the fact
that the series could be thought of as an estimation of the area under the curve of f(x)
where f ( n ) = an . We can do something similar with the remainder.
First, let’s recall that the remainder is,
Rn =

∞

∑a

i = n +1

i

= an +1 + an + 2 + an +3 + an + 4 +

Now, if we start at x=n+1, take rectangles of width 1 and use the left endpoint as the
height of the rectangle we can estimate the area under f(x) on the interval [ n + 1, ∞ ) as
shown in the sketch below.


219


Calculus II

We can see that the remainder, Rn, is exactly this area estimation and it will over estimate
the exact area. So, we have the following inequality.
Rn ≥ ∫

∞

n +1

f ( x ) dx

(2)

Next, we could also estimate the area by starting at x=n, taking rectangles of width 1
again and then using the right endpoint as the height of the rectangle. This will give an
estimation of the area under f(x) on the interval [ n, ∞ ) . This is shown in the following
sketch.

Again, we can see that the remainder, Rn, is again this estimation and in this case it will
underestimate the area. This leads to the following inequality,
∞

Rn ≤ ∫ f ( x ) dx
n

(3)

Combining (2) and (3) gives,


220


Calculus II

∫

∞

n +1

∞

f ( x ) dx ≤ Rn ≤ ∫ f ( x ) dx
n

So, provided we can do these integrals we can get both an upper and lower bound on the
remainder. This will in turn give us an upper bound and a lower bound on just how good
the partial sum, sn, is as an estimation of the actual value of the series.
In this case we can also use these results to get a better estimate for the actual value of the
series as well.
First, we’ll start with the fact that
s = sn + Rn
Now, if we use (2) we get,
s = sn + Rn ≥ sn + ∫

∞

n +1

f ( x ) dx

Likewise if we use (3) we get,
∞

s = sn + Rn ≤ sn + ∫ f ( x ) dx
n

Putting these two together gives us,

sn + ∫

∞

n +1

∞

f ( x ) dx ≤ s ≤ sn + ∫ f ( x ) dx
n

(4)

This gives an upper and a lower bound on the actual value of the series. We could then
use as an estimate of the actual value of the series the average of the upper and lower
bound.
Let’s work an example with this.
∞

Example 1 Using n=15 to estimate the value of

1

∑n
n =1

2

.

Solution
First, for comparison purposes, we’ll note that the actual value of this series is known to
be,
∞
1 π2
∑ n2 = 6 = 1.644934068
n =1

Using n=15 let’s first get the partial sum.
15
1
s15 = ∑ 2 = 1.580440283
i =1 i


221


Calculus II

Note that this is “close” to the actual value in some sense, but isn’t really all that close
either.
Now, let’s compute the integrals. These are fairly simple integrals so we’ll leave it to
you to verify the values.
∞
∞
⌠ 1 dx = 1
⌠ 1 dx = 1
⎮ 2
⎮ 2
⌡ 15 x
⌡16 x
15
16
Plugging these into (4) gives us,
1
1
≤ s ≤ 1.580440283 +
16
15
1.642940283 ≤ s ≤ 1.647106950

1.580440283 +

Both the upper and lower bound are now very close to the actual value and if we take the
average of the two we get the following estimate of the actual value.
s ≈ 1.6450236165

That is pretty darn close to the actual value.
So, that is how we can use the Integral Test to estimate the value of a series. Let’s move
on to the next test.

Comparison Test
In this case, unlike with the integral test, we may or may not be able to get an idea of how
good a particular partial sum will be as an estimate of the exact value of the series. Much
of this will depend on how the comparison test is used.
First, let’s remind ourselves on how the comparison test actually works. Given a series
∑ an let’s assume that we’ve used the comparison test to show that it’s convergent.
Therefore, we found a second series

∑b

n

that converged and an ≤ bn for all n.

What we want to do is determine how good of a job the partial sum,
n

sn = ∑ ai
i =1

will do in estimating the actual value of the series

∑a

n

. Again, we will use the

remainder to do this. Let’s actually write down the remainder for both series.
Rn =

∞

∑

i = n +1

Tn =

ai

∞

∑b

i = n +1

i

Now, since an ≤ bn we also know that
Rn ≤ Tn

222


Calculus II

When using the comparison test it is often the case that the bn are fairly nice terms and
that we might actually be able to get an idea on the size of Tn. For instance, if our second
series is a p-series we can use the results from above to get an upper bound on Tn as
follows,
Rn ≤ Tn ≤ ∫

∞

n +1

g ( x ) dx

where g ( n ) = bn

Also, if the second series is a geometric series then we will be able to compute Tn exactly.
If we are unable to get an idea of the size of Tn then using the comparison test to help
with estimates won’t do us much good.
∞


2n
∑ 4n + 1 .
n=0

Solution
To do this we’ll first need to go through the comparison test so we can get the second
series. So,

2n
2n ⎛ 1 ⎞
≤ n =⎜ ⎟
4n + 1 4 ⎝ 2 ⎠

n

and
∞

⎛1⎞
∑⎜ 2 ⎟
⎠
n=0 ⎝

n

is a geometric series and converges because r =

1
< 1.
2

Now that we’ve gotten our second series let’s get the estimate.
15
2n
s15 = ∑ n
= 1.383062486
n=0 4 + 1
So, how good is it? Well we know that,
∞

n

⎛1⎞
∑⎜2⎟
⎠
n =16 ⎝
will be an upper bound for the error between the actual value and the estimate. Since our
second series is a geometric series we can compute this directly as follows.
R15 ≤ T15 =

∞

n

n

n

15
∞
⎛1⎞
⎛1⎞
⎛1⎞
= ∑⎜ ⎟ + ∑ ⎜ ⎟
∑ ⎜ 2 ⎟ n=0 ⎝ 2 ⎠ n=16 ⎝ 2 ⎠
⎠
n=0 ⎝
The series on the left is in the standard form and so we can compute that directly. The
first series on the right has a finite number of terms and so can be computed exactly and


223


Calculus II

the second series on the right is the one that we’d like to have the value for. Doing the
work gives,
n
n
n
∞
∞
15
⎛1⎞
⎛1⎞
⎛1⎞
∑ ⎜ ⎟ = ∑⎜ 2 ⎟ − ∑⎜ 2 ⎟
⎠ n=0 ⎝ ⎠
n =16 ⎝ 2 ⎠
n =0 ⎝
1
=
− 1.999969482
1− ( 1 )
2

= 0.000030518
So, according to this if we use

s ≈ 1.383062486
as an estimate of the actual value we will be off from the exact value by no more than
0.000030518 and that’s not too bad.

In this case it can be shown that
∞

2n
∑ 4n + 1 = 1.383093004
n=0
and so we can see that the actual error in our estimation is,
Error = Actual − Estimate = 1.383093004 − 1.383062486 = 0.000030518

Note that in this case the estimate of the error is actually fairly close (and in fact exactly
the same) as the actual error. This will not always happen and so we shouldn’t expect
that to happen in all cases. The error estimate above is simply the upper bound on the
error and the actual error will often be less than this value.
Before moving on to the final part of this section let’s again note that we will only be able
to determine how good the estimate is using the comparison test if we can easily get our
hands on the remainder of the second term. The reality is that we won’t always be able to
do this.

Both of the methods that we’ve looked at so far have required the series to contain only
positive terms. If we allow series to have negative terms in it the process is usually more
difficult. However, with that said there is one case where it isn’t too bad. That is the
case of an alternating series.
Once again we will start of with a convergent series

∑ a = ∑ ( −1)
n

n

bn which in this

case happens to be an alternating series, so we know that bn ≥ 0 for all n. Also note that
we could have any power on the “-1” we just used n for the sake of convenience. We
want to know how good of an estimation of the actual value will the partial sum sn be. As
with the prior cases we know that the remainder, Rn, will be the error in the estimation
and so if we can get a handle on that we’ll know approximately how good the estimation
is.


224


Calculus II

In this case the work is a little harder to justify so we’ll just give the relevant fact. It can
be shown that,
Rn = s − sn ≤ bn +1

We needed absolute value bars because we won’t know ahead of time if the estimation is
larger or smaller than the actual value and we know that the bn’s are positive.
∞


∑
n =1

( −1)
n2

n

.

Solution
This is an alternating series and is does converge. In this case the exact value is known
and so for comparison purposes,
∞

∑
n =1

( −1)

n

n2

=−

π2
12

= −0.8224670336

Now, the estimation is,
15

s15 = ∑
n =1

( −1)
n2

n

= −0.8245417574

From the fact above we know that
R15 = s − s15 ≤ b16 =

1
= 0.00390625
162

So, our estimation will have an error of no more than 0.00390625. In this case the exact
value is known and so the actual error is,
R15 = s − s15 = 0.0020747238
In the pervious example the estimation had only half the estimated error. It will often be
the case the actual error will be less than the estimated error. Remember that this is only
an upper bound for the actual error.

Ratio Test
This will be the final case that we’re going to look at for estimating series values and we
are going to have to put a couple of fairly stringent restriction on the series terms in order
to do the work. One of the main restrictions we’re going to make is to assume that the
series terms are positive. Well also be adding on another restriction in a bit.
In this case we’ve used the ratio test to show that

∑a

n

is convergent. To do this we

computed


225


Calculus II

L = lim

n →∞

an +1
an

and found that L < 1 .
As with the previous cases we are going to use the remainder, Rn, to determine how good
of an estimation of the actual value the partial sum, sn, is.

To get an estimate of the remainder let’s first define the following sequence,
a
rn = n +1
an
We now have two possible cases.
1. If {rn } is a decreasing sequence and rn +1 < 1 then,
Rn ≤

an +1
1 − rn +1

2. If {rn } is a increasing sequence then,
Rn ≤

an +1
1− L

Note that there are some restrictions on the sequence {rn } and at least one of its terms in
order to use these formulas. If the restrictions aren’t met then the formulas can’t be used.
∞


n

∑3
n=0

n

.

Solution
First, let’s use the ratio test to verify that this is a convergent series.
n + 1 3n
n +1 1
L = lim n +1
= lim
= <1
n →∞ 3
n →∞ 3n
n
3
So, it is convergent. Now let’s get the estimate.
15
n
s15 = ∑ n 0.7499994250
n=0 3

To determine an estimate on the remainder, and hence the error, let’s first get the
sequence {rn } .


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Calculus II

rn =

n + 1 3n n + 1 1 ⎛ 1 ⎞
=
= ⎜1 + ⎟
3n +1 n
3n
3⎝ n ⎠

The last rewrite was just to simplify some of the computations a little. Now, notice that,
1⎛ 1⎞
1
f ( x ) = ⎜1 + ⎟
f ′( x) = − 2 < 0
3⎝ x ⎠
3x
Since this function is always decreasing and f ( n ) = rn and so, this sequence is

1
decreasing. Also note that r16 = 1 (1 + 16 ) < 1 . Therefore we can use the first case from
3

the fact above to get,
R15 ≤

16
a16
316
=
= 0.0000005755187
1
1 − r16 1 − 1 (1 + 16 )
3

So, it looks like our estimate is probably quite good. In this case the exact value is
known.
∞
n 3
∑ 3n = 4
n=0
and so we can compute the actual error.
R15 = s − s15 = 0.000000575
This is less than the upper bound, but unlike in the previous example this actual error is
quite close to the upper bound.
In the last two examples we’ve seen that the upper bound computations on the error can
sometimes be quite close to the actual error and at other times they can be off by quite a
bit. There is usually no way of knowing ahead of time which it will be and without the
exact value in hand there will never be a way of determining which it will be.
Notice that this method did require the series terms to be positive, but that doesn’t mean
that we can’t deal with ratio test series if they have negative terms. Often series that we
used ratio test on are also alternating series and so if that is the case we can always resort
to the previous material to get an upper bound on the error in the estimation, even if we
didn’t use the alternating series test to show convergence.
Note however that if the series does have negative terms, but doesn’t happen to be an
alternating series then we can’t use any of the methods discussed in this section to get an
upper bound on the error.

Power Series
We’ve spent quite a bit of time talking about series now and with only a couple of
exceptions we’ve spent most of that time talking about how to determine if a series will

227


Calculus II

converge or not. It’s now time to start looking at some specific kinds of series and we’ll
eventually reach the point where we can talk about applications of series.
In this section we are going to start talking about power series. A power series about a,
or just power series, is any series that can be written in the form,
∞

∑ c ( x − a)
n=0

n

n

where a and cn are numbers. The cn’s are often called the coefficients of the series. The
first thing to notice about a power series is that it is a function of x. That is different from
any other kind of series that we’ve looked at to this point. In all the prior sections we’ve
only allowed numbers in the series and now we are allowing variables to be in the series
as well. This will not change how things work however. Everything that we know about
series still holds.
In the discussion of power series convergence is still a major question that we’ll be
dealing with. The difference is that the convergence of the series will now depend upon
the value of x that we put into the series. A power series may converge for some values
of x and not for other values of x.
Before we get too far into power series there is some terminology that we need to get out
of the way.
First, as we will see in our examples, we will be able to show that there is a number R so
that the power series will converge for, x − a < R and will diverge for x − a > R . This
number is called the radius of convergence for the series. Note that the series may or
may not converge if x − a = R . What happens at these points will not change the radius
of convergence.
Secondly, the interval of all x’s, including the end points if need be, for which the power
series converges is called the interval of convergence of the series.
These two concepts are fairly closely tied together. If we know that the radius of
convergence of a power series is R then we have the following.
a−R< x<a+R
x < a − R and x > a + R

power series converges
power series diverges

The interval of convergence must then contain the interval a − R < x < a + R since we
know that the power series will converge for these values. We also know that the interval
of convergence can’t contain x’s in the ranges x < a − R and x > a + R since we know
the power series diverges for these value of x. Therefore, to completely identify the
interval of convergence all that we have to do is determine if the power series will
converge for x = a − R or x = a + R . If the power series converges for one or both of
these values then we’ll need to include those in the interval of validity.


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Calculus II

Before getting into some examples let’s take a quick look at the convergence of a power
series for the case of x = a . In this case the power series becomes,
∞

∞

∑ c ( a − a ) = ∑ c ( 0)
n=0

n

n

n=0

n

n

∞

∞

= c0 ( 0 ) + ∑ cn ( 0 ) = c0 + ∑ 0 = c0 + 0 = c0
0

n

n =1

n =1

and so the power series converges. Note that we had to strip out the first term since it
was the only non-zero term in the series.
It is important to note that no matter what else is happening in the power series we are
guaranteed to get convergence for x = a . The series may not converge for any other
value of x, but it will always converge for x = a .
Let’s work some examples. We’ll put quite a bit of detail into the first example and then
not put quite as much detail in the remaining examples.
Example 1 Determine the radius of convergence and interval of convergence for the
following power series.
∞

∑
n =1

( −1)
4

n

n

n

( x + 3)

n

Solution
Okay, we know that this power series will converge for x = −3 , but that’s it at this point.
To determine the remainder of the x’s for which we’ll get convergence we can use any of
the tests that we’ve discussed to this point. After application of the test that we choose to
work with we will arrive at condition(s) on x that we can use to determine which values
of x for which the power series will converge and which values of x for which the power
series will diverge. From this we can get the radius of convergence and most of the
interval of convergence (with the possible exception of the endpoints.

With all that said, the best tests to use here are almost always the ratio or root test. Most
of the power series that we’ll be looking at are set up for one or the other. In this case
we’ll use the ratio test.

( −1) ( n + 1)( x + 3)
L = lim
n +1

n →∞

= lim

n →∞

4n +1

n +1

4n

( −1) ( n )( x + 3)
n

n

− ( n + 1)( x + 3)
4n

Before going any farther with the limit let’s notice that since x is not dependent on the
limit and so it can be factored out of the limit. Notice as well that in doing this well need
to keep the absolute value bars on it since we need to make sure everything stays positive
and x could well be a value that will make things negative. The limit is then,


229


Calculus II

L = x + 3 lim

n →∞

n +1
4n

1
x+3
4
So, the ratio test tells us that if L < 1 the series will converge, if L > 1 the series will
diverge, and if L = 1 we don’t know what will happen. So, we have,
1
x +3 <1
x+3 < 4
series converges
⇒
4
1
x +3 >1
x+3 > 4
series diverges
⇒
4

=

We’ll deal with the L = 1 case in a bit. Notice that we now have the radius of
convergence for this power series. These are exactly the conditions required for the
radius of convergence. The radius of convergence for this power series is R = 4 .
Now, let’s get the interval of convergence. We’ll get most (if not all) of the interval by
solving the first inequality from above.
−4 < x + 3 < 4
−7 < x < 1
So, most of the interval of validity is given by −7 < x < 1 . All we need to do is determine
if the power series will converge or diverge at the endpoints of this interval. Note that
these values of x will correspond to the value of x that will give L = 1 .
The way to determine convergence at these points is to simply plug them into the original
power series and see if the series converges or diverges using any test necessary.
x = −7 :
In this case the series is,
n
n
∞
( −1) n −4 n = ∞ ( −1) n −1 n 4n
∑ 4n ( ) ∑ 4n ( )
n =1
n =1
∞

= ∑ ( −1) ( −1) n
n

( −1) ( −1)

n

n

n

= ( −1) = 1
2n

n =1
∞

= ∑n
n =1

This series is divergent by the Divergence Test since lim n = ∞ ≠ 0 .
n →∞

x =1:
In this case the series is,
∞

∑
n =1


( −1)
4

n

n

n

( 4)

n

∞

= ∑ ( −1) n
n

n =1

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Calculus II

This series is also divergent by the Divergence Test since lim ( −1) n doesn’t exist.
n

n →∞

So, in this case the power series will not converge for either endpoint. The interval of
convergence is then,
−7 < x < 1
In the previous example the power series didn’t converge for either end point of the
interval. Sometimes that will happen, but don’t always expect that to happen. The power
series could converge at either both of the end points or only one of the end points.
∞
2n
n
∑ n ( 4 x − 8)
n =1
Solution
Let’s jump right into the ratio test.
L = lim

2n +1 ( 4 x − 8 )
n +1

n →∞

= lim

n →∞

n +1

n
2 ( 4 x − 8)
n

n

2n ( 4 x − 8 )
n +1

2n
n →∞ n + 1
= 2 4x − 8
= 4 x − 8 lim

So we will get the following convergence/divergence information from this.
2 4x − 8 < 1
series converges
2 4x − 8 > 1

series diverges

We need to be careful here in determining the interval of convergence. The interval of
convergence requires x − a < R and x − a > R . In other words, we need to factor a 4 out
of the absolute value bars in order to get the correct radius of convergence. Doing this
gives,
1
x−2 <
8 x − 2 <1
series converges
⇒
8
1
x−2 >
8 x − 2 >1
series diverges
⇒
8
1
So, the radius of convergence for this power series is R = .
8


231


Calculus II

Now, let’s find the interval of convergence. Again, we’ll first solve the inequality that
gives convergence above.
1
1
− < x−2<
8
8
15
17
<x<
8
8
Now check the end points.
15
:
8
The series here is,
x=

∞

n

∞
2n ⎛ 15 ⎞
2n ⎛ 1 ⎞
− 8⎟ = ∑ ⎜ − ⎟
∑ n ⎜ 2 ⎠ n=1 n ⎝ 2 ⎠
⎝
n =1

2n ( −1)
=∑
2n
n =1 n
∞

∞

=∑

( −1)

n =1

n

n

n

n

This is the alternating harmonic series and we know that it converges.
17
:
8
The series here is,
x=

∞

n

∞
2n ⎛ 17
2n ⎛ 1 ⎞
⎞
− 8⎟ = ∑ ⎜ ⎟
∑ n ⎜ 2 ⎠ n=1 n ⎝ 2 ⎠
⎝
n =1

n

∞

2n 1
n
n =1 n 2

=∑
∞

1
n =1 n

=∑

This is the harmonic series and we know that it diverges.
So, the power series converges for one of the end points, but not the other. This will
often happen so don’t get excited about it when it does. The interval of convergence for
this power series is then,
15
17
≤x<
8
8
We now need to take a look at a couple of special cases with radius and intervals of
convergence.


