CalculusStudyGuide

1
Table of Content
Topic Page Number
Continuity ,Differentiability And
Limits
1
Mean Value Theorem 11
Derivatives 13
Rates 22
Sketching f(x) given f’(x)graph 25
Sketching f’(x) given f(x) 27
Modeling and Optimization 30
Maximum, Minimum and Points
Of Inflection
33
Riemann Sum 40
Trapezoidal Method 43
Integral 45
Areas between 2 graphs 54
Volume 57
Exponential Growth And Decay 62
Non-Traditional problems 65

2
Continuity and Differtiability
Example1: f(x) = IxI
a. Show that f(x) is differentiable and continuous at all x except 0.
– Solution –
Draw the graph y = IxI
Y = IxI = X if x ≥ 0
Y = IxI = - x if x ˂ 0
if x ˃ 0 f(x) = x fˊ(x) =1
if x ˂ 0 f(x) = - x fˊ(x) = -1
Thus f(x) is continuous and differentiable at all x except 0.
b. Is the function continuous and differentiable at x = 0?
– Solution –
x 0⁺ f(x) = 0+ = 0
x 0⁻ f(x) = 0⁻ = 0
Therefore it is continuous at x = 0
Now: h 0⁺
Derivative fˊ(x) =
=
h 0⁻
fˊ(x) = -1
f(0+h) – f(0)
h
I0+hI – I0I
h
I h I
h
h
h
= = 1
f(0+h) – f(0)
h
-h
h
=
= =

3
Therefore it is not differentiable at x = 0.
Example2: a. f(x) = I x2
+2x-3 I
Show that it is continuous everywhere.
f(x) = I (x+3)(x-1) I
f(x) = (x+3)(x-1) x=-3, x=1
−∝ -3 1 ∝
+ _ +
f(x) = x2+2x-3 if x ≤ -3 or x ≥ 1
f(x) = - x2-2x+3 if -3<X<1
at x=-3+ f(-3+)= - x2-2x+3
= -(-3)2-2(-3)+3
=-9+6+3=0
at x=-3- f(-3-)= x2+2x-3
= (-3)2+2(-3)-3
=9-6-3=0
at x=1+ f(x)= x2+2x-3 =(1)2+2(1)-3=0
at x=1- f(x)= - x2-2x+3 =-(1)2-2(1)+3=0
Therefore f(x) is continuous everywhere.

4
b. Is f(x) differentiable everywhere?
– Solution –
Graph of f(x) f(x)
Since at x =-3 and x=1 are vertices, it is not differentiable at x =-3
and x=1
-3 1
-3 1 x

5
-limits-
Example1: Find the limit of
– Solution –
e4(eh-1)
h
as h 0 ,
Take derivative of the numerator and denominator.
=
eh
Answer =e4
lim
h 0
e4
eh
-e4
h
= e4
eh-1
h
eh-1
h
0
eh
1
as h 0
1 ,
eh
-1
h
1
lim

6
Example2: Given that:
Find
– Solution –
Rewrite the problem as
Talce the derivative of the numerator of denominator.
lim
x→ 0
Sin x
x
=1
Limit x+4x2+sinx
3x
X+4x2
sinx
3x 3x
+
lim
x→ 0
Sinx
3x
1
3
= X 1=
1
3
lim
x→ 0
X+4x2
3x
=
0
0
1+8x
3
And as x 0
1 1
3 3
+ =
2
3
Answer is:
1
3
=

7
Example3:
– Solution –
tan x =
sin x
cos x
But
lim
x→ 0
tan x
x
sin x