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Calculus II

∞

∑ n!( 2 x + 1)

n

n=0

Solution
Well start this example with the ratio test as we have for the previous ones.

( n + 1)!( 2 x + 1)
L = lim
n
n →∞
n !( 2 x + 1)
( n + 1) n ! ( 2 x + 1)
= lim

n +1

n →∞

n!

= 2 x + 1 lim ( n + 1)
n →∞

At this point we need to be careful. The limit is infinite, but there is that term with the x’s
1
in front of the limit. We’ll have L = ∞ > 1 provided x ≠ − .
2
1
So, this power series will only converge if x = − . If you think about it we actually
2
already knew that however. From our initial discussion we know that every power series
1
n
will converge for x = a and in this case a = − . Remember that we get a from ( x − a ) ,
2
and notice the coefficient of the x must be a one!.
In this case we say the radius of convergence is R = 0 and the interval of convergence is
1
x = − , and yes we really did mean interval of convergence even though it’s only a
2
point.
∞

( x − 6)

n =1

nn

∑

n

Solution
In this example the root test seems more appropriate. So,


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Calculus II

L = lim

( x − 6)

n →∞

nn

= lim

1
n n

x−6
n

n →∞

1
n →∞ n

= x − 6 lim
=0

So, since L = 0 < 1 regardless of the value of x this power series will converge for every
x.
In these cases we say that the radius of convergence is R = ∞ and interval of
convergence is −∞ < x < ∞ .
So, let’s summarize the last two examples. If the power series only converges for x = a
then the radius of convergence is R = 0 and the interval of convergence is x = a .
Likewise if the power series converges for every x the radius of convergence is R = ∞
and interval of convergence is −∞ < x < ∞ .
Let’s work one more example.
∞
x2n

∑
n =1

( −3)

n

Solution
First notice that a=0 in this problem. That’s not really important to the problem, but it’s
worth pointing out so people don’t get excited about it.

The important difference in this problem is the exponent on the x. In this case it is 2n
rather than the standard n. As we will see some power series will have exponents other
than an n and so we still need to be able to deal with these kinds of problems.
This one seems set up for the root test again so let’s use that.
L = lim

n →∞

= lim

n →∞

x

1
n

2n

( −3)

n

x2
−3

x2
=
3


234


Calculus II

So, we will get convergence if
x2
<1
⇒
x2 < 3
3
The radius of convergence is NOT 3 however. The radius of convergence requires an
exponent of 1 on the x. Therefore,
x2 < 3
x< 3

Be careful with the absolute value bars! In this case it looks like the radius of
convergence is R = 3 . Notice that we didn’t bother to put down the inequality for
divergence this time. The inequality for divergence is just the interval for convergence
that the test gives with the inequality switched and generally isn’t needed. We will
usually skip that part.
Now let’s get the interval of convergence. First from the inequality we get,
− 3<x< 3
Now check the end points.
x=− 3:
Here the power series is,
∞

(− 3)

n =1

( −3)

∑

2n

∞

=∑

((

− 3

))

2 n

( −3)
n
∞
( 3)
=∑
n
n
n =1 ( −1) ( 3 )

n

n

n =1

∞

= ∑ ( −1)

n

n =1

This series is divergent by the Divergence Test since lim ( −1) doesn’t exist.
n

n→∞

x = 3:
Because we’re squaring the x this series will be the same as the previous step.
∞

( 3)

n =1

( −3)

∑

2n
∞

n

= ∑ ( −1)

n

n =1

which is divergent.
The interval of convergence is then,

− 3<x< 3


235


Calculus II

Power Series and Functions
We opened the last section by saying that we were going to start thinking about
applications of series and then promptly spent the section talking about convergence
again. It’s now time to actually start with the applications of series.
With this section we will start talking about how to represent functions with power series.
The natural question of why we might want to do this will be answered in a couple of
sections once we actually learn how to do this.
Let’s start off with one that we already know how to do, although when we first ran
across this series we didn’t think of it as a power series nor did we acknowledge that it
represented a function.
Recall that the geometric series is
∞

a
provided r < 1
1− r
n=0
Don’t forget as well that if r ≥ 1 the series diverges.

∑ ar

n

=

Now, if we take a=1 and r=x this becomes,
∞
1
∑ xn = 1 − x
n=0

provided x < 1

Turning this around we can see that we can represent the function
1
f ( x) =
1− x
with the power series
∞

∑x

n

provided x < 1

(1)

(2)

(3)

n=0

This provision is important. We can clearly plug any number other than x=1 into the
function, however, we will only get a convergent power series if x < 1 . This means the
equality in (1) will only hold if x < 1 . For any other value of x the equality won’t hold.
Note as well that we can also use this to acknowledge that the radius of convergence of
this power series is R = 1 and the interval of convergence is x < 1 .
This idea of convergence is important here. We will be representing many functions as
power series and it will be important to recognize that the representations will often only
be valid for a range of x’s and that there may be values of x that we can plug into the
function that we can’t plug into the power series representation.


236


Calculus II

In this section we are going to concentrate on representing functions with power series
where the functions can be related back to (2).
In this way we will hopefully become familiar with some of the kinds of manipulations
that we will sometimes need to do when working with power series.
So, let’s jump into a couple of examples.
Example 1 Find a power series representation for the following function and determine
it’s interval of convergence.
1
g ( x) =
1 + x3
Solution
What we need to do here is to relate this function back to (2). This is actually easier than
it might look. Recall that the x in (2) is simply a variable and can represent anything. So,
a quick rewrite of g(x) gives,
1
g ( x) =
1 − ( − x3 )

and so the − x3 in g(x) holds the same place as the x in (2). Therefore, all we need to do
is replace the x in (3) and we’ve got a power series representation for g(x).
∞

g ( x ) = ∑ ( − x3 )

n

provided − x3 < 1

n=0

Notice that we replaced both the x in the power series and in the interval of convergence.
All we need to do now is a little simplification.
∞

g ( x ) = ∑ ( −1) x 3n
n

provided x < 1
3

⇒

x <1

n=0

So, in this case the interval of convergence is the same as the original power series. This
usually won’t happen. More often than not the new interval of convergence will be
different form the original interval of convergence.
2 x2
h ( x) =
1 + x3
Solution
This function is similar to the previous function. The difference is the numerator and at
first glance that looks to be an important difference. Since (2) doesn’t have an x in the
numerator it appears that we can’t relate this function back to that.

However, now that we’ve worked the first example this one is actually very simple since
we can use the result of the answer from that example. To see how to do this let’s first
rewrite the function a little.


237


Calculus II

1
1 + x3
Now, from the first example we’ve already got a power series for the second term so let’s
use that to write the function as,
h ( x ) = 2 x2

∞

h ( x ) = 2 x 2 ∑ ( −1) x 3n

provided x < 1

n

n=0

Notice that the presence of x’s outside of the series will NOT affect its convergence and
so the interval of convergence remains the same.
The last step is to bring the coefficient into the series and we’ll be done. When we do
this make sure and combine the x’s as well. We typically only want a single x in a power
series.
∞

h ( x ) = ∑ 2 ( −1) x 3n + 2
n

provided x < 1

n =0

As we saw in the previous example we can often use previous results to help us out. This
is an important idea to remember as it can often greatly simplify our work.
x
f ( x) =
5− x
Solution
So, again, we’ve got an x in the numerator. So, as with the last example let’s factor that
out and see what we’ve got left.
1
f ( x) = x
5− x
If we had a power series representation for
1
g ( x) =
5− x
we could get a power series representation for f(x).

So, let’s find one. We’ll first notice that in order to use (4) we’ll need the number in the
denominator to be a one. That’s easy enough to get.
1 1
g ( x) =
5 1− x
5
Now all we need to do to get a power series representation is to replace the x in (3) with
x
. Doing this gives,
5
n
1 ∞ ⎛x⎞
x
<1
g ( x) = ∑ ⎜ ⎟
provided
5 n =0 ⎝ 5 ⎠
5


238


Calculus II

Now let’s do a little simplification on the series.
1 ∞ xn
g ( x) = ∑ n
5 n=0 5
∞

xn
n +1
n =0 5

=∑

The interval of convergence for this series is,
x
1
x <1
<1
⇒
5
5

⇒

x <5

Okay, this was the work for the power series representation for g(x) let’s now find a
power series representation for the original function. All we need to do for this is to
multiply the power series representation for g(x) by x and we’ll have it.
1
f ( x) = x
5− x
∞
xn
= x ∑ n +1
n=0 5

x n +1
n +1
n=0 5
∞

=∑

The interval of convergence doesn’t change and so it will be x < 5 .
So, hopefully we now have an idea on how to find the power series representation for
some functions. Admittedly all of the functions could be related back to (2) but it’s a
start.
We now need to look at some further manipulation of power series that we will need to
do on occasion. We need to discuss differentiation and integration of power series.
Let’s start with differentiation of the power series,
∞

f ( x ) = ∑ cn ( x − a ) = c0 + c1 ( x − a ) + c2 ( x − a ) + c3 ( x − a ) +
n

2

3

n=0

Now, we know that if we differentiate a finite sum of terms all we need to do is
differentiate each of the terms and then add them back up. With infinite sums there are
some subtleties involved that we need to be careful with, but are somewhat beyond the
scope of this course. While, we can always just differentiate all the terms in an infinite
series is it not always guaranteed to be the power series representation of the derivative of
the original function.
Nicely enough for us however, it is known that if the power series representation of f(x)
has a radius of convergence of R > 0 then the term by term differentiation of the power


239


Calculus II

series will also have a radius of convergence of R and (more importantly) will in fact be
the power series representation of f ′ ( x ) provided x is in the interval of convergence of
the original function.
In other words,
∞

= ∑ ncn ( x − a )

f ′ ( x ) = c1 + 2c2 ( x − a ) + 3c3 ( x − a ) +
2

n −1

n =1

Note the initial value of this series. It has been changed from n=0 to n=1. This is an
acknowledgement of the fact that the derivative of the first term is zero and hence isn’t in
the derivative. Notice however, that since the n=0 term of the above series is also zero,
we could start the series at n=0 if it was required for a particular problem. In general
however, this won’t be done in this class.
We can now find formulas for higher order derivatives as well now.
∞

f ′′ ( x ) = ∑ n ( n − 1) cn ( x − a )

n−2

n=2
∞

f ′′′ ( x ) = ∑ n ( n − 1)( n − 2 ) cn ( x − a )

n −3

n =3

etc.
Once again, notice that the initial value of n changes with each differentiation in order to
acknowledge that a term from the original series differentiated to zero.
Let’s now briefly talk about integration. Just as with the differentiation, when we’ve got
an infinite series we need to be careful about just integration term by term. As long as we
are in the interval of convergence for the original function we can do the integration. In
this case we get,
∞

∫ f ( x ) dx = C + ∑ c
n=0

n

( x − a)

n +1

n +1

Notice that we pick up a constant of integration, C, that is outside the series here.
Let’s summarize the differentiation and integration ideas before moving on to an example
or two.

Fact
∞

If f ( x ) = ∑ cn ( x − a ) has a radius of convergence of R > 0 then,
n

n =0


240


Calculus II

∞

f ′ ( x ) = ∑ ncn ( x − a )

n −1

n =1

∞

∫ f ( x ) dx = C + ∑ c

( x − a)

n +1

n +1
and both of these also have a radius of convergence of R.
n=0

n

Now, let’s see how we can use these facts to generate some more power series
representations of functions.
1
g ( x) =
2
(1 − x )
Solution
To do this problem let’s notice that
1

(1 − x )

2

=

d ⎛ 1 ⎞
⎜
⎟
dx ⎝ 1 − x ⎠

Then since we’ve got a power series representation for
1
1− x
all that we’ll need to do is differentiate that power series to get a power series
representation for g(x).
1
g ( x) =
2
(1 − x )
=

d ⎛ 1 ⎞
⎜
⎟
dx ⎝ 1 − x ⎠

=

d ⎛ ∞ n⎞
∑x ⎠
dx ⎜ n =0 ⎟
⎝
∞

= ∑ nx n −1
n =1

Then since the original power series had a radius of convergence of R = 1 the derivative,
and hence g(x), will also have a radius of convergence of R = 1 .
h ( x ) = ln ( 5 − x )
Solution
In this case we need to notice that


241


Calculus II

⌠ 1 dx = − ln ( 5 − x )
⎮
⌡ 5− x
and the recall that we have a power series representation for
1
5− x
Remember we found a representation for this in Example 3. So,
1
ln ( 5 − x ) = −⌠
dx
⎮
⌡ 5− x
n
⌠ ∞ x
= −⎮ ∑ n +1 dx
⌡ n=0 5
x n +1
=C −∑
n +1
n = 0 ( n + 1) 5
∞

We can find the constant of integration, C, by plugging in a value of x. A good choice is
x=0 since that will make the series easy to evaluate.
∞
0n +1
ln ( 5 − 0 ) = C − ∑
n +1
n = 0 ( n + 1) 5
ln ( 5 ) = C

So, the final answer is,
x n +1
n +1
n = 0 ( n + 1) 5
∞

ln ( 5 − x ) = ln ( 5 ) − ∑

Note that it is okay to have the constant sitting outside of the series like this. In fact,
there is no way to bring it into the series so don’t get excited about it.

Taylor Series
In the previous section we started looking at writing down a power series representation
of a function. The problem with the approach in that section is that everything came
down to needing to be able to relate the function in some way to
1
1− x
and while there are many functions out there that can be related to this function there are
many more that simply can’t be related to this.
So, without taking anything away from the process we looked at in the previous section,
we need to do is come up with a more general method for writing a power series
representation for a function.


242


Calculus II

So, for the time being, let’s make two assumptions. First, lets assume that the function
f(x) does in fact have a power series representation about x=a,
∞

f ( x ) = ∑ cn ( x − a ) = c0 + c1 ( x − a ) + c2 ( x − a ) + c3 ( x − a ) + c4 ( x − a ) +
2

n

3

4

n=0

Next, we will need to assume that the function, f(x), has derivatives of every order and
that we can in fact find them all.
Now that we’ve assumed that a power series representation exists we need to determine
what the coefficients, cn, are. This is easier than it might at first appear to be. Let’s first
just evaluate everything at x=a. This gives,
f ( a ) = c0

So, all the terms except the first are zero and we now know what c0 is. Unfortunately,
there isn’t any other value of x that we can plug into the function that will allow us to
quickly find any of the other coefficients. However, if we take the derivative of the
function (and its power series) then plug in x=a we get,
2
3
f ′ ( x ) = c1 + 2c2 ( x − a ) + 3c3 ( x − a ) + 4c4 ( x − a ) +
f ′ ( a ) = c1
and we now know c1.
Lets’ continue with this idea and find the second derivative.
2
f ′′ ( x ) = 2c2 + 3 ( 2 ) c3 ( x − a ) + 4 ( 3) c4 ( x − a ) +
f ′′ ( a ) = 2c2
So, it looks like,
c2 =

f ′′ ( a )
2

Using the third derivative gives,
f ′′′ ( x ) = 3 ( 2 ) c3 + 4 ( 3)( 2 ) c4 ( x − a ) +
f ′′′ ( a ) = 3 ( 2 ) c3

⇒

c3 =

f ′′′ ( a )
3 ( 2)

Using the fourth derivative gives,
f ( 4) ( x ) = 4 ( 3)( 2 ) c4 + 5 ( 4 )( 3)( 2 ) c5 ( x − a )
f

( 4)

( a ) = 4 ( 3)( 2 ) c4


⇒

243

f ( 4) ( a )
c4 =
4 ( 3)( 2 )


Calculus II

Hopefully by this time you’ve seen the pattern here. It looks like, in general, we’ve got
the following formula for the coefficients.
n
f ( ) (a)
cn =
n!
This even works for n=0 if you recall that 0! = 1 and define f ( 0) ( x ) = f ( x ) .
So, provided a power series representation for the function f(x) about x=a exists it will
have the form,
∞

f ( x) = ∑
n=0

f ( n) ( a )
n
( x − a)
n!

= f ( a ) + f ′ ( a )( x − a ) +

f ′′ ( a )
f ′′′ ( a )
2
3
( x − a) +
( x − a) +
2!
3!

This is called the Taylor Series for f(x) about x=a. In the case that a=0, so the Taylor
Series about x=0, we have,
∞

f ( x) = ∑
n=0

f ( n) ( 0 ) n
x
n!

= f ( 0) + f ′ ( 0) x +

f ′′ ( 0 ) 2 f ′′′ ( 0 ) 3
x +
x +
2!
3!

This is called a Maclaurin Series for f(x).
Before working any examples of Taylor Series we first need to address the assumption
that a Taylor Series will in fact exist for a given function. Let’s first get some notation
out of the way first.
We’ll first split the Taylor Series formula as follows,
∞
∞
n
f (n) ( a )
f (i ) ( a )
f (i ) ( a )
n
i
i
x − a) = ∑
x − a) + ∑
(
( x − a)
∑ n! (
i!
i!
n=0
i =0
i = n +1
= Tn ( x ) + Rn ( x )

where,

f (i ) ( a )
i
( x − a)
i!
i =0
is called the nth degree Taylor Polynomial of f(x) and
∞
f (i ) ( a )
i
Rn ( x ) = ∑
( x − a)
i!
i = n +1
is called the remainder.
n

Tn ( x ) = ∑


244


Calculus II

Note that the Taylor polynomial really is a polynomial and its degree will be at most n.
We’ll see a nice application of Taylor polynomials in the next section.
We now have the following Theorem.

Theorem
Suppose that f ( x ) = Tn ( x ) + Rn ( x ) . Then if,
lim Rn ( x ) = 0
n →∞

for x − a < R then,
∞

f ( x) = ∑
n=0

f ( n) ( a )
n
( x − a)
n!

on x − a < R .
In general showing that lim Rn ( x ) = 0 is a somewhat difficult process and so we will be
n →∞

assuming that this can be done for some R in all of the examples that we’ll be looking at.
Now let’s look at some examples.
Example 1 Find the Taylor Series for f ( x ) = e x about x=0.
Solution
This is actually one of the easier Taylor Series that we’ll be asked to compute. To find
the Taylor Series for a function we will need to determine a general formula for f ( n ) ( a ) .

This is one of the few functions where this is easy to do right from the start.
To get a formula for f ( n ) ( 0 ) all we need to do is recognize that,
f (n) ( x ) = e x
and so,

n = 0,1, 2,3,…

f ( n ) ( 0 ) = e0 = 1

n = 0,1, 2,3,…

Therefore, the Taylor series for f ( x ) = e x about x=0 is,
∞

1 n ∞ xn
x =∑
n=0 n !
n=0 n !

ex = ∑

Example 2 Find the Taylor Series for f ( x ) = e − x about x=0.
Solution
There are two ways to do this problem. Both are fairly simple, however one of them
requires significantly less work. We’ll work both solutions since the longer one has some
nice ideas that we’ll see in other examples.


245


Calculus II

Solution 1
As with the first example we’ll need to get a formula for f ( n ) ( 0 ) . However, unlike the
first one we’ve got a little more work to do. Let’s first take some derivatives and
evaluate them at x=0.
f ( 0) ( x ) = e− x
f ( 0) ( 0 ) = 1
f (1) ( x ) = −e− x

f (1) ( 0 ) = −1

f ( 2) ( x ) = e− x

f ( 2) ( 0 ) = 1

f ( 3 ) ( x ) = −e − x

f (3) ( 0 ) = −1

f ( n ) ( x ) = ( −1) e− x

f ( n ) ( 0 ) = ( −1)

n

n = 0,1, 2,3

n

After a couple of computations we were able to get general formulas for both f ( n ) ( x )
and f ( n ) ( 0 ) . We often won’t be able to get a general formula for f ( n ) ( x ) so don’t get
too excited about getting that formula. Also, as we will see it won’t always be easy to get
a general formula for f ( n ) ( a ) .
So, in this case we’ve got general formulas so all we need to do is plug these into the
Taylor Series formula and be done with the problem.
e

−x

∞

=∑

( −1)

n

xn

n!

n=0

Solution 2
The previous solution wasn’t too bad and we often have to do things in that manner.
However, in this case there is a much shorter solution method. In the previous section we
used series that we’ve already found to help us find a new series. Let’s do the same thing
with this one. We already know a Taylor Series for e x about x=0 and in this case the
only difference is we’ve got a “-x” in the exponent instead of just an x.
So, all we need to do is replace the x in the Taylor Series that we found in the first
example with “-x”.
e

−x

∞

=∑
n=0

(−x)
n!

n

∞

=∑

( −1)

n

xn

n!

n=0

This is a much shorter method of arriving at the same answer so don’t forget about using
previously computed series where possible (and allowed of course).
Example 3 Find the Taylor Series for f ( x ) = x 4e −3 x about x=0.
2


246


Calculus II

Solution
For this example we will take advantage of the fact that we already have a Taylor Series
for e x about x=0. In this example, unlike the previous example, doing this directly
would be significantly longer and more difficult.
4 −3 x 2

xe

∞

= x4 ∑

( −3x )

2 n

n!

n=0

=x

∞

4

∑

( −3 )

∞

x2n

n!

n=0

=∑

n

( −3 )

x 2n+4
n!

n=0

n

To this point we’ve only looked at Taylor Series about x=0 (also known as Maclaurin
Series) so let’s take a look at a Taylor Series that isn’t about x=0. Also, we’ll pick on the
exponential function one more time since it makes some of the work easier. This will be
the final Taylor Series for exponentials in this section.
Example 4 Find the Taylor Series for f ( x ) = e − x about x = −4 .
Solution
Finding a general formula for f ( n ) ( −4 ) is fairly simple.

f ( n ) ( x ) = ( −1) e − x

f ( n ) ( −4 ) = ( −1) e 4

n

n

The Taylor Series is then,
e

−x

∞

=∑
n =0

( −1)

n

n!

e4

( x + 4)

n

Okay, we now need to work some examples that don’t involve the exponential function
since these will tend to require a little more work.
Example 5 Find the Taylor Series for f ( x ) = cos ( x ) about x=0.
Solution
First we’ll need to take some derivatives of the function and evaluate them at x=0.


247


Calculus II

( x ) = cos x
1
f ( ) ( x ) = − sin x
2
f ( ) ( x ) = − cos x
3
f ( ) ( x ) = sin x
4
f ( ) ( x ) = cos x
5
f ( ) ( x ) = − sin x
6
f ( ) ( x ) = − cos x
f(

(0) = 1
1
f ( ) ( 0) = 0
2
f ( ) ( 0 ) = −1
3
f ( ) ( 0) = 0
4
f ( ) ( 0) = 1
5
f ( ) ( 0) = 0
6
f ( ) ( 0 ) = −1

0)

f(

0)

In this example, unlike the previous ones, there is not an easy formula for either then
general derivative or the evaluation of the derivative. However, there is a clear pattern to
the evaluations. So, let’s plug what we’ve got into the Taylor series and see what we get,
∞
f (n) ( 0 ) n
cos x = ∑
x
n!
n=0
f ′′ ( 0 ) 2 f ′′′ ( 0 ) 3 f ( 4) ( 0 ) 4 f ( 5) ( 0 ) 5
= f ( 0) + f ′ ( 0) x +
x +
x +
x +
x +
2!
3!
4!
5!
1
1
1
= 1 + 0 − x2 + 0 + x4 + 0 − x6 +
2!
4!
6!
n =0
n =1
n =3
n =5
n=2

n=4

n=6

So, we only pick up terms with even powers on the x’s. This doesn’t really help us to get
a general formula for the Taylor Series. However, let’s drop the zeroes and “renumber”
the terms as follows to see what we can get.
1
1
1
cos x = 1 − x 2 + x 4 − x 6 +
2!
4!
6!
n=0
n =1

n=2

n =3

By renumbering the terms as we did we can actually come up with a general formula for
the Taylor Series and here it is,

( −1) x 2 n
cos x = ∑
( 2n ) !
n=0
∞

n

This idea of renumbering the series terms as we did in the previous example isn’t used all
that often, but occasionally is very useful. There is one more series where we need to do
it so let’s take a look at that so we can get one more example down of renumbering series
terms.
Example 6 Find the Taylor Series for f ( x ) = sin ( x ) about x=0.
Solution
As with the last example we’ll start off in the same manner.


248


Calculus II

( x ) = sin x
1
f ( ) ( x ) = cos x
2
f ( ) ( x ) = − sin x
3
f ( ) ( x ) = − cos x
4
f ( ) ( x ) = sin x
5
f ( ) ( x ) = cos x
6
f ( ) ( x ) = − sin x
f(

(0) = 0
1
f ( ) (0) = 1
2
f ( ) ( 0) = 0
3
f ( ) ( 0 ) = −1
4
f ( ) ( 0) = 0
5
f ( ) ( 0) = 1
6
f ( ) (0) = 0

0)

f(

0)

So, we get a similar pattern for this one. Let’s plug the numbers into the Taylor Series.
∞
f (n) ( 0 ) n
sin x = ∑
x
n!
n =0
1
1
1
1
= x − x3 + x5 − x 7 +
1!
3!
5!
7!
In this case we only get terms that have an odd exponent on x and as with the last
problem once we ignore the zero terms there is a clear pattern and formula. So
renumbering the terms as we did in the previous example we get the following Taylor
Series.

( −1) x 2 n+1
sin x = ∑
n = 0 ( 2n + 1) !
∞

n

We really need to work another example or two in which isn’t about x=0.
Example 7 Find the Taylor Series for f ( x ) = ln ( x ) about x=2.
Solution
Here are the first few derivatives and the evaluations.


249


Calculus II

f ( 0) ( x ) = ln ( x )
f (1) ( x ) =

f ( 0) ( 2 ) = ln 2

1
x

f ( 2) ( x ) = −

f (1) ( 2 ) =
1
x2

f ( 2) ( 2 ) = −

2
x3
2 ( 3)
f ( 4) ( x ) = − 4
x
2 ( 3)( 4 )
f ( 5) ( x ) =
x5

f

( x)

f ( 3) ( 2 ) =

( −1) ( n − 1)!
=
n +1

xn

1
22

2
23
2 ( 3)
f ( 4) ( 2 ) = − 4
2
2 ( 3)( 4 )
f (5) ( 2 ) =
25

f ( 3) ( x ) =

(n)

1
2

f

(n)

( 2)

( −1) ( n − 1)!
=
n +1

n = 1, 2,3,…

2n

Note that while we got a general formula here it doesn’t work for n=0. This will happen
on occasion so don’t worry about it when it does.
In order to plug this into the Taylor Series formula we’ll need to strip out the n=0 term
first.
∞
f ( n) ( 2 )
n
ln ( x ) = ∑
( x − 2)
n!
n=0
∞

= f ( 2) + ∑
n =1
∞

= ln ( 2 ) + ∑

f (n) ( 2)
n
( x − 2)
n!

( −1) ( n − 1)!
n +1

n! 2

n =1
∞

= ln ( 2 ) + ∑
n =1

( −1)
n2

n +1
n

n

( x − 2)

( x − 2)

n

n

Notice that we simplified the factorials in this case. You should always simplify them if
there are more than one and it’s possible to simplify them.
Also, do not get excited about the term sitting in front of the series. Sometimes we need
to do that when we can’t get a general formula that will hold for all values of n.
Example 8 Find the Taylor Series for f ( x ) =

1
about x = −1 .
x2

Solution
Again, here are the derivatives and evaluations.


250


Calculus II

f ( 0) ( x ) =

1
x2

f (1) ( x ) = −
f ( 2) ( x ) =

f

( x)

2
x3

1

( −1)

f (1) ( −1) = −

2

=1

2

=2

( −1)
2 ( 3)
f ( 2) ( −1) =
= 2 ( 3)
4
( −1)
2 ( 3)( 4 )
f (3) ( −1) = −
= 2 ( 3)( 4 )
5
( −1)

2 ( 3)
x4

f ( 3) ( x ) = −

(n)

f ( 0) ( −1) =

2 ( 3)( 4 )
x5

( −1) ( n + 1)!
=
n

f

xn+2

(n)

3

( −1) ( n + 1)! = n + 1 !
( −1) =
( )
n+2
( −1)
n

Notice that all the negatives signs will cancel out in the evaluation. Also, this formula
will work for all n, unlike the previous example.
Here is the Taylor Series for this function.
∞
f ( n ) ( −1)
1
n
=∑
( x + 1)
2
x
n!
n =0
∞

( n + 1)!

n =0

n!

=∑
∞

( x + 1)

= ∑ ( n + 1)( x + 1)

n

n

n =0

Now, let’s work one of the easier examples in this section. The problem for most
students is that it may not appear to be that easy (or maybe it will appear to be too easy)
at first glance.
Example 9 Find the Taylor Series for f ( x ) = x 3 − 10 x 2 + 6 about x = 3 .
Solution
Here are the derivatives for this problem.
f ( 0) ( x ) = x3 − 10 x 2 + 6

f ( 0 ) ( 3) = −57

f (1) ( x ) = 3 x 2 − 20 x

f (1) ( 3) = −33

f ( 2) ( x ) = 6 x − 20

f ( 2) ( 3) = −2

f ( 3) ( x ) = 6

f ( 3) ( 3 ) = 6

f (n) ( x ) = 0
f ( 4 ) ( 3) = 0
n≥4
This Taylor series will terminate after n=3. This will always happen when we are finding
the Taylor Series of a polynomial. Here is the Taylor Series for this one.


251


Calculus II

∞

x3 − 10 x 2 + 6 = ∑
n =0

f ( n ) ( 3)
n
( x − 3)
n!

= f ( 3) + f ′ ( 3)( x − 3) +

f ′′ ( 3)
f ′′′ ( 3)
2
3
( x − 3) +
( x − 3) + 0
2!
3!

= −57 − 33 ( x − 3) − ( x − 3) + ( x − 3)
2

3

When finding the Taylor Series of a polynomial we don’t do any simplification of the
right hand side. We leave it like it is. In fact, if we were to multiply everything out we
just get back to the original polynomial!
While it’s not apparent that writing the Taylor Series for a polynomial is useful there are
times where this needs to be done. The problem is that they are beyond the scope of this
course and so aren’t covered here. For example, there is one application to series in the
field of Differential Equations where this needs to be done on occasion.
So, we’ve seen quite a few examples of Taylor Series to this point and in all of them we
where able to find general formulas for the series. This won’t always be the case. To see
an example of one that doesn’t have a general formula check out the last example in the
next section.
Before leaving this section there are three important Taylor Series that we’ve derived in
this section that we should summarize up in one place. In my class I will assume that you
know these formulas from this point on.
∞

ex = ∑
n=0

xn
n!

( −1) x 2 n
cos x = ∑
( 2n ) !
n=0
n
∞
( −1) x 2 n +1
sin x = ∑
n = 0 ( 2n + 1) !
∞

n

Applications of Series
Now, that we know how to represent function as power series we can now talk about at
least a couple of applications of series.
There are in fact many applications of series, unfortunately most of them are beyond the
scope of this course. One application of power series (with the occasional use of Taylor
Series) is in the field of Ordinary Differential Equations when finding Series Solutions to
Differential Equations. If you are interested in seeing how that works you can check out
that chapter of my Differential Equations notes.


252


Calculus II

Another application of series arises in the study of Partial Differential Equations. One of
the more commonly used methods in that subject makes use of Fourier Series.
Many of the applications of series, especially those in the differential equations fields,
rely on the fact that functions can be represented as a series. In these applications it is
very difficult, if not impossible, to find the function itself. However, there are methods
of determining the series representation for the unknown function.
While the differential equations applications are beyond the scope of this course there are
some applications from a Calculus setting that we can look at.
Example 1 Determine a Taylor Series about x=0 for the following integral.
⌠ sin x dx
⎮
⌡ x
Solution
To do this we will first need to find a Taylor Series about x=0 for the integrand. This
however isn’t terribly difficult. We already have a Taylor Series for sine about x=0 so
we’ll just use that as follows,
2 n +1
∞
( −1) x 2 n
sin x 1 ∞ ( −1) x
= ∑
=∑
x
x n =0 ( 2n + 1) ! n =0 ( 2n + 1) !
n

n

We can now do the problem.

2n
∞
⌠ sin x dx = ⌠ ( −1) x dx
⎮
⎮∑
⌡ x
⌡ n =0 ( 2n + 1) !
n

( −1) x 2 n+1
=C+∑
n = 0 ( 2n + 1) ( 2n + 1) !
∞

n

So, while we can’t integrate this function in terms of known functions we can come up
with a series representation for the integral.
This idea of deriving a series representation for a function instead of trying to find the
function itself is used quite often in several fields. In fact, there are some fields where
this is one of the main ideas used and without this idea it would be very difficult to
accomplish anything in those fields.
Another application of series isn’t really an application of infinite series. It’s more an
application of partial sums. In fact, we’ve already seen this application in use once in this
chapter. In the Estimating the Value of a Series we used a partial sum to estimate the
value of a series. We can do the same thing with power series and series representations
of functions. The main difference is that we will now be using the partial sum to
approximate a function instead of a single value.


253


Calculus II

We will look at Taylor series for our examples, but we could just as easily use any series
representation here. Recall that the nth degree Taylor Polynomial of f(x) is given by,
n
f (i ) ( a )
i
Tn ( x ) = ∑
( x − a)
i!
i =0

Let’s take a look at example of this.
Example 2 For the function f ( x ) = cos ( x ) plot the function as well as T2 ( x ) , T4 ( x ) ,

and T8 ( x ) on the same graph for the interval [-4,4].

Solution
Here is the general formula for the Taylor polynomials for cosine.

( −1) x 2i
Tn ( x ) = ∑
( 2i )!
i =0
n

i

The three Taylor polynomials that we’ve got are then,
x2
T2 ( x ) = 1 −
2
2
x
x4
T4 ( x ) = 1 − +
2 24
2
4
x
x
x6
x8
T8 ( x ) = 1 − + −
+
2 24 720 40320
Here is the graph of these three Taylor polynomials as well as the graph of cosine.


254


Calculus II

As we can see from this graph as we increase the degree of the Taylor polynomial it starts
to look more and more like the function itself. In fact by the time we get to T8 ( x ) the
only difference is right at the ends. The higher the degree of the Taylor polynomial the
better it approximates the function.
Also the larger the interval the higher degree Taylor polynomial we need to get a good
approximation for the whole interval.
Before moving on let’s notice write down a couple more Taylor polynomials from the
previous example. Notice that because the Taylor series for cosine doesn’t contain any
terms with odd powers on x we get the following Taylor polynomials.
T3 ( x ) = 1 −

x2
2

x2 x4
+
2 24
x2 x4 x6
x8
T9 ( x ) = 1 − + −
+
2 24 720 40320
T5 ( x ) = 1 −

These are identical to those used in the example. Sometimes this will happen although
that was not really the point of this. The point is to notice that the nth degree Taylor
polynomial may actually have a degree that is less than n. It will never be more than n,
but it can be less than n.
The final example in this section really isn’t an application of series and probably
belonged in the previous section. However, the previous section was getting too long so
the example is in this section. This is an example of how to multiply series together and
while this isn’t an application of series it is something that does have to be done on
occasion in the applications. So, in that sense it does belong in this section.
Example 3 Find the first three non-zero terms in the Taylor Series for f ( x ) = e x cos x

about x=0.
Solution
Before we start let’s acknowledge that the easiest way to do this problem is to simply
compute the first 3-4 derivatives, evaluate them at x=0, plug into the formula and we’d
be done. However, as we noted prior to this example we want to use this example to
illustrate how we multiply series.
We will make use of the fact that we’ve got Taylor Series for each of these so we can use
them in this problem.
n
⎛ ∞ x n ⎞ ⎛ ∞ ( −1) x 2 n ⎞
x
e cos x = ⎜ ∑ ⎟ ⎜ ∑
⎟
⎝ n = 0 n ! ⎠ ⎜ n = 0 ( 2n ) ! ⎟
⎝
⎠


255


Calculus II

We’re not going to completely multiply out these series. We’re going to enough of the
multiplication to get an answer. The problem statement says that we want the first three
non-zero terms. That non-zero bit is important as it is possible that some of the terms
will be zero. If none of the terms are zero this would mean that the first three non-zero
terms would be the constant term, x term, and x2 term. However, because some might be
zero let’s assume that if we get all the terms up through x4 we’ll have enough to get the
answer. If we’ve assumed wrong it will be very easy to fix so don’t worry about that.
Now, let’s write down the first few terms of each series and we’ll stop at the x4 term in
each.
⎛
⎞ ⎛ x2 x4
⎞
x 2 x3 x 4
x
e cos x = ⎜ 1 + x + + +
+ ⎟ ⎜1 − + + ⎟
2 6 24
2 24
⎝
⎠⎝
⎠
Note that we do need to acknowledge that these series don’t stop. That’s the purpose of
the “ + ” at the end of each. Just for a second however, let’s suppose that each of these
did stop and ask ourselves how we would multiply each out. If this were the case we
would take every term in the second and multiply by every term in the first. In other
words, we would first multiply every term in the second series by 1, then every term in
the second series by x, then by x2 etc.
By stopping each series at x4 we have now guaranteed that we’ll get all terms that have an
exponent of 4 or less. Do you see why?
Each of the terms that we neglected to write down have an exponent of at least 5 and so
multiplying by 1 or any power of x will result in a term with an exponent that is at a
minimum 5. Therefore, none of the neglected terms will contribute terms with an
exponent of 4 or less and so weren’t needed.
So, let’s start the multiplication process.
⎛
x 2 x3 x 4
x
e cos x = ⎜1 + x + + + +
2 6 24
⎝
= 1−

x2 x4
+ +
2 24

Second Series × 1

+

⎞ ⎛ x2 x4
⎞
⎟ ⎜1 − + + ⎟
2 24
⎠⎝
⎠
3
5
2
4
x
x
x
x
x6
+ x− + + + − + +
2 24
2 4 48
Second Series × x

x3 x5 x7
− +
+
6 12 144
3
Second Series × x 6

+

2

x4 x6
x8
− +
+
24 48 576

+

4

Now, collect like terms ignoring everything with an exponent of 5 or more since we
won’t have all those terms and don’t want them either. Doing this gives,


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Calculus II

⎛ 1 1⎞
⎛ 1 1⎞
⎛ 1 1 1 ⎞
e x cos x = 1 + x + ⎜ − + ⎟ x 2 + ⎜ − + ⎟ x 3 + ⎜ − + ⎟ x 4 +
⎝ 2 2⎠
⎝ 2 6⎠
⎝ 24 4 24 ⎠
3
4
x x
= 1+ x − − +
3 6

There we go. It looks like we over guessed and ended up with four non-zero terms, but
that’s okay. If we had under guessed and it turned out that we needed terms with x5 in
them all we would need to do at this point is go back and add in those terms to the
original series and do a couple quick multiplications. In other words, there is no reason
to completely redo all the work.

Binomial Series
In this final section of this chapter we are going to look at another series representation
for a function. Before we do this let’s first recall the following theorem.

Binomial Theorem
If n is any positive integer then,
n
⎛n⎞
n
( a + b ) = ∑ ⎜ ⎟ a n −i bi
i =0 ⎝ i ⎠
= a n + na n −1b +
where,

⎛ n ⎞ n ( n − 1)( n − 2 )
⎜ ⎟=
i!
⎝i⎠
⎛n⎞
⎜ ⎟ =1
⎝0⎠

n ( n − 1) n − 2 2
a b +
2!

( n − i + 1)

+ nab n −1 + b n

i = 1, 2,3,… n

This is useful for expanding ( a + b ) for large n when straight forward multiplication
n

wouldn’t be easy to do. Let’s take a quick look at an example.
Example 1 Use the Binomial Theorem to expand ( 2 x − 3)

4

Solution
There really isn’t much to do other than plugging into the theorem.


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Calculus II

( 2 x − 3)

4

4
⎛ 4⎞
4 −i
i
= ∑ ⎜ ⎟ ( 2 x ) ( −3)
i =0 ⎝ i ⎠
⎛ 4⎞
⎛ 4⎞
⎛ 4⎞
⎛ 4⎞
⎛ 4⎞
4
3
2
2
3
4
= ⎜ ⎟ ( 2 x ) + ⎜ ⎟ ( 2 x ) ( −3) + ⎜ ⎟ ( 2 x ) ( −3) + ⎜ ⎟ ( 2 x )( −3) + ⎜ ⎟ ( −3)
⎝0⎠
⎝1⎠
⎝ 2⎠
⎝ 3⎠
⎝ 4⎠

4 ( 3)
2
2
3
4
( 2 x ) ( −3) + 4 ( 2 x )( −3) + ( −3)
2
4
3
2
= 16 x − 96 x + 216 x − 216 x + 81
= ( 2 x ) + 4 ( 2 x ) ( −3) +
4

3

Now, the Binomial Theorem required that n be a positive integer. There is an extension
to this however that allows for any number at all.

Binomial Series
If k is any number and x < 1 then,

(1 + x )

where,

k

∞
⎛k ⎞
= ∑ ⎜ ⎟ xn
n=0 ⎝ n ⎠
k ( k − 1) 2 k ( k − 1)( k − 2 ) 3
= 1 + kx +
x +
x +
2!
3!

⎛ k ⎞ k ( k − 1)( k − 2 )
⎜ ⎟=
n!
⎝n⎠
⎛k ⎞
⎜ ⎟ =1
⎝0⎠

( k − n + 1)

n = 1, 2,3,…

So, similar to the binomial theorem except that it’s an infinite series and we must have
x < 1 in order to get convergence.
Let’s check out an example of this.
Example 2 Write down the first four terms in the binomial series for

9− x

Solution

So, in this case k =

1
and we’ll need to rewrite the term a little to put it into the form
2

required.
1

1

⎛ ⎛ x ⎞ ⎞2
⎛ x ⎞2
9 − x = 3 ⎜1 − ⎟ = 3 ⎜1 + ⎜ − ⎟ ⎟
⎝ 9⎠
⎝ ⎝ 9 ⎠⎠
The first four terms in the binomial series is then,


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Calculus II

1

⎛ ⎛ x ⎞ ⎞2
9 − x = 3 ⎜1 + ⎜ − ⎟ ⎟
⎝ ⎝ 9 ⎠⎠
∞
⎛ 1 ⎞⎛ x ⎞
= 3∑ ⎜ 2 ⎟ ⎜ − ⎟
9⎠
n=0 ⎝ n ⎠ ⎝

n

⎡ ⎛ 1 ⎞ ⎛ x ⎞ 1 ( − 1 ) ⎛ x ⎞2
= 3 ⎢1 + ⎜ ⎟ ⎜ − ⎟ + 2 2 ⎜ − ⎟ +
2 ⎝ 9⎠
⎢ ⎝ 2 ⎠⎝ 9 ⎠
⎣
x x2
x3
= 3− −
−
−
6 216 3888

1
2

( − 1 ) ( − 3 ) ⎛ − x ⎞3 +
2
2
6

⎜
⎟
⎝ 9⎠

⎤
⎥
⎥
⎦

Vectors
Introduction
This is a fairly short chapter. We will be taking a brief look at vectors and some of their
properties. We will need some of this material in the next chapter and those of you
heading on towards Calculus III will use a fair amount of this there as well.

Vectors – The Basics – In this section we will introduce some of the basic concepts
about vectors.
Vector Arithmetic – Here we will give the basic arithmetic operations for vectors.
Dot Product – We will discuss the dot product in this section as well as an application or
two.
Cross Product – In this section we’ll discuss the cross product and see a quick
application.

Vectors – The Basics
Let’s start this section off with a quick discussion on what vectors are used for. Vectors
are used to represent quantities that have both a magnitude and a direction. Good
examples of quantities that can be represented by vectors are force and velocity. Both of
these have a direction and a magnitude.
Let’s consider force for a second. A force of say 5 Newtons that is applied in a particular
direction can be applied at any point in space. In other words, the point where we apply


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Calculus II

the force does not change the force itself. Forces are independent of the point of
application. To define a force all we need to know is the magnitude of the force and the
direction that the force is applied in.
The same idea holds more generally with vectors. Vectors only impart magnitude and
direction. They don’t impart any information about where the quantity is applied. This is
an important idea to always remember in the study of vectors.
In a graphical sense vectors are represented by directed line segments. The length of the
line segment is the magnitude of the vector and the direction of the line segment is the
direction of the vector. However, because vectors don’t impart any information about
where the quantity is applied any directed line segment with the same length and
direction will represent the same vector.
Consider the sketch below.

Each of the directed line segments in the sketch represents the same vector. In each case
the vector starts at a specific point then moves 2 units to the left and 5 units up. The
notation that we’ll use for this vector is,
v = −2,5
and each of the directed line segments in the sketch are called representations of the
vector.
Be careful to distinguish vector notation, −2,5 , from the notation we use for points,

( −2,5) .

The vector denotes a magnitude and a direction of a quantity while the point

denotes a location in space. So don’t mix the notations up!
A representation of the vector v = a1 , a2 in two dimensional space is any directed line
segment, AB , from the point A = ( x, y ) to the point B = ( x + a1 , y + a2 ) . Likewise a


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Calculus II

representation of the vector v = a1 , a2 , a3 in three dimensional space is any directed line
segment, AB , from the point A = ( x, y, z ) to the point B = ( x + a1 , y + a2 , z + a3 ) .
Note that there is very little difference between the two dimensional and three
dimensional formulas above. To get from the three dimensional formula to the two
dimensional formula all we did is take out the third component/coordinate. Because of
this most of the formulas here are given only in their three dimensional version. If we
need them in their two dimensional form we can easily modify the three dimensional
form.
There is one representation of a vector that is special in some way. The representation of
the vector v = a1 , a2 , a3 that starts at the point A = ( 0, 0, 0 ) and ends at the point
B = ( a1 , a2 , a3 ) is called the position vector of the point ( a1 , a2 , a3 ) . So, when we talk

about position vectors we are specifying the initial and final point of the vector. These
are useful if we ever need to (and we will on occasion need to) represent a point as a
vector.
Next we need to discuss briefly how to generate a vector given the initial and final points
of the representation. Given the two points A = ( a1 , a2 , a3 ) and B = ( b1 , b2 , b3 ) the vector
with the representation AB is,
v = b1 − a1 , b2 − a2 , b3 − a3

Note that we have to be very careful with direction here. This is the vector that starts at A
and ends at B. The vector that starts at B and ends at A, i.e. with representation BA is,
w = a1 − b1 , a2 − b2 , a3 − b3
When determining the vector from the initial and final points we always subtract the
initial point from the final point.
Example 1 Give the vector for each of the following.
(a) The vector from ( 2, −7, 0 ) to (1, −3, −5 ) .

(b) The vector from (1, −3, −5 ) to ( 2, −7, 0 ) .
(c) The position vector for ( −90, 4 )
Solution
(a) Remember that to construct this vector we subtract coordinates of the starting point
from the ending point.
1 − 2, −3 − ( −7 ) , −5 − 0 = −1, 4, −5

(b) Same thing here.

2 − 1, −7 − ( −3) , 0 − ( −5 ) = 1, −4,5

Notice that the only difference between the first two is the signs are all opposite. This
difference is important as it is this difference that tells us that the two vectors point in
opposite directions.


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Calculus II

(c) Not much to this one other than acknowledging that the position vector of a point is
nothing more than a vector with the points coordinates as its components.
−90, 4
We now need to start discussing some of the basic concepts that we will run into on
occasion.

Magnitude
The magnitude, or length, of the vector v = a1 , a2 , a3 is given by,
2
2
v = a12 + a2 + a3

Example 2 Determine the magnitude of each of the following vectors.
(a) a = 3, −5,10

1
2
,−
5
5
(c) w = 0, 0

(b) u =

(d) i = 1, 0, 0
Solution
There isn’t too much to these other than plug into the formula.

(a) a = 9 + 25 + 100 = 134
(b) u =

1 4
+ = 1 =1
5 5

(c) w = 0 + 0 = 0
(d) i = 1 + 0 + 0 = 1
We also have the following fact about the magnitude.
If a = 0 then a = 0
This should make sense. Because we square all the components the only way we can get
zero out of the formula was for the components to be zero in the first place.

Unit Vector
Any vector with magnitude of 1, i.e. u = 1 , is called a unit vector.


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Calculus II

Example 3 Which of the vectors from the first example are unit vectors?
Solution
Both the second and fourth vectors had a length of 1 and so they are the only unit vectors
from the first example.

Zero Vector
The vector w = 0, 0 that we saw in the first example is called a zero vector since its
components are all zero. Zero vectors are often denoted by 0 . Be careful to distinguish
0 (the number) from 0 (the vector). The number 0 denote the origin in space, while the
vector 0 denotes a vector that has no magnitude or direction.

Standard Basis Vectors
The fourth vector from the second example, i = 1, 0, 0 , is called a standard basis
vector. In three dimensional space there are three standard basis vectors,
i = 1, 0, 0
j = 0,1, 0
k = 0, 0,1
In two dimensional space there are two standard basis vectors,
i = 1, 0
j = 0,1
Note that standard basis vectors are also unit vectors.

Warning
We are pretty much done with this section however, before proceeding to the next section
we should point out that vectors are not restricted to two dimensional or three
dimensional space. Vectors can exist in general n-dimensional space. The general
notation for a n-dimensional vector is,
v = a1 , a2 , a3 ,… , an
and each of the ai’s are called components of the vector.
Because we will be working almost exclusively with two and three dimensional vectors
in this course most of the formulas will be given for the two and/or three dimensional
cases. However, most of the concepts/formulas will work with general vectors and the
formulas are easily (and naturally) modified for general n-dimensional vectors. Also,
because it is easier to visualize things in two dimensions most of the figures related to
vectors will be two dimensional figures.
So, we need to be careful to not get too locked into the two or three dimensional cases
from our discussions in this chapter. We will be working in these dimensions either
because it’s easier to visualize the situation or because physical restrictions of the
problems will enforce a dimension upon us.


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Calculus II

Vector Arithmetic
In this section we need to have a brief discussion of vector arithmetic.
We’ll start with addition of two vectors. So, given the vectors a = a1 , a2 , a3 and
b = b1 , b2 , b3 the addition of the two vectors is given by,
a + b = a1 + b1 , a2 + b2 , a3 + b3
The following figure gives the geometric interpretation of the addition of two vectors.

This is sometimes called the parallelogram law or triangle law.
Computationally, subtraction is very similar. Given the vectors a = a1 , a2 , a3 and
b = b1 , b2 , b3 the difference of the two vectors is given by,
a − b = a1 − b1 , a2 − b2 , a3 − b3
Here is the geometric interpretation of the difference of two vectors.


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Calculus II

Note that we can’t add or subtract two vectors unless they a have the same number of
components. If they don’t have the same number of components then addition and
subtraction can’t be done.
The next arithmetic operation that we want to look at is scalar multiplication. Given the
vector a = a1 , a2 , a3 and any number c the scalar multiplication is,
ca = ca1 , ca2 , ca3

So, we multiply all the components by the constant c. To see the geometric interpretation
of scalar multiplication let’s take a look at an example.
Example 1 For the vector a = 2, 4 compute 3a ,

1
a and −2a . Graph all four vectors
2

on the same axis system.
Solution
Here are the three scalar multiplications.
1
3a = 6,12
a = 1, 2
2
Here is the graph.


265

−2a = −4, −8


Calculus II

In the previous example we can see that if c is positive all scalar multiplication will do is
stretch (if c > 1 ) or shrink (if c < 1 ) the original vector, but it won’t change the direction.
Likewise, if c is negative scalar multiplication will switch the direction so that the vector
will point in exactly the opposite direction and it will again stretch or shrink the
magnitude of the vector depending upon the size of c.
There are several nice applications of scalar multiplication that we should now take a
look at.
The first is parallel vectors. This is a concept that we will see quite a bit over the next
couple of sections. Two vectors are parallel if they have the same direction or are in
exactly opposite directions. Now, recall again the geometric interpretation of scalar
multiplication. When we performed scalar multiplication we generated new vectors that
were parallel to the original vectors (and each other for that matter).
So, let’s suppose that a and b are parallel vectors. If they are parallel then there must
be a number c so that,
a = cb
So, two vectors are parallel if one is a scalar multiple of the other.
Example 2 Determine if the sets of vectors are parallel or not.
(a) a = 2, −4,1 , b = −6,12, −3


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Calculus II

(b) a = 4,10 , b = 2, −9
Solution
(a) These two vectors are parallel since,
b = −3a

⎛1⎞
(b) These two vectors aren’t parallel. This can be seen by noticing that 4 ⎜ ⎟ = 2 and yet
⎝2⎠
⎛1⎞
10 ⎜ ⎟ = 5 ≠ −9 . In other words we can’t make a be a scalar multiple of b .
⎝2⎠
The next application is best seen in an example.
Example 3 Find a unit vector that points in the same direction as w = −5, 2,1 .
Solution
Okay, what we’re asking for is a new parallel vector (points in the same direction) that
happens to be a unit vector. We can do this with a scalar multiplication since all scalar
multiplication does is change the length of the original vector (along with possibly
flipping the direction to the opposite direction).

Here’s what we’ll do. First let’s determine the magnitude of w .
w = 25 + 4 + 1 = 30
Now, let’s form the following new vector,
1
1
5
2
1
u=
w=
−5, 2,1 = −
,
,
w
30
30 30 30
The claim is that this is a unit vector. That’s easy enough to check
25 4
1
30
+ +
=
=1
u =
30 30 30
30
This vector also points in the same direction as w since it is only a scalar multiple of w
and we used a positive multiple.
So, in general, given a vector w , u =

w
will be a unit vector that points in the same
w

direction as w .

Standard Basis Vectors Revisited
In the previous section we introduced the idea of standard basis vectors without really
discussing why they were important. We can now do that. Let’s start with the vector
a = a1 , a2 , a3


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Calculus II

We can use the addition of vectors to break this up as follows,
a = a1 , a2 , a3
= a1 , 0, 0 + 0, a2 , 0 + 0, 0, a3
Using scalar multiplication we can further rewrite the vector as,
a = a1 , 0, 0 + 0, a2 , 0 + 0, 0, a3
= a1 1, 0, 0 + a2 0,1, 0 + a3 0, 0,1
Finally, notice that these three new vectors are simply the three standard basis vectors for
three dimensional space.
a1 , a2 , a3 = a1i + a2 j + a3 k
So, we can take any vector and write it in terms of the standard basis vectors. From this
point on we will use the two notations interchangeably so make sure that you can deal
with both notations.
Example 4 If a = 3, −9,1 and w = −i + 8k compute 2a − 3w .
Solution
In order to do the problem we’ll convert to one notation and then perform the indicated
operations.
2a − 3w = 2 3, −9,1 − 3 −1, 0,8

= 6, −18, 2 − −3, 0, 24
= 9, −18, −22

We will leave this section with some basic properties of vector arithmetic.

Properties
If v , w and u are vectors (each with the same number of components) and a and b are
two numbers then we have the following properties.
v + w = w+v
v +0=v

1v = v

a ( v + w ) = av + aw


u + ( v + w) = (u + v ) + w

( a + b ) v = av + bv

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Calculus II

Dot Product
The next topic for discussion is that of the dot product. Let’s jump right into the
definition of the dot product. Given the two vectors a = a1 , a2 , a3 and b = b1 , b2 , b3 the
dot product is,
a ib = a1b1 + a2b2 + a3b3

(1)

Sometimes the dot product is called the scalar product. The dot product is also an
example of an inner product and so on occasion you may hear it called an inner product.
Example 1 Compute the dot product for each of the following.
(a) v = 5i − 8 j , w = i + 2 j

(b) a = 0,3, −7 , b = 2,3,1
Solution
Not much to do with these other than use the formula.
(a) v i w = 5 − 16 = −11

(b) a ib = 0 + 9 − 7 = 2

Here are some properties of the dot product.

Properties

u i( v + w ) = u iv + u i w

( cv )i w = v i( cw ) = c ( v i w )

v i w = wiv

v i0 = 0

v iv = v

If v iv = 0 then v = 0

2

There is also a nice geometric interpretation to the dot product. First suppose that θ is the
angle between a and b such that 0 ≤ θ ≤ π as shown in the image below.


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Calculus II

then,
a ib = a b cos θ

(2)

This is, more often than not, used to determine the angle between two vectors as shown in
the following example.
Example 2 Determine the angle between a = 3, −4, −1 and b = 0,5, 2 .
Solution
We will need the dot product as well as the magnitudes of each vector.
a ib = −22
a = 26
b = 29

The angle is then,
cos θ =

a ib
a b

=

−22
= −0.8011927
26 29

θ = cos1 ( −0.8011927 ) = 2.5 radians=143.24 degrees
The dot product gives us a very nice method for determining if two vectors are
perpendicular and it will give another method for determining when two vectors are
parallel. Note as well that often we will use the term orthogonal in place of
perpendicular.
Now, if two vectors are orthogonal then we know that the angle between them is 90
degrees. From (2) this tells us that if two vectors are orthogonal then,
a ib = 0


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Calculus II

Likewise, if two vectors are parallel then the angle between them is either 0 degrees
(pointing in the same direction) or 180 degrees (pointing in the opposite direction). Once
again using (2) this would mean that one of the following would have to be true.
a ib = a b (θ = 0° )
OR
a ib = − a b (θ = 180° )
Example 3 Determine if the following vectors are parallel, orthogonal, or neither.
(a) a = 6, −2, −1 , b = 2,5, 2

1
1
(b) u = 2i − j , v = − i + j
2
4
Solution
(a) First get the dot product to see if they are orthogonal.
a ib = 12 − 10 − 2 = 0
The two vectors are orthogonal.

(b) Again, let’s get the dot product first.
1
5
=−
4
4
So, they aren’t orthogonal. Let’s get the magnitudes and see if they are parallel.
5
5
u = 5
v =
=
16
4
u iv = −1 −

Now, notice that,
u iv = −

⎛ 5⎞
5
= − 5⎜
⎜ 4 ⎟=− u v
⎟
4
⎝
⎠

So, the two vectors are parallel.
There are several nice applications of the dot product as well that we should look at.

Projections
The best way to understand projections is to see a couple of sketches. So, given two
vectors a and b we want to determine the projection of b onto a . The projection is
denoted by proja b . Here are a couple of sketches illustrating the projection.


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Calculus II

So, to get the projection of b onto a we drop straight down from the end of b until we
hit (and form a right angle) with the line that is parallel to a . The projection is then the
vector that is parallel to a , starts at the same point both of the original vectors started at
and ends where the dashed line hits the line parallel to a .
There is an nice formula for finding the projection of b onto a . Here it is,
proja b =

a ib
2

a

a

Note that we also need to be very careful with notation here. The projection of a onto
b is given by
a ib
projb a = 2 b
b
We can see that this will be a totally different vector. This vector is parallel to b , while
proja b is parallel to a . So, be careful with notation and make sure you are finding the
correct projection.
Here’s an example.
Example 4 Determine the projection of b = 2,1, −1 onto a = 1, 0, −2 .
Solution
We need the dot product and the magnitude of a .

a ib = 4

a =5
2

The projection is then,


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Calculus II

proja b =

a ib
a

2

a

4
1, 0, −2
5
4
8
= , 0, −
5
5
=

For comparison purposes let’s do it the other way around as well.
Example 5 Determine the projection of a = 1, 0, −2 onto b = 2,1, −1 .
Solution
We need the dot product and the magnitude of b .
a ib = 4

b

2

=6

The projection is then,
projb a =

a ib
b

2

b

4
2,1, −1
6
4 2 2
= , ,−
3 3 3
=

As we can see form the previous two examples the two projections are different so be
careful.

Direction Cosines
This application of the dot product requires that we be in three dimensional space unlike
all the other applications we’ve looked at to this point.
Let’s start with a vector, a , in three dimensional space. This vector will form angles
with the x-axis (α ), the y-axis (β ), and the z-axis (γ ). These angles are called direction
angles and the cosines of these angles are called direction cosines.
Here is a sketch of a vector and the direction angles.


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Calculus II

The formulas for the direction cosines are,
cos α =

a ii
a
= 1
a
a

cos β =

a i j a2
=
a
a

cos γ =

a ik a3
=
a
a

where i , j and k are the standard basis vectors.
Let’s verify the first dot product above. We’ll leave the rest to you to verify.
a ii = a1 , a2 , a3 i 1, 0, 0 = a1
Here are a couple of nice facts about the direction cosines.
1. The vector u = cos α , cos β , cos γ is a unit vector.
2. cos 2 α + cos 2 β + cos 2 γ = 1
3. a = a cos α , cos β , cos γ
Example 6 Determine the direction cosines and direction angles for a = 2,1, −4 .
Solution
We will need the magnitude of the vector.
a = 4 + 1 + 16 = 21

The direction cosines and angles are then,


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Calculus II

2
21
1
cos β =
21
−4
cos γ =
21
cos α =

α = 1.119 radians = 64.123 degrees
β = 1.351 radians = 77.396 degrees
γ = 2.632 radians = 150.794 degrees

Cross Product
In this final section of this chapter we will look at the cross product of two vectors. We
should note that the cross product requires both of the vectors to be three dimensional
vectors.
Also, before getting into how to compute these we should point out a major difference
between dot products and cross products. The result of a dot product is a number and the
result of a cross product is a vector! Be careful not to confuse the two.
So, let’s start with the two vectors a = a1 , a2 , a3 and b = b1 , b2 , b3 then the cross
product is given by the formula,
a × b = a2b3 − a3b2 , a3b1 − a1b3 , a1b2 − a2b1
This is not an easy formula to remember. There are two ways to derive this formula.
Both of them use the fact that the cross product is really the determinant of a 3x3 matrix.
If you don’t know what this is that is don’t worry about it. You don’t need to know
anything about matrices or determinants to use either of the methods. The notation for
the determinant is as follows,
i
a × b = a1
b1

j
a2
b2

k
a3
b3

The first row is the standard basis vectors and must appear in the order given here. The
second row is the components of a and the third row is the components of b . Now, let’s
take a look at the different methods for getting the formula.
The first method uses the Method of Cofactors. If you don’t know the method of
cofactors that is fine, the result is all that we need. Here is the formula.
a a3
a a
a a2
a ×b = 2
i − 1 3 j+ 1
k
b2 b3
b1 b3
b1 b2


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where,
a b
= ad − bc
c d
This formula is not as difficult to remember as it might at first appear to be. First, the
terms alternate in sign and notice that the 2x2 is missing the column below the standard
basis vector that multiplies it.
The second method is slightly easier; however, many textbooks don’t cover this method
as it will only work on 3x3 determinants. This method says to take the determinant as
listed above and then copy the first two columns onto the end as shown below.
i
j k i
j
a × b = a1 a2 a3 a1 a2
b1 b2 b3 b1 b2
We now have three diagonals that move from left to right and three diagonals that move
from right to left. We multiply along each diagonal and add those that move from left to
right and subtract those that move from right to left.
This is best seen in an example. We’ll also use this example to illustrate a fact about
cross products.
Example 1 If a = 2,1, −1 and b = −3, 4,1 compute each of the following.

(a) a × b
(b) b × a
Solution
(a) Here is the computation for this one.
i
j k i
j
a × b = 2 1 −1 2 1
−3 4 1 −3 4
= i (1)(1) + j ( −1)( −3) + k ( 2 )( 4 ) − j ( 2 )(1) − i ( −1)( 4 ) − k (1)( −3)
= 5i + j + 11k
(b) And here is the computation for this one.
i
j k i
j
b × a = −3 4 1 −3 4
2 1 −1 2 1
= i ( 4 )( −1) + j (1)( 2 ) + k ( −3)(1) − j ( −3)( −1) − i (1)(1) − k ( 4 )( 2 )
= −5i − j − 11k


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Calculus II

Notice that switching the order of the vectors in the cross product simply changed all the
signs in the result. Note as well that this means that the two cross products will point in
exactly opposite directions since they only differ by a sign. We’ll formalize up this fact
shortly when we list several facts.
There is also a geometric interpretation of the cross product. First we will let θ be the
angle between the two vectors a and b and assume that 0 ≤ θ ≤ π , then we have the
following fact,
a × b = a b sin θ
(1)
and the following figure.

There should be a natural question at this point. How did we know that the cross product
pointed in the direction that we’ve given it here?
First, as this figure, implies the cross product is orthogonal to both of the original vectors.
This will always be the case with one exception that we’ll get to in a second.
Second, we knew that it pointed in the upward direction (in this case) by the “right hand
rule”. This says that if we take our right hand, start at a and rotate our fingers towards
b our thumb will point in the direction of the cross product. Therefore, if we’d sketched
in b × a above we would have gotten a vector in the downward direction.
Example 2 A plane is defined by any three points that are in the plane. If a plane
contains the points P = (1, 0, 0 ) , Q = (1,1,1) and R = ( 2, −1,3) find a vector that is

orthogonal to the plane.
Solution
The one way that we know to get an orthogonal vector is to take a cross product. So, if
we could find two vectors that we knew where in the plane and took the cross product of
these two vectors we know that the cross product would be orthogonal to both the
vectors. However, since both the vectors are in the plane the cross product would then
also be orthogonal to the plane.


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Calculus II

So, we need two vectors that are in the plane. This is where the points come into the
problem. Since all three points lie in the plane any vector between them must also be in
the plane. There are many ways to get two vectors between these points. We will use the
following two,
PQ = 1 − 1,1 − 0,1 − 0 = 0,1,1
PR = 2 − 1, −1 − 0,3 − 0 = 1, −1,3
The cross product of these two vectors will be orthogonal to the plane. So, let’s find the
cross product.
i
j k i
j
PQ × PR = 0 1 1 0 1
1 −1 3 1 −1
= 4i + j − k

So, the vector 4i + j − k will be orthogonal to the plane containing the three points.
Now, let’s address the one time where the cross product will not be orthogonal to the
original vectors. If the two vectors, a and b , are parallel then the angle between them is
either 0 or 180 degrees. From (1) this implies that,
a ×b = 0
From a fact about the magnitude we saw in the first section we know that this implies
a ×b = 0
In other words, it won’t be orthogonal to the original vectors since we have the zero
vector. This does give us another test for parallel vectors however.

Fact
If a × b = 0 then a and b will be parallel vectors.
Let’s also formalize up the fact about the cross product being orthogonal to the original
vectors.

Fact
Provided a × b ≠ 0 then a × b is orthogonal to both a and b .
Here are some nice properties about the cross product.

Properties
If u , v and w are vectors and c is a number then,


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Calculus II

u × v = −v × u
u × ( v + w) = u × v + u × w
u1
u i( v × w ) = v1
w1

u2
v2
w2

( cu ) × v = u × ( cv ) = c ( u × v )
u i( v × w ) = ( u × v )i w

u3
v3
w3

The determinant in the last fact is computed in the same way that the cross product is
computed. We will see an example of this computation shortly.
There are a couple of geometric applications to the cross product as well. Suppose we
have three vectors a , b and c and we form the three dimensional figure shown below.

The area of the parallelogram (two dimensional front of this object) is given by,
Area = a × b
and the volume of the parallelpiped (the whole three dimensional object) is given by,

(

Volume = a i b × c

)

Note that the absolute value bars are required since the quantity could be negative and
volume isn’t negative.
We can use this volume fact to determine if three vectors lie in the same plane or not. If
three vectors lie in the same plane then the volume of the parallelpiped will be zero.
Example 3 Determine if the three vectors a = 1, 4, −7 , b = 2, −1, 4 and c = 0, −9,18


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Calculus II

lie in the same plane or not.
Solution
So, as we noted prior to this example all we need to do is compute the volume of the
parallelpiped formed by these three vectors. If the volume is zero they lie in the same
plane and if the volume isn’t zero they don’t lie in the same plane.

1 4 −7 1 4
a i b × c = 2 −1 4 2 −1
0 −9 18 0 −9

(

)

= (1)( −1)(18 ) + ( 4 )( 4 )( 0 ) + ( −7 )( 2 )( −9 ) −

( 4 )( 2 )(18) − (1)( 4 )( −9 ) − ( −7 )( −1)( 0 )
= −18 + 126 − 144 + 36
=0
So, the volume is zero and so they lie in the same plane.

Three Dimensional Space
Introduction
In this chapter we will start taking a more detailed look at three dimensional space (3-D
space or 3 ). This is a very important topic in Calculus III since a good portion of
Calculus III is done in three (or higher) dimensional space.
We will be looking at the equations of graphs in 3-D space as well as vector valued
functions and how we do calculus with them. We will also be taking a look at a couple of
new coordinate systems for 3-D space.
This is the only chapter that exists in two places in my notes. When I originally wrote
these notes all of these topics were covered in Calculus II however, we have since moved
several of them into Calculus III. So, rather than split the chapter up I have kept it in the
Calculus II notes and also put a copy in the Calculus III notes. Many of the sections not
covered in Calculus III will be used on occasion there anyway and so they serve as a
quick reference for when we need them.

The 3-D Coordinate System – We will introduce the concepts and notation for the three
dimensional coordinate system in this section.


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Calculus II

Equations of Lines – In this section we will develop the various forms for the equation
of lines in three dimensional space.
Equations of Planes – Here we will develop the equation of a plane.
Quadric Surfaces – In this section we will be looking at some examples of quadric
surfaces.
Functions of Several Variables – A quick review of some important topics about
functions of several variables.
Vector Functions – We introduce the concept of vector functions in this section. We
concentrate primarily on curves in three dimensional space. We will however, touch
briefly on surfaces as well.
Calculus with Vector Functions – Here we will take a quick look at limits, derivatives,
and integrals with vector functions.
Tangent, Normal and Binormal Vectors – We will define the tangent, normal and
binormal vectors in this section.
Arc Length with Vector Functions – In this section we will find the arc length of a
vector function.
Velocity and Acceleration – In this section we will revisit a standard application of
derivatives. We will look at the velocity and acceleration of an object whose position
function is given by a vector function.
Curvature – We will determine the curvature of a function in this section.
Cylindrical Coordinates – We will define the cylindrical coordinate system in this
section. The cylindrical coordinate system is an alternate coordinate system for the three
dimensional coordinate system.
Spherical Coordinates – In this section we will define the spherical coordinate system.
The spherical coordinate system is yet another alternate coordinate system for the three
dimensional coordinate system.

The 3-D Coordinate System
We’ll start the chapter off with a fairly short discussion introducing the 3-D coordinate
system and the conventions that we’ll be using. We will also take a brief look at how the
different coordinate systems can change the graph of an equation.


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Calculus II

Let’s first get some basic notation out of the way. The 3-D coordinate system is often
denoted by 3 . Likewise the 2-D coordinate system is often denoted by 2 and the 1-D
coordinate system is denoted by . Also, as you might have guessed then a general n
dimensional coordinate system is often denoted by n .
Next, let’s take a quick look at the basic coordinate system.

This is the standard placement of the axes in this class. It is assumed that only the
positive directions are shown by the axes. If we need the negative axis for any reason we
will put them in as needed.
Also note the various points on this sketch. The point P is the general point sitting out in
3-D space. If we start at P and drop straight down until we reach a z-coordinate of zero
we arrive that the point Q. We say that Q sits in the xy-plane. The xy-plane corresponds
to all the points which have a zero z-coordinate. We can also start at P and move in the
other two directions as shown to get points in the xz-plane (this is S with a y-coordinate of
zero) and the yz-plane (this is R with an x-coordinate of zero).
Collectively, the xy, xz, and yz-planes are sometimes called the coordinate planes. In the
remainder of this class you will need to be able to deal with the various coordinate planes
so make sure that you can.
Also, the point Q is often referred to as the projection of P in the xy-plane. Likewise, R is
the projection of P in the yz-plane and S is the projection of P in the xz-plane.
Many of the formulas that you are used to working with in 2 have natural extensions in
3
. For instance the distance between two points in 2 is given by,
d ( P , P2 ) =
1

( x2 − x1 ) + ( y2 − y1 )
2

While the distance between any two points in
d ( P , P2 ) =
1


3

2

is given by,

( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
2

282

2

2


Calculus II

Likewise, the general equation for a circle with center ( h, k ) and radius r is given by,

( x − h) + ( y − k ) = r 2
and the general equation for a sphere with center ( h, k , l ) and radius r is given by,
2
2
2
( x − h) + ( y − k ) + ( z − l ) = r 2
2

2

With that said we do need to be careful about just translating everything we know about
2
into 3 and assuming that it will work the same way. A good example of this is in
graphing to some extent. Consider the following example.
Example 1 Graph x = 3 in

,

2

and

3

.

Solution
In
we have a single coordinate system and so x = 3 is a point in a 1-D coordinate
system.

In

2

the equation x = 3 tells us to graph all the points that are in the form ( 3, y ) . This

is a vertical line in a 2-D coordinate system.
In

3

the equation x = 3 tells us to graph all the points that are in the form ( 3, y, z ) . If

you go back and look at the coordinate plane points this is very similar to the coordinates
for the yz-plane except this time we have x = 3 instead of x = 0 . So, in a 3-D coordinate
system this is a plane that will be parallel to the yz-plane
Here are the graphs of each of these.


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Calculus II

Note that at this point we can now write down the equations for each of the coordinate
planes as well using this idea.
z=0
xy − plane
y=0
xz − plane
x=0

yz − plane

Let’s take a look at a slightly more general example.
Example 2 Graph y = 2 x − 1 in

2

and

3

.

Solution
Of course we had to throw out for this example since there are two variables which
means that we can’t be in a 1-D space.

In

2

this is a line with slope 2 and a y intercept of -1.

However, in 3 this is not necessarily a line. Because we have not specified a value of z
we are forced to let z take any value. This means that at any particular value of z we will
get a copy of this line. So, the graph is then a vertical plane that lies over the line given
by y = 2 x − 1 in the xy-plane.
Here are the graphs for this example.


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Calculus II

Notice that if we look to where the plane intersect the xy-plane we will get the graph of
the line in 2 as noted in the above graph.
Let’s take a look at one more example of the difference between graphs in the different
coordinate systems.
Example 3 Graph x 2 + y 2 = 4 in

2

and

3

.

Solution
As with the previous example this won’t have a 1-D graph since there are two variables.

In

2

this is a circle centered at the origin with radius 2.

In 3 however, as with the previous example, this may or may not be a circle. Since we
have not specified z in any way we must assume that z can take on any value. In other
words, at any value of z this equation must be satisfied and so at any value z we have a
circle of radius 2 centered on the z-axis. This means that we have a cylinder of radius 2


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Calculus II

centered on the z-axis.
Here are the graphs for this example.

Notice that again, if we look to where the cylinder intersects the xy-plane we will again
get the circle from 3 .
We need to be careful with the last two examples. It would be tempting to take the
results of these and say that we can’t graph lines or circles in 3 and yet that doesn’t
really make sense. There is no reason for the graph of a line or a circle in 3 . Let’s
think about the example of the circle. To graph a circle in 3 we would need to do
something like x 2 + y 2 = 4 at z = 5 . This would be a circle of radius 2 centered on the zaxis at the level of z = 5 . So, as long as we specify a z we will get a circle and not a
cylinder. We will see an easier way to specify circles in a later section.
We could do the same thing with the line from the second example. However, we will be
looking at line in more generality in the next section and so we’ll see a better way to deal
with lines in 3 there.
The point of the examples in this section is to make sure that we are being careful with
graphing equations and making sure that we always remember which coordinate system
that we are in.
Another quick point to make here is that, as we’ve seen in the above examples, many
graphs of equations in 3 are surfaces. That doesn’t mean that we can’t graph curves in
3
. We can and will graph curves in 3 as well as we’ll see later in this chapter.

Equations of Lines
In this section we need to take a look at the equation of a line in 3 . As we saw in the
previous section the equation y = mx + b does not describe a line in 3 , instead it
describes a plane.


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Calculus II

This doesn’t mean however that we can’t write down an equation for a line in 3-D space.
To see how to do this let’s think about what we need to write down the equation of a line
in 2 . In two dimensions we need the slope (m) and a point that was on the line in order
to write down the equation.
In 3 that is still all that we need except in this case the “slope” won’t be a simple
number as it was in two dimensions. In this case we will need to acknowledge that a line
can have a three dimensional slope. So, we need something that will allow us to describe
a direction that is potentially in three dimensions. We already have a quantity that will
do this for us. Vectors give directions and can be three dimensional objects.
So, let’s start with the following information. Suppose that we know a point that is on
the line, P0 = ( x0 , y0 , z0 ) , and that v = a, b, c is some vector that is parallel to the line.
Note, in all likelihood, v will not be on the line itself. We only need v to be parallel to
the line. Finally, let P = ( x, y, z ) be any point on the line.
Now, since our “slope” is a vector let’s also turn the two points into vectors as well. Of
course, we don’t actually turn them into vectors, we instead use position vectors to
represent them. So, let r0 and r be the position vectors for P0 and P respectively. Also,
for no apparent reason, let’s define a to be the vector with representation P0 P .
We now have the following sketch with all these vectors.


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Calculus II

At this point, notice that we can write r as follows,
r = r0 + a
If you’re not sure about this go back and check out the sketch for vector addition in the
vector arithmetic section. Now, notice that the vectors a and v are parallel. Therefore
there is a number, t, such that
a =tv
We now have,
r = r0 + t v = x0 , y0 , z0 + t a, b, c
This is called the vector form of the equation of a line. The only part of this equation
that is not known is the t. Notice that t v will be a vector that lies along the line and it
tells us how far from the original point that we should move. If t is positive we move to
the right of the original point and if t is negative we move to the left of the original point.
As t varies over all possible values we will completely cover the line.

There are several other forms of the equation of a line. To get the first alternate form
let’s start with the vector form and do a slight rewrite.
r = x0 , y0 , z0 + t a, b, c
x, y, z = x0 + ta, y0 + tb, z0 + tc
The only way for two vectors to be equal is for the components to be equal. In other
words,


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Calculus II

x = x0 + ta
y = y0 + tb
z = z0 + tc
This set of equations is called the parametric form of the equation of a line. Notice as
well that this is really nothing more than an extension of the parametric equations we’ve
seen previously. The only difference is that we are now working in three dimensions
instead of two dimensions.
To get a point on the line all we do is pick a t and plug into either form of the line. In the
vector form of the line we get a position vector for the point and in the parametric form
we get the actual coordinates of the point.
There is one more form of the line that we want to look at. If we assume that a, b, and c
are all non-zero numbers we can solve each of the equations in the parametric form of the
line for t. We can then set all of them equal to each other since t will be the same number
in each. Doing this gives the following,
x − x0 y − y0 z − z0
=
=
a
b
c
This is called the symmetric equations of the line.
If one of a, b, or c does happen to be zero we can still write down the symmetric
equations. To see this let’s suppose that b=0. In this case t will not exist in the
parametric equation for y and so we will only solve the parametric equations for x and z
for t. We then set those equal and acknowledge the parametric equation for y as follows,
x − x0 z − z0
=
y = y0
a
c
Example 1 Write down the equation of the line that passes through the points ( 2, −1,3)

and (1, 4, −3) . Write down all three forms of the equation of the line.
Solution
To do this we need the vector v that will be parallel to the line. This can be any vector
as long as it’s parallel to the line. In general, v won’t lie on the line itself. However, in
this case it will. All we need to do is let v be the vector that starts at the second point
and ends at the first point. Since these two points are one the line the vector between
them will also lie on the line and will hence be parallel to the line. So,
v = 1, −5, 6

Note that the order of the points was chosen to reduce the number of minus signs in the


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Calculus II

vector. We just have easily gone the other way.
Once we’ve got v there really isn’t anything else to do. To use the vector form we’ll
need a point on the line. We’ve got two and so we can use either one. We’ll use the first
point. Here is the vector form of the line.
r = 2, −1,3 + t 1, −5, 6 = 2 + t , −1 − 5t ,3 + 6t
Once we have this equation the other two forms follow. Here are the parametric
equations of the line.
x = 2+t
y = −1 − 5t
z = 3 + 6t
Here is the symmetric form.

x − 2 y +1 z − 3
=
=
1
−5
6

Example 2 Determine if the line that passes through the point ( 0, −3,8 ) and is parallel to

the line given by x = 10 + 3t , y = 12 t and z = −3 − t passes through the xz-plane. If it
does give the coordinates of that point.
Solution
To answer this we will first need to write down the equation of the line. We know a point
on the line and just need a parallel vector. We know that the new line must be parallel to
the line given by the parametric equations in the problem statement. That means that any
vector that is parallel to the given line must also be parallel to the new line.

Now recall that in the parametric form of the line the numbers multiplied by t are the
components of the vector that is parallel to the line. Therefore, the vector,
v = 3,12, −1
is parallel to the given line and so must also be parallel to the new line.
The equation of new line is then,
r = 0, −3,8 + t 3,12, −1 = 3t , −3 + 12t ,8 − t
If this line passes through the xz-plane then we know that the y-coordinate of that point
must be zero. So, let’s set the y component of the equation equal to zero and see if we
can solve for t. If we can, this will give the value of t for which the point will pass
through the xz-plane.
1
−3 + 12t = 0
⇒
t=
4
So, the line does pass through the xz-plane. To get the complete coordinates of the point


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Calculus II

1
into any of the equations. We’ll use the vector form.
4
1
3 31
⎛1⎞
⎛1⎞
r = 3 ⎜ ⎟ , −3 + 12 ⎜ ⎟ ,8 −
= , 0,
4
4
4
⎝4⎠
⎝4⎠

all we need to do is plug t =

Recall that this vector is the position vector for the point on the line and so the
⎛ 3 31 ⎞
coordinates of the point here the line will pass through the xz-plane are ⎜ , 0, ⎟ .
4⎠
⎝4

Equations of Planes
In the first section of this chapter we saw some equations of planes. However, none of
those equations had three variables in them and were really extensions of graphs that we
could look at in two dimensions. We would like a more general equation for planes.
So, let’s start by assuming that we know a point that is on the plane, P0 = ( x0 , y0 , z0 ) .
Let’s also suppose that we have a vector that is orthogonal (perpendicular) to the plane,
n = a, b, c . This vector is called the normal vector. Now, assume that P = ( x, y, z ) is
any point in the plane. Finally, since we are going to be working with vectors initially
we’ll let r0 and r be the position vectors for P0 and P respectively.
Here is a sketch of all these vectors.


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Calculus II

Notice that we added in the vector r − r0 which will lie completely in the plane. Also
notice that we put the normal vector on the plane, but there is actually no reason to expect
this to be the case. We put it here to illustrate the point. It is completely possible that the
normal vector does not touch the plane in any way.
Now, because n is orthogonal to the plane, it’s also orthogonal to any vector that lies in
the plane. In particular it’s orthogonal to r − r0 . Recall from the Dot Product section
that two orthogonal vectors will have a dot product of zero. In other words,
n i r − r0 = 0
⇒
n ir = n ir0

(

)

This is called the vector equation of the plane.
The vector equation of the plane is not a very useful equation in some ways. Let’s get a
much more useful form of the equations. Let’s start with the first form of the vector
equation.
a, b, c i( x, y, z − x0 , y0 , z0 ) = 0
a, b, c i x − x0 , y − y0 , z − z0 = 0
Now, actually compute the dot product.
a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0
This is called the scalar equation of plane. Often this will be written as,
ax + by + cz = d
where d = ax0 + by0 + cz0 .
This second form is often how we are given equations of planes. Notice that if we are
given the equation of a plane in this form we can quickly get a normal vector for the
plane. A normal vector is,
n = a , b, c
Example 1 Determine the equation of the plane that contains the points P = (1, −2, 0 ) ,
Q = ( 3,1, 4 ) and R = ( 0, −1, 2 ) .

Solution
In order to write down the equation of plane we need a point (we’ve got three so we’re
cool there) and a normal vector. We need to find a normal vector. Recall however, that
we saw how to do this in the Cross Product section.

We can form the following two vectors from the given points.
PQ = 2,3, 4
PR = −1,1, 2


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Calculus II

These two vectors will lie completely in the plane since we formed them from points that
were in the plane. Notice as well that there are many possible vectors to use here, we just
chose two of the possibilities.
Now, we know that the cross product of two vectors will be orthogonal to both of these
vectors. Since both of these are in the plane any vector that is orthogonal to both of these
will also be orthogonal to the plane. Therefore, we can use the cross product as the
normal vector.
i
j k i
j
n = PQ × PR = 2 3 4 2 3 = 2i − 8 j + 5k
−1 1 2 −1 1
The equation of the plane is then,
2 ( x − 1) − 8 ( y + 2 ) + 5 ( z − 0 ) = 0
2 x − 8 y + 5 z = 18
We used P for the point, but could have used any of the three points.
Example 2 Determine if the plane given by − x + 2 z = 10 and the line given by
r = 5, 2 − t ,10 + 4t are orthogonal, parallel or neither.
Solution
This is not as difficult a problem as it may at first appear to be. We can pick off a vector
that is normal to the plane. This is n = −1, 0, 2 . We can also get a vector that is parallel

to the line. This is v = 0, −1, 4 .
Now, if these two vectors are parallel then the line and the plane will be orthogonal. If
you think about it this makes some sense. If n and v are parallel, then v is orthogonal
to the plane, but v is also parallel to the line. So, if the two vectors are parallel the line
and plane will be orthogonal.
Let’s check this.
i
j k i
j
n × v = −1 0 2 −1 0 = 2i + 4 j + k ≠ 0
0 −1 4 0 −1
So, the vectors aren’t parallel and so the plane and the line are not orthogonal.
Now, let’s check to see if the plane and line are parallel. If the line is parallel to the plane
then any vector parallel to the line will be orthogonal to the normal vector of the plane.
In other words, if n and v are orthogonal then the line and the plane will be parallel.


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Calculus II

Let’s check this.

n iv = 0 + 0 + 8 = 8 ≠ 0

The two vectors aren’t orthogonal and so the line and plane aren’t parallel.
So, the line and the plane are neither orthogonal nor parallel.

Quadric Surfaces
In the previous two sections we’ve looked at lines and planes in three dimensions (or 3 )
and while these are used quite heavily at times in a Calculus class there are many other
surfaces that are also used fairly regularly and so we need to take a look at those.
In this section we are going to be looking at quadric surfaces. Quadric surfaces are the
graphs of any equation that can be put into the general form
Ax 2 + By 2 + Cz 2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0
where A,…J are constants.
There is no way that we can possibly list all of them, but there are some standard
equations so here is a list of some of the more common quadric surfaces.

Ellipsoid
Here is the general equation of an ellipsoid.
x2 y 2 z 2
+
+ =1
a 2 b2 c2
Here is a sketch of a typical ellipsoid.

If a=b=c then we will have a sphere.

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Calculus II

Notice that we only gave the equation for the ellipsoid that has been centered on the
origin. Clearly ellipsoids don’t have to be centered on the origin. However, in order to
make the discussion in this section a little easier we have chosen to concentrate on
surfaces that are “centered” on the origin in one way or another.

Cone
Here is the general equation of a cone.
x2 y 2 z 2
+
=
a 2 b2 c2
Here is a sketch of a typical cone.

Note that this is the equation of a cone that will open along the z-axis. To get the
equation of a cone that opens along one of the other axes all we need to do is make a
slight modification of the equation. This will be the case for the rest of the surfaces that
we’ll be looking at in this section as well.
In the case of a cone the variable that sits by itself on one side of the equal sign will
determine the axis that the cone opens up along. For instance, a cone that opens up along
the x-axis will have the equation,
y 2 z 2 x2
+ =
b2 c2 a 2
For most of the following surfaces we will not give the other possible formulas. We will
however acknowledge how each formula needs to be changed to get a change of
orientation for the surface.

Cylinder
Here is the general equation of a cylinder.


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Calculus II

x2 + y 2 = r 2
Here is a sketch of typical cylinder.

The cylinder will be centered on the axis corresponding to the variable that does not
appear in the equation.
Be careful to not confuse this with a circle. In two dimensions it is a circle, but in three
dimensions it is a cylinder.

Hyperboloid of One Sheet
Here is the equation of a hyperboloid of one sheet.
x2 y 2 z 2
+
− =1
a 2 b2 c 2
Here is a sketch of a typical hyperboloid of one sheet.


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Calculus II

The variable with the negative in front of it will give the axis along which the graph is
centered.

Hyperboloid of Two Sheets
Here is the equation of a hyperboloid of two sheets.
x2 y 2 z 2
− 2 − 2 + 2 =1
a b
c
Here is a sketch of a typical hyperboloid of two sheets.

The variable with the positive in front of it will give the axis along which the graph is
centered.
Notice that the only difference between the hyperboloid of one sheet and the hyperboloid
of two sheets is the signs in front of the variables. They are exactly the opposite signs.

Elliptic Paraboloid
Here is the equation of an elliptic paraboloid.
x2 y 2 z
+
=
a 2 b2 c
Here is a sketch of a typical elliptic paraboloid.


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Calculus II

In this case the variable that isn’t squared determines the axis upon which the paraboloid
opens up. Also, the sign of c will determine the direction that the paraboloid opens. If c
is positive then it opens up and if c is negative then it opens down.

Hyperbolic Paraboloid
Here is the equation of a hyperbolic paraboloid.
x2 y 2 z
−
=
a 2 b2 c
Here is a sketch of a typical hyperbolic paraboloid.

As with the elliptic paraoloid the sign of c will determine the direction in which the
surface “opens up”. The graph above is shown for c positive.
With the both of the paraboloids the surface can be easily moved up or down by
adding/subtracting a constant from the left side.


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Calculus II

For instance
z = − x2 − y2 + 6
is an elliptic paraboloid that opens downward and starts at z=6 instead of z=0.
Here is a sketch of this surface.

Functions of Several Variables
In this section we want to go over some of the basic ideas about functions of more than
one variable.
First, remember that graphs of functions of two variables, z = f ( x, y ) are surfaces in
three dimensional space. For example here is the graph of z = x 2 + y 2 − 6 .


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Calculus II

This is an elliptic parabaloid and is an example of a quadric surface. We saw several of
these in the previous section. We will be seeing quadric surfaces fairly regularly later on
in the semester.
Another common graph that we’ll be seeing quite a bit in this course is the graph of a
plane. We have a convention for graphing planes that will make them a little easier to
graph and hopefully visualize.
Recall that the equation of a plane is given by
ax + by + cz = d
or in terms of function notation this would be given by,
f ( x, y ) = ax + by + c

To graph a plane we will generally find the intersection points with the three axes and the
graph the triangle that connects those three points. This triangle will be a portion of the
plane and it will give us a fairly decent idea on what the plane itself should look like. For
example let’s graph the plane given by,
f ( x, y ) = 12 − 3 x − 4 y

For purposes of graphing this it would probably be easier to write this as,
z = 12 − 3x − 4 y
⇒
3 x + 4 y + z = 12
Now, each of the intersection points with the three main coordinate axes is defined by the
fact that two of the coordinates are zero. For instance, the intersection with the z-axis is
defined by x = y = 0 . So, the three intersection points are,

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Calculus II

x − axis : ( 4, 0, 0 )
y − axis : ( 0,3, 0 )
z − axis : ( 0, 0,12 )
Here is the graph of the plane.

Now, to extend this out, graphs of functions of the form w = f ( x, y, z ) would be four
dimensional surfaces. Of course we can’t graph them, but it doesn’t hurt to point this out.
We next want to talk about the domains of functions of more than one variable. Recall
that domains of functions of a single variable, y = f ( x ) , consisted of all the values of x
that we could plug into the function and get back a real number. Now, if we think about
it, this means that the domain of a function of a single variable is an interval (or intervals)
of values from the number line, or one dimensional space.
The domain of functions of two variables, y = f ( x, y ) , are regions from two dimensional
space and consist of all the coordinate pairs, ( x, y ) , that we could plug into the function
and get back a real number.
Example 1 Determine the domain of each of the following.
(a) f ( x, y ) = x + y

(b) f ( x, y ) = x + y

(c) f ( x, y ) = ln ( 9 − x 2 − 9 y 2 )
Solution
(a) In this case we know that we can’t take the square root of a negative number so this
means that we must require,
x+ y ≥0


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Calculus II

Here is a sketch of the graph of this region.

(b) This function is different from the function in the previous part. Here we must
require that,
x ≥ 0 and y ≥ 0
and they really do need to be separate inequalities. There is one for each square root in
the function. Here is the sketch of this region.

(c) In this final part we know that we can’t take the logarithm of a negative number or
zero. Therefore we need to require that,
x2
9 − x2 − 9 y2 > 0
⇒
+ y2 < 1
9
and upon rearranging we see that we need to stay interior to an ellipse for this function.
Here is a sketch of this region.


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Calculus II

Note that domains of functions of three variables, w = f ( x, y, z ) , will be regions in three
dimensional space.
Example 2 Determine the domain of the following function,
1
f ( x, y , z ) =
x 2 + y 2 + z 2 − 16
Solution
In this case we have to deal with the square root and division by zero issues. These will
require,
x 2 + y 2 + z 2 − 16 > 0
⇒
x 2 + y 2 + z 2 > 16

So, the domain for this function is the set of points that lies completely outside a sphere
of radius 4 centered at the origin.
The next topic that we should look at is that of level curves or contour curves. The
level curves of the function f ( x, y ) are two dimensional curves with equation
f ( x, y ) = k where k is any number.

You’ve probably seen level curves (or contour curves, whatever you want to call them)
before. If you’ve ever seen the elevation map for a piece of land, this is nothing more
than the contour curves for the function that gives the elevation of the land in that area.
Of course, we probably don’t have the function that gives the elevation, but we can at
least graph the contour curves.
Example 3 Identify the level curves of f ( x, y ) = x 2 + y 2 . Sketch a few of them.
Solution
First, for the sake of practice, let’s identify what this surface given by f ( x, y ) is. To do

this let’s rewrite it as,


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Calculus II

z = x2 + y2
Now, this equation is not listed in the Quadric Surfaces section, but if we square both
sides we get,
z 2 = x2 + y2
and this is listed in that section. So, we have a cone, or at least a portion of a cone. Since
we know that square roots will only return positive numbers, it looks like we’ve only got
the upper half of a cone.
Note that this was not required for this problem. It was done for the practice of
identifying the surface and this may come in handy down the road.
Now on to the real problem. The level curves (or contour curves) for this surface are
given by the equation,
k = x2 + y2
⇒
x2 + y2 = k 2
where k is any number. This is the equation of a circle of radius k with center at the
origin.
We can graph these in one of two ways. We can either graph them on the surface itself or
we can graph them in a two dimensional axis system. Here is each graph for some values
of k.


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Calculus II

Note that we can think of contours in terms of the intersection of the surface that is given
by z = f ( x, y ) and the plane z = k . The contour will represent the intersection of the
surface and the plane.
For functions of the form f ( x, y, z ) we will occasionally look at level surfaces. The
equations of level surfaces are given by f ( x, y, z ) = k where k is any number.
The final topic in this section is that of traces. In some ways these are similar to
contours. As noted above we can think of contours as the intersection of the surface
given by z = f ( x, y ) and the plane z = k . Traces of surfaces are curves that represent
the intersection of the surface and the plane given by x = a or y = b .
Let’s take a quick look at an example of traces.
Example 4 Sketch the traces of f ( x, y ) = 10 − 4 x 2 − y 2 for the plane x = 1 and y = 2 .
Solution
We’ll start with x = 1 . We can get an equation for the trace by plugging x = 1 into the
equation. Doing this gives,
2
z = f (1, y ) = 10 − 4 (1) − y 2
⇒
z = 6 − y2

and this will be graphed in the plane given by x = 1 .
Below are two graphs. The graph on the left is a graph showing the intersection of the
surface and the plane given by x = 1 . On the right is a graph of the surface and the trace
that we are after in this part.


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Calculus II

For y = 2 we will do pretty much the same thing that we did with the first part. Here is
the equation of the trace,
2
z = f ( x, 2 ) = 10 − 4 x 2 − ( 2 )
⇒
z = 6 − 4x2
and here are the sketches for this case.

Vector Functions
To this point, with the exception of lines, we only looked at graphing surfaces in 3 .
However, as we saw with lines, not every graph in 3 needs to be a surface. We can
graph curves (sometimes called space curves) that are three dimensional as well. To do
this we use vector-valued function or vector functions.
Note that we can also use vector functions to represent surfaces as well as we’ll see at the
end of this section. With that being said however we will spend most of this section
talking about curves instead of surfaces.
The vector form of the equation of a line is a good example a vector function.


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Calculus II

r ( t ) = r0 + tv
Vector functions take real numbers as arguments, t in this case, and return vectors that are
the position vector for points on the curve (or surface). The general form of a three
dimensional vector function for a curve is,
r (t ) = f (t ) , g (t ) , h (t )
where f(t), g(t), and h(t) are sometimes called the component functions.
The domain of a vector function is the set of all t’s for which all the component functions
are defined.
Example 1 Determine the domain of the following function.
r ( t ) = cos t , ln ( 4 − t ) , t + 1
Solution
The first component is defined for all t’s. The second component is only defined for
t < 4 . The third component is only defined for t ≥ −1 . Putting all of these together gives
the following domain.
[ −1, 4 )

This is the largest possible interval for which all three components are defined.
We now need to think about how to get the graph of a space curve from a vector function.
There are two ways to do this. The first is to think of the graph as the set of points
( x, y, z ) where,
x = f (t )

y = g (t )

z = h (t )

Note that these are also the parametric equations for the curve. We’ve seen parametric
equations before and the only difference is that we are now working with them in three
dimensions instead of the two dimensions that we used the last time we worked with
them.
The second way of thinking of the graph is to think of r ( t ) = f ( t ) , g ( t ) , h ( t ) as the
position vector of the point ( f ( t ) , g ( t ) , h ( t ) ) .
Either of these two ways of thinking of the graph will work and each has its uses. We
will mostly use the first way of thinking of the graph of a vector function.
Let’s take a look at a couple of graphs of vector functions.
Example 2 Sketch the graph of the following vector function.
r ( t ) = 2 − 4t ,1 + 2t , −3 − t
Solution
Notice that this is nothing more than a line. It might help if we rewrite it a little.
r ( t ) = 2,1, −3 + t −4, 2, −1


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Calculus II

In this form we can see that this is the equation of a line that goes through the point
( 2,1, −3) and is parallel to the vector v = −4, 2, −1 .
To graph this line all that we need to do is plot the point and then sketch in the parallel
vector. In order to get the sketch will assume that the vector is on the line and will start
at the point in the line. To sketch in the line all we do this is extend the parallel vector
into a line.
Here is a sketch.

r ( t ) = 2 cos t , 2sin t ,5
Solution
In this case to see what we’ve got for a graph let’s get the parametric equations for the
curve.
x = 2 cos t
y = 2sin t
z =5

If we ignore the z equation for a bit we’ll recall (hopefully) that the parametric equations
for x and y give a circle of radius 2 centered on the origin (or about the z-axis since we
are in 3 ).
Now, all the parametric equations here tell us is that no matter what is going on in the
graph all the z coordinates must be 5. So, we get a circle of radius 2 centered on the zaxis and at the level of z=5.
Here is a sketch.


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Calculus II

Note that it is very easy to modify the above vector function to get a circle centered on
the x or y-axis as well. For instance,
r ( t ) = 10sin t , −3,10 cos t
will be a circle of radius 10 centered on the y-axis and at y=-3.
In other words, as long as two of the terms are a sine and a cosine (with the same
coefficient) and the other is a fixed number then we will have a circle that is centered on
the axis that is given by the fixed number.
r ( t ) = 4 cos t , 4sin t , t
Solution
If this one had a constant in the z component we would have another circle. However, in
this case we don’t have a constant. Instead we’ve got a t and that will change the curve.
However, because the x and y component functions are still a circle in parametric
equations our curve should have a circular nature to it in some way.

In fact, the only change is in the z component and as t increases the z coordinate will
increase. Also, as t increases the x and y coordinates will continue to form a circle
centered on the z-axis. Putting these two ideas together tells us that at we increase t the
circle that is being traced out in the x and y directions should be also be rising.
Here is a sketch of this curve.


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Calculus II

So, we’ve got a helix here.
As with circles the component that has the t will determine the axis that the helix rotates
about. For instance,
r ( t ) = t , 6 cos t , 6sin t
is a helix that rotates around the x-axis.
Also note that if we allow the coefficients on the sine and cosine for both the circle and
helix to be different we will get ellipses.
For example,

r ( t ) = 9sin t , t , 2sin t

will be a helix that rotates about the y-axis and is in the shape of an ellipse.
There is a nice formula that we should derive before moving onto representing surfaces
with vector functions.
Example 5 Determine the vector equation for the line segment starting at the point
P = ( x1 , y1 , z1 ) and ending at the point Q = ( x2 , y2 , z2 ) .
Solution
It is important to note here that we only want the equation of the line segment that starts
at P and ends at Q. We don’t want any other portion of the line and we do want the
direction of the line segment preserved as we increase t. With all that said, let’s not
worry about that and just find the vector equation of the line that passes through the two
points. Once we have this we will be able to get what we’re after.

So, we need a point on the line. We’ve got two and we will use P. We need a vector that
is parallel to the line and since we’ve got two points we can find the vector between

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them. This vector will lie on the line and hence be parallel to the line. Also, let’s
remember that we want to preserve the starting and ending point of the line segment so
let’s construct the vector using the same “orientation”.
v = x2 − x1 , y2 − y1 , z2 − z1

Using this vector and the point P we get the following vector equation of the line.
r ( t ) = x1 , y1 , z1 + t x2 − x1 , y2 − y1 , z2 − z1

While this is the vector equation of the line, let’s rewrite the equation slightly.
r ( t ) = x1 , y1 , z1 + t x2 , y2 , z2 − t x1 , y1 , z1
= (1 − t ) x1 , y1 , z1 + t x2 , y2 , z2
This is the equation of the line that contains the points P and Q. We of course just want
the line segment that starts at P and ends at Q. We can get this by simply restricting the
values of t.
Notice that

r ( 0 ) = x1 , y1 , z1

r (1) = x2 , y2 , z2

So, if we restrict t to be between zero and one we will cover the line segment and we will
start and end at the correct point.
So the vector equation of the line segment that starts at P = ( x1 , y1 , z1 ) and ends at
Q = ( x2 , y2 , z2 ) is,
r ( t ) = (1 − t ) x1 , y1 , z1 + t x2 , y2 , z2

0 ≤ t ≤1

As noted at the beginning of this section we can also use vector functions for surfaces as
well. So, to make sure that we don’t forget that let’s work an example with that as well.
Example 6 Identify the surface that is described by r ( x, y ) = x i + y j + ( x 2 + y 2 ) k .
Solution
First, notice that in this case the vector function will in fact be a function of two
variables. This will always be the case when we are using vector functions to represent
surfaces.

To identify the surface let’s go back to parametric equations.


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Calculus II

x=x
y=y
z = x2 + y2
The first two are really only acknowledging that we are picking x and y for free and then
determining z form our choices of these two. The last equation is the one that we want.
We should recognize that function from the section on quadric surfaces. The third
equation is the equation of an elliptic paraboloid and so the vector function represents an
elliptic paraboloid.
As a final topic for this section let’s generalize the idea from the previous example and
note that given any function of one variable ( y = f ( x ) or x = h ( y ) ) or any function of
two variables ( z = g ( x, y ) , x = g ( y, z ) , or y = g ( x, z ) ) we can always write down a
vector form of the equation.
For a function of one variable this will be,
r ( x) = x i + f ( x) j
r ( y) = h( y)i + y j
and for a function of two variables the vector form will be,
r ( x, y ) = x i + y j + g ( x, y ) k
r ( y, z ) = g ( y, z ) i + y j + z k
r ( x, z ) = x i + g ( x , z ) j + z k
depending upon the original form of the function.
For example the hyperbolic paraboloid y = 2 x 2 − 5 z 2 can be written as the following
vector function.
r ( x, z ) = x i + ( 2 x 2 − 5 z 2 ) j + z k
This is a fairly important idea and we will be doing quite a bit of this kind of thing in the
later portions of this class.

Calculus with Vector Functions
In this section we need to talk briefly about limits, derivatives and integrals of vector
functions. As you will see, these behave in a fairly predictable manner. We will be
doing all of the work in 3 but we can naturally extend the formulas/work in this section
to n (i.e. n-dimensional space).
Let’s start with limits. Here is the limit of a vector function.


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Calculus II

lim r ( t ) = lim f ( t ) , g ( t ) , h ( t )
t →a

t →a

= lim f ( t ) , lim g ( t ) , lim h ( t )
t →a

t →a

t →a

= lim f ( t ) i + lim g ( t ) j + lim h ( t ) k
t →a

t →a

t →a

So, all that we do is take the limit of each of the components functions and leave it as a
vector.
Example 1 Compute lim r ( t ) where r ( t ) = t 3 ,
t →1

Solution
There really isn’t all that much to do here.

lim r ( t ) = lim t 3 , lim
t →1

t →1

t →1

= lim t 3 , lim
t →1

t →1

sin ( 3t − 3)
t −1

, e2t .

sin ( 3t − 3)
, lim e 2t
t →1
t −1
3cos ( 3t − 3)
, lim e 2t
t →1
1

= 1,3, e 2
Notice that we had to use L’Hospital’s Rule on the y component.
Now let’s take care of derivatives and after seeing how limits work it shouldn’t be to
surprising that we have the following for derivatives.
r ′ ( t ) = f ′ ( t ) , g ′ ( t ) , h′ ( t )
= f ′ ( t ) i + g ′ ( t ) j + h′ ( t ) k
Example 2 Compute r ′ ( t ) for r ( t ) = t 6 i + sin ( 2t ) j − ln ( t + 1) k .
Solution
There really isn’t too much to this problem other than taking the derivatives.
1
r ′ ( t ) = 6t 5 i + 2 cos ( 2t ) j −
k
t +1

Most of the basic facts that we know about derivatives still hold however, just to make it
clear here are some facts about derivatives of vector functions.
Facts


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Calculus II

d
( u + v ) = u ′ + v′
dt
( cu )′ = c u′
d
( f ( t ) u ( t ) ) = f ′ ( t ) u ( t ) + f ( t ) u′
dt
d
( u iv ) = u′iv + u iv′
dt
d
( u × v ) = u ′ × v + u × v′
dt
d
u ( f ( t ) ) = f ′ ( t ) u′ ( f ( t ) )
dt

(

)

There is also one quick definition that we should get out of the way so that we can use it
when we need to.
A smooth curve is any curve for which r ′ ( t ) is continuous and r ′ ( t ) ≠ 0 for any t. A
helix is a smooth curve, for example.
Finally, we need to discuss integrals of vector functions. Using both limits and
derivatives as a guide it shouldn’t be too surprising that we also have the following for
integration for indefinite integrals

∫ r ( t ) = ∫ f ( t ) dt , ∫ g ( t ) dt , ∫ h ( t ) dt + c
∫ r ( t ) = ∫ f ( t ) dt i + ∫ g ( t ) dt j + ∫ h ( t ) dt k + c
and the following for definite integrals.

∫ r ( t ) dt = ∫ f ( t ) dt , ∫ g ( t ) dt , ∫ h ( t ) dt
b

b

a

a

b

b

a

a

∫ r ( t ) dt = ∫ f ( t ) dt i + ∫ g ( t ) dt
b

b

b

a

a

a

j + ∫ h ( t ) dt k
b

a

With the indefinite integrals we put in a constant of integration to make sure that it was
clear that the constant in this case needs to be a vector instead of a regular constant.
Also, for the definite integrals we will sometimes write it as follows,

∫
∫


b
a
b
a

r ( t ) dt =
r ( t ) dt =

(∫
(∫

f ( t ) dt , ∫ g ( t ) dt , ∫ h ( t ) dt

)

b
a

f ( t ) dt i + ∫ g ( t ) dt j + ∫ h ( t ) dt k

314

)

b
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Calculus II

In other words, we will do the indefinite integral and then do the evaluation of the vector
as a whole instead of on a component by component basis.
Example 3 Compute ∫ r ( t ) dt for r ( t ) = sin ( t ) , 6, 4t .
Solution
All we need to do is integrate each of the components and be done with it.
2
∫ r ( t ) dt = − cos ( t ) , 6t , 2t + c
Example 4 Compute

∫ r ( t ) dt for r ( t ) =
1

0

sin ( t ) , 6, 4t .

Solution
In this case all that we need to do is reuse the result from the previous example and then
do the evaluation.

∫ r ( t ) dt = (
1

0

− cos ( t ) , 6t , 2t 2

)

1
0

= − cos (1) , 6, 2 − −1, 0, 0
= 1 − cos (1) , 6, 2

Tangent, Normal and Binormal Vectors
In this section we want to look at an application of derivatives for vector functions.
Actually, there are a couple of applications, but they all come back to needing the first
one.
In the past we’ve used the fact that the derivative of a function was the slope of the
tangent line. With vector functions we get exactly the same result, with one exception.
Given the vector function, r ( t ) , we call r ′ ( t ) the tangent vector provided it exists and
provided r ′ ( t ) ≠ 0 . The tangent line to r ( t ) at P is then the line that passes through the
point P and is parallel to the tangent vector, r ′ ( t ) . Note that we really do need to require
r ′ ( t ) ≠ 0 in order to have a tangent vector. If we had r ′ ( t ) = 0 we would have a vector
that had not magnitude and so couldn’t give us the direction of the tangent.
Also, provided r ′ ( t ) ≠ 0 , the unit tangent vector to the curve is given by,
T (t ) =


r′ (t )
r′ (t )

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Calculus II

While, the components of the unit tangent vector can be somewhat messy on occasion
there are times when we will need to use the unit tangent vector instead of the tangent
vector.
Example 1 Find the general formula for the tangent vector and unit tangent vector to the
curve given by r ( t ) = t 2 i + 2sin t j + 2 cos t k .
Solution
First, by general formula we mean that we won’t be plugging in a specific t and so we
will be finding a formula that we can use at a later date if we’d like to find the tangent at
any point on the curve. With that said there really isn’t all that much to do at this point
other than to do the work.

Here is the tangent vector to the curve.
r ′ ( t ) = 2t i + 2 cos t j − 2sin t k
To get the unit tangent vector we need the length of the tangent vector.
r ′ ( t ) = 4t 2 + 4 cos 2 t + 4sin 2 t

= 4t 2 + 4

The unit tangent vector is then,
T (t ) =
=

1
4t 2 + 4
2t
4t + 4
2

( 2t i + 2 cos t j − 2sin t k )
2 cos t

i+

4t + 4
2

j−

2sin t
4t 2 + 4

k

Example 2 Find the vector equation of the tangent line to the curve given by

r ( t ) = t 2 i + 2sin t j + 2 cos t k at t =

π

3

.

Solution
First we need the tangent vector and since this is the function we were working with in
the previous example we can just reuse the tangent vector from that example and plug in

t=

π

3

.
2π
⎛ π ⎞ 2π
⎛π ⎞
⎛π ⎞
r′ ⎜ ⎟ =
i + 2 cos ⎜ ⎟ j − 2sin ⎜ ⎟ k =
i + j − 3k
3
⎝3⎠ 3
⎝3⎠
⎝3⎠

We’ll also need the point on the line at t =

π
3

so,

⎛π ⎞ π
r⎜ ⎟=
i + 3 j +k
⎝3⎠ 9
2


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Calculus II

The vector equation of the line is then,
r (t ) =

π2
9

, 3,1 + t

2π
,1, − 3
3

Before moving on let’s note a couple of things about the previous example. First, we
could have used the unit tangent vector had we wanted to for the parallel vector.
However, that would have made for a more complicated equation for the tangent line.
Second, notice that we used r ( t ) to represent the tangent line despite the fact that we
used that as well for the function. Do not get excited about that. The r ( t ) here is much
like y is with normal functions. With normal functions, y is the generic letter that we
used to represent functions and r ( t ) tends to be used in the same way with vector
functions.
Next we need to talk about the unit normal and the binormal vectors.
The unit normal vector is defined to be,
N (t ) =

T ′ (t )

T ′ (t )

The unit normal is orthogonal (or normal) to the unit tangent vector and hence to the
curve as well.
The binormal vector is defined to be,
B (t ) = T (t ) × N (t )
The binormal vector is orthogonal to both the tangent vector and the normal vector.
Example 3 Find the normal and binormal vectors for r ( t ) = t ,3sin t ,3cos t .
Solution
We first need the unit tangent vector so first get the tangent vector and its magnitude.
r ′ ( t ) = 1,3cos t , −3sin t
r ′ ( t ) = 1 + 9 cos 2 t + 9sin 2 t = 10

The unit tangent vector is then,
1
3
3
,
cos t , −
sin t
10 10
10
The unit normal vector will now require the derivative of the unit tangent and its
magnitude.
T (t ) =


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Calculus II

3
3
sin t , −
cos t
10
10

T ′ ( t ) = 0, −
T ′ (t ) =

9
9
9
3
sin 2 t + cos 2 t =
=
10
10
10
10

The unit normal vector is then,
10
3
3
N (t ) =
0, −
sin t , −
cos t = 0, − sin t , − cos t
3
10
10
Finally, the binormal vector is,
B (t ) = T (t ) × N (t )
i

=

1
10
0

j

k

3
cos t
10
− sin t

−

i

3
sin t
10
− cos t

j

1
10
0

3
cos t
10
− sin t

3
1
1
3
cos 2 t i −
sin t k +
cos t j −
sin 2 t i
10
10
10
10
3
1
1
i+
cos t j −
sin t k
=−
10
10
10
=−

Arc Length with Vector Functions
In this section we’ll recast an old formula into terms of vector functions. We want to
determine the length of a vector function,
r (t ) = f (t ) , g (t ) , h (t )
on the interval a ≤ t ≤ b .
We actually already know how to do this. Recall that we can write the vector function
into the parametric form,
x = f (t )
y = g (t )
z = h (t )
Also, recall that with two dimensional parametric curves the arc length is given by,
b

L = ⌠ ⎡ f ′ ( t ) ⎤ + ⎡ g ′ ( t ) ⎤ dt
⎦ ⎣
⎦
⌡a ⎣
2

2

There is a natural extension of this to three dimensions. So, the length of the curve r ( t )
on the interval a ≤ t ≤ b is,
b

L = ⌠ ⎡ f ′ ( t ) ⎤ + ⎡ g ′ ( t ) ⎤ + ⎡ h′ ( t ) ⎤ dt
⎦ ⎣
⎦ ⎣
⎦
⌡a ⎣


2

2

318

2


Calculus II

There is a nice simplification that we can make for this. Notice that the integrand (the
function we’re integrating) is nothing more than the magnitude of the tangent vector,
r ′ ( t ) = ⎡ f ′ ( t ) ⎤ + ⎡ g ′ ( t ) ⎤ + ⎡ h′ ( t ) ⎤
⎣
⎦ ⎣
⎦ ⎣
⎦
2

2

2

Therefore, the arc length can be written as,
L = ∫ r ′ ( t ) dt
b

a

Example 1 Determine the length of the curve r ( t ) = 2t ,3sin ( 2t ) ,3cos ( 2t ) on the

interval 0 ≤ t ≤ 2π .
Solution
We will first need the tangent vector and its magnitude.
r ′ ( t ) = 2, 6 cos ( 2t ) , −6sin ( 2t )

r ′ ( t ) = 4 + 36 cos 2 ( 2t ) + 36sin 2 ( 2t ) = 4 + 36 = 2 10
The length is then,
L = ∫ r ′ ( t ) dt
b

a

=∫

2π
0

2 10 dt

= 4π 10
We need to take a quick look at another concept here. We define the arc length function
as,
s ( t ) = ∫ r ′ ( t ) du
t

0

Before we look at why this might be important let’s work a quick example.
Example 2 Determine the arc length function for r ( t ) = 2t ,3sin ( 2t ) ,3cos ( 2t ) .
Solution
From the previous example we know that,
r ′ ( t ) = 2 10

The arc length function is then,

(

s ( t ) = ∫ 2 10 du = 2 10 u
t

0


319

)

t

= 2 10 t

0


Calculus II

Okay, just why would we want to do this? Well let’s take the result of the example above
and solve it for t.
s
t=
2 10
Now, taking this and plugging it into the original vector function and we can
reparameterize the function into the form, r ( t ( s ) ) . For our function this is,
s
⎛ s ⎞
⎛ s ⎞
,3sin ⎜
⎟ ,3cos ⎜
⎟
10
⎝ 10 ⎠
⎝ 10 ⎠

r (t ( s )) =

So, why would we want to do this? Well with the reparameterization we can now tell
where we are on the curve after we’ve traveled a distance of s along the curve. Note as
well that we will start the measurement of distance from where we are at t=0.
Example 3 Where on the curve r ( t ) = 2t ,3sin ( 2t ) ,3cos ( 2t ) are we after traveling for

a distance of

π 10
3

?

Solution
To determine this we need the reparameterization, which we have from above.
s
⎛ s ⎞
⎛ s ⎞
,3sin ⎜
⎟ ,3cos ⎜
⎟
10
⎝ 10 ⎠
⎝ 10 ⎠

r (t ( s )) =

Then, to determine where we are all that we need to do is plug in s =

π 10
3

into this and

we’ll get our location.
⎛ ⎛ π 10 ⎞ ⎞
π
π 3 3 3
⎛π ⎞
⎛π ⎞
r ⎜t ⎜
,3sin ⎜ ⎟ ,3cos ⎜ ⎟ =
,
,
⎟⎟ =
⎜ ⎜ 3 ⎟⎟
3
3 2 2
⎝3⎠
⎝3⎠
⎝
⎠⎠
⎝
So, after traveling a distance of

π 10
3

⎛π 3 3 3⎞
along the curve we are at the point ⎜ ,
⎜ 3 2 ,2⎟.
⎟
⎝
⎠

Velocity and Acceleration
In this section we need to take a look at the velocity and acceleration of a moving object.


320


Calculus II

From Calculus I we know that given the position function of an object that the velocity of
the object is the first derivative of the position function and the acceleration of the object
is the second derivative of the position function.
So, given this it shouldn’t be too surprising that if the position function of an object is
given by the vector function r ( t ) then the velocity and acceleration of the object is given
by,

v (t ) = r′ (t )

a ( t ) = r ′′ ( t )

Notice that the velocity and acceleration are also going to be vectors as well.
In the study of the motion of objects the acceleration is often broken up into a tangential
component, aT, and a normal component, aN. The tangential component is the part of
the acceleration that is tangential to the curve and the normal component is the part of the
acceleration that is normal (or orthogonal) to the curve. If we do this we can write the
acceleration as,
a = aT T + aN N
where T and N are the unit tangent and unit normal for the position function.
If we define v = v ( t ) then the tangential and normal components of the acceleration are
given by,
aT = v′ =

r ′ ( t )ir ′′ ( t )
r′ (t )

aN = κ v 2 =

r ′ ( t ) × r ′′ ( t )
r′ (t )

where κ is the curvature for the position function. The curvature is actually given in the
next section, but isn’t really needed for the computations here. We just wanted to
acknowledge the first form for the normal acceleration.
There are two formulas to use here for each acceleration and while the second formula
may seem overly complicated it is often the easier of the two. In the tangential
component, v, may be messy and computing the derivative may be unpleasant. In the
normal component we will already be computing both of these quantities in order to get
the curvature and so the second formula in this case is definitely the easier of the two.
Example 1 If the acceleration of an object is given by a = i + 2 j + 6tk find the objects

velocity and position functions given that the initial velocity is v ( 0 ) = j − k and the
initial position is r ( 0 ) = i − 2 j + 3k .
Solution
We’ll first get the velocity. To do this all (well almost all) we need to do is integrate the
acceleration.


321


Calculus II

v ( t ) = ∫ a ( t ) dt
= ∫ i + 2 j + 6tk dt
= t i + 2t j + 3t 2 k + c
To completely get the velocity we will need to determine the “constant” of integration.
We can use the initial velocity to get this.
j − k = v ( 0) = c
The velocity of the object is then,
v ( t ) = t i + 2t j + 3t 2 k + j − k

= t i + ( 2t + 1) j + ( 3t 2 − 1) k

We will find the position function by integrating the velocity function.
r ( t ) = ∫ v ( t ) dt
= ∫ t i + ( 2t + 1) j + ( 3t 2 − 1) k dt

1
= t 2 i + (t 2 + t ) j + (t 3 − t ) k + c
2

Using the initial position gives us,
i − 2 j + 3k = r ( 0 ) = c
So, the position function is,
⎛1
⎞
r ( t ) = ⎜ t 2 + 1⎟ i + ( t 2 + t − 2 ) j + ( t 3 − t + 3) k
⎝2
⎠
Example 2 For the object in the previous example determine the tangential and normal
components of the acceleration.
Solution
There really isn’t much to do here other than plug into the formulas. To do this we’ll
need to notice that,
r ′ ( t ) = t i + ( 2t + 1) j + ( 3t 2 − 1) k
r ′′ ( t ) = i + 2 j + 6tk

Let’s first compute the dot product and cross product that we’ll need for the formulas.
r ′ ( t )ir ′′ ( t ) = t + 2 ( 2t + 1) + 6t ( 3t 2 − 1) = 18t 3 − t + 2


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Calculus II

i
j
k
i
j
r ′ ( t ) × r ′′ ( t ) = t 2t + 1 3t 2 − 1 t 2t + 1
1
2
6t
1
2
= ( 6t )( 2t + 1) i + ( 3t 2 − 1) j + 2tk − 6t 2 j − 2 ( 3t 2 − 1) i − ( 2t + 1) k
= ( 6t 2 + 6t + 2 ) i − ( 3t 2 + 1) j − k
Next, we also need a couple of magnitudes.
r ′ ( t ) = t 2 + ( 2t + 1) + ( 3t 2 − 1) = 9t 4 − t 2 + 4t + 2
2

2

r ′ ( t ) × r ′′ ( t ) =

( 6t

2

+ 6t + 2 ) + ( 3t 2 + 1) + 1 = 45t 4 + 72t 3 + 66t 2 + 24t + 6
2

2

The tangential component of the acceleration is then,
18t 3 − t + 2
aT =
9t 4 − t 2 + 4t + 2
The normal component of the acceleration is,
aN =

45t 4 + 72t 3 + 66t 2 + 24t + 6
9t 4 − t 2 + 4t + 2

=

45t 4 + 72t 3 + 66t 2 + 24t + 6
9t 4 − t 2 + 4t + 2

Curvature
In this section we want to briefly discuss the curvature of a smooth curve (recall that for
a smooth curve we require r ′ ( t ) is continuous and r ′ ( t ) ≠ 0 ). The curvature measures
how fast a curve is changing direction at a given point.
There are several formulas for determining the curvature for a curve. The formal
definition of curvature is,
dT
κ=
ds
where T is the unit tangent and s is the arc length. Recall that we saw in a previous
section how to reparameterize a curve to get it into terms of the arc length.
In general the formal definition of the curvature is not easy to use so there are two
alternate formulas that we can use. Here they are.


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Calculus II

κ=

T ′ (t )

κ=

r′ (t )

r ′ ( t ) × r ′′ ( t )
r′ (t )

3

These may not be particularly easy to deal with either, but at least we don’t need to
reparameterize the unit tangent.
Example 1 Determine the curvature for r ( t ) = t ,3sin t ,3cos t .
Solution
Back in the section when we introduced the tangent vector we computed the tangent and
unit tangent vectors for this function. These were,
r ′ ( t ) = 1,3cos t , −3sin t
T (t ) =

1
3
3
,
cos t , −
sin t
10 10
10

The derivative of the unit tangent is,
T ′ ( t ) = 0, −

3
3
sin t , −
cos t
10
10

The magnitudes of the two vectors are,
r ′ ( t ) = 1 + 9 cos 2 t + 9sin 2 t = 10
T ′ (t ) =

9 9
9
3
+
=
=
10 10
10
10

The curvature is then,

κ=

T ′ (t )
r′ (t )

3
=

10 = 3
10
10

In this case the curvature is constant. This means that the curve is changing direction at
the same rate at every point along it. Recalling that this curve is a helix this result makes
sense.
Example 2 Determine the curvature of r ( t ) = t 2 i + t k .
Solution
In this case the second form of the curvature would probably be easiest. Here are the first
couple of derivatives.
r ′ ( t ) = 2t i + k
r ′′ ( t ) = 2 i

Next, we need the cross product.


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Calculus II

i
r ′ ( t ) × r ′′ ( t ) = 2t
2

j k
0 1
0 0

i
2t
2

j
0
0

=2j
The magnitudes are,
r ′ ( t ) × r ′′ ( t ) = 2

r ′ ( t ) = 4t 2 + 1

The curvature at any value of t is then,

κ=

2

( 4t

3

2

+ 1) 2

There is a special case that we can look at here as well. Suppose that we have a curve
given by y = f ( x ) and we want to find its curvature.
As we saw when we first looked at vector functions we can write this as follows,
r ( x) = x i + f ( x) j
If we then use the second formula for the curvature we will arrive at the following
formula for the curvature.
f ′′ ( x )
κ=
3

(1 + ⎡⎣ f ′ ( x )⎤⎦ )

2 2

Cylindrical Coordinates
As with two dimensional space the standard ( x, y, z ) coordinate system is called the
Cartesian coordinate system. In the last two sections of this chapter we’ll be looking at
some alternate coordinates systems for three dimensional space.
We’ll start off with the cylindrical coordinate system. This one is fairly simple as it is
nothing more than an extension of polar coordinates into three dimensions. Not only is it
an extension of polar coordinates, but we extend it into the third dimension just as we
extend Cartesian coordinates into the third dimension. All that we do is add a z on as the
third coordinate. The r and θ are the same as with polar coordinates.
Here is a sketch of a point in


3

.

325


Calculus II

The conversions are the same conversions that we used back in when we were looking at
polar coordinates. So, if we have a point in cylindrical coordinates the Cartesian
coordinates can be found by using the following conversions.
x = r cos θ
y = r sin θ
z=z
The third equation is just an acknowledgement that the z-coordinate of a point in
Cartesian and polar coordinates is the same.
Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be
found by using the following conversions.
r = x2 + y 2

OR

r 2 = x2 + y 2

⎛ y⎞
⎝ ⎠

θ = tan −1 ⎜ ⎟
x
z=z
Let’s take a quick look at some surfaces in cylindrical coordinates.
Example 1 Identify the surface for each of the following equations.
(a) r = 5
(b) r 2 + z 2 = 100
(c) z = r
Solution


326


Calculus II

(a) In two dimensions we know that this is a circle of radius 5. Since we are now in
three dimensions and there is no z in equation this means it is allowed to vary freely. So,
for any given z we will have a circle of radius 5 centered on the z-axis.

In other words, we will have a cylinder of radius 5 centered on the z-axis.
(b) This equation will be easy to identify once we convert back to Cartesian coordinates.
r 2 + z 2 = 100
x 2 + y 2 + z 2 = 100

So, this is a sphere centered at the origin with radius 10.
(c) Again, this one won’t be too bad if we convert back to Cartesian. For reasons that
will be apparent eventually, we’ll first square both sides, then convert.
z2 = r2
z 2 = x2 + y2

From the section on quadric surfaces we know that this is the equation of a cone.

Spherical Coordinates
In this section we will introduce spherical coordinates. Spherical coordinates can take a
little getting used to. It’s probably easiest to start things off with a sketch.

Spherical coordinates consist of the following three quantities.


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Calculus II

First there is ρ . This is the distance from the origin to the point and we will require
ρ ≥ 0.
Next there is θ . This is the same angle that we saw in polar/cylindrical coordinates. It is
the angle between the positive x-axis and the line above denoted by r (which is also the
same r as in polar/cylindrical coordinates). There are no restrictions on θ .
Finally there is ϕ . This is the angle between the positive z-axis and the line from the
origin to the point. We will require 0 ≤ ϕ ≤ π .
In summary, ρ is the distance from the origin of the point, ϕ is the angle that we need to
rotate down from the positive z-axis to get to the point and θ is how much we need to
rotate around the z-axis to get to the point.
We should first derive some conversion formulas. Let’s first start with a point in
spherical coordinates and ask what the cylindrical coordinates of the point are. So, we
know ( ρ , θ , ϕ ) and what to find ( r , θ , z ) . Of course we really only need to find r and z
since θ is the same in both coordinate systems.
We will be able to do all of our work by looking at the right triangle shown above in our
sketch. With a little geometry we see that the angle between z and ρ is ϕ and so we can
see that,
z = ρ cos ϕ
r = ρ sin ϕ
and these are exactly the formulas that we were looking for. So, given a point in
spherical coordinates the cylindrical coordinates of the point will be,
r = ρ sin ϕ

θ =θ
z = ρ cos ϕ
Note as well that,
r 2 + z 2 = ρ 2 cos 2 ϕ + ρ 2 sin 2 ϕ = ρ 2 ( cos 2 ϕ + sin 2 ϕ ) = ρ 2
Or,

ρ 2 = r2 + z2
Next, let’s find the Cartesian coordinates of the same point. To do this we’ll start with
the cylindrical conversion formulas from the previous section.
x = r cos θ
y = r sin θ
z=z


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Calculus II

Now all that we need to do is use the formulas from above for r and z to get,
x = ρ sin ϕ cos θ
y = ρ sin ϕ sin θ
z = ρ cos ϕ
Also note that since we know that r 2 = x 2 + y 2 we get,

ρ 2 = x2 + y2 + z 2
Converting points from Cartesian or cylindrical coordinates into spherical coordinates is
usually done with the same conversion formulas. To see how this is done let’s work an
example of each.
Example 1 Perform each of the following conversions.
π
⎛
⎞
(a) Convert the point ⎜ 6, , 2 ⎟ from cylindrical to spherical coordinates.
4
⎝
⎠

(

)

(b) Convert the point −1,1, − 2 from Cartesian to spherical coordinates.
Solution
(a) We’ll start by acknowledging that θ is the same in both coordinate systems and so we
don’t need to do anything with that.

Next, let’s find ρ .

ρ = r2 + z2 = 6 + 2 = 8 = 2 2
Finally, let’s get ϕ . To do this we can use either the conversion for r or z. We’ll use the
conversion for z.
z
2
⎛1⎞ π
⇒
⇒
ϕ = cos −1 ⎜ ⎟ =
cos ϕ = =
z = ρ cos ϕ
ρ 2 2
⎝2⎠ 3
1
Notice that there are many possible values of ϕ that will give cos ϕ = , however, we
2
have restricted ϕ to the range 0 ≤ ϕ ≤ π and so this is the only possible value in that
range.

π π⎞
⎛
So, the spherical coordinates of this point will are ⎜ 2 2, , ⎟ .
4 3⎠
⎝
(b) The first thing that we’ll do here is find ρ .

ρ = x2 + y2 + z 2 = 1 + 1 + 2 = 2


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Calculus II

Now we’ll need to find ϕ . We can do this using the conversion for z.
z = ρ cos ϕ

⇒

cos ϕ =

z

ρ

=

− 2
2

⇒

⎛ − 2 ⎞ 3π
⎟=
2 ⎟ 4
⎝
⎠

ϕ = cos −1 ⎜
⎜

As with the last parts this will be the only possible ϕ in the range allowed.

Finally, let’s find θ . To do this we can use the conversion for x or y. We will use the
conversion for y in this case.
y
1
1
2
π
3π
=
=
=
⇒
sin θ =
θ = or θ =
2
4
4
ρ sin ϕ
⎛ 2⎞
2
2⎜
⎟
⎝ 2 ⎠
Now, we actually have more possible choices for θ but all of them will reduce down to
one of the two angles above since they will just be one of these two angles with one or
more complete rotations around the unit circle added on.
We will however, need to decide which one is the correct angle since only one will be.
To do this let’s notice that, in two dimensions, the point with coordinates x = −1 and
y = 1 lies in the second quadrant. This means that θ must be angle that will put the point
3π
into the second quadrant. Therefore, the second angle, θ =
, must be the correct one.
4
⎛ 3π 3π ⎞
The spherical coordinates of this point are then ⎜ 2, , ⎟ .
⎝ 4 4 ⎠
Now, let’s take a look at some equations and identify the surfaces that they represent.
Example 2 Identify the surface for each of the following equations.
(a) ρ = 5

(b) ϕ =

π

3
2π
(c) θ =
3
(d) ρ sin ϕ = 2
Solution
(a) There are a couple of ways to think about this one.
First, think about what this equation is saying. This equation says that, no matter what θ
and ϕ are, the distance from the origin must be 5. So, we can rotate as much as we want
away from the z-axis and around the z-axis, but we must always remain at a fixed
distance from the origin. This is exactly what a sphere is. So, this is a sphere of radius 5
centered at the origin.


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Calculus II

The other way to think about it is to just convert to Cartesian coordinates.
ρ =5

ρ 2 = 25
x 2 + y 2 + z 2 = 25
Sure enough a sphere of radius 5 centered at the origin.
(b) In this case there isn’t an easy way to convert to Cartesian coordinates so we’ll just
need to think about this one a little. This equation says that no matter how far away from
the origin that we move and no matter how much we rotate around the z-axis the point

must always be at an angle of

π

3

from the z-axis.

This is exactly what happens in a cone. All of the points on a cone are a fixed angle from
the z-axis. So, we have a cone whose points are all at an angle of

π

3

from the z-axis.

(c) As with the last part we won’t be able to easily convert to Cartesian coordinates here.
In this case no matter how far from the origin we get or how much we rotate down from
2π
the positive z-axis the points must always form an angle of
with the x-axis.
3

Points in a vertical plane will do this. So, we have a vertical plane that forms an angle of
2π
with the positive x-axis.
3
(d) In this case we can convert to Cartesian coordinates so let’s do that. There are
actually two ways to do this conversion. We will look at both since both will be used on
occasion.

Solution 1
In this solution method we will convert directly to Cartesian coordinates. To do this we
will first need to square both sides of the equation.
ρ 2 sin 2 ϕ = 4
Now, for no apparent reason add ρ 2 cos 2 ϕ to both sides.

ρ 2 sin 2 ϕ + ρ 2 cos 2 ϕ = 4 + ρ 2 cos 2 ϕ
ρ 2 ( sin 2 ϕ + cos 2 ϕ ) = 4 + ρ 2 cos 2 ϕ
ρ 2 = 4 + ( ρ cos ϕ )

2

Now we can convert to Cartesian coordinates.


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Calculus II

x2 + y 2 + z 2 = 4 + z 2
x2 + y2 = 4

So, we have a cylinder of radius 2 centered on the z-axis.
This solution method wasn’t too bad, but it did require some not so obvious steps to
complete.
Solution 2
This method is much shorter, but also involves something that you may not see the first
time around. In this case instead of going straight to Cartesian coordinates we’ll first
convert to cylindrical coordinates.
This won’t always work, but in this case all we need to do is recognize that r = ρ sin ϕ
and we will get something we can recognize. Using this we get,
ρ sin ϕ = 2
r=2
At this point we know this is a cylinder (remember that we’re in three dimensions and so
this isn’t a circle!). However, let’s go ahead and finish the conversion process out.
r2 = 4
x2 + y 2 = 4

So, as we saw in the last part of the previous example it will sometimes be easier to
convert equations in spherical coordinates into cylindrical coordinates before converting
into Cartesian coordinates. This won’t always be easier, but it can make some of the
conversions quicker and easier.
The last thing that we want to do in this section is generalize the first three parts of the
previous example.

ρ =a
ϕ =α
θ =β


sphere of radius a centered at the origin
cone that makes and angle of α with the positive z − axis
vertical plane that makes and angle of β with the positive x − axis

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Calc ii complete

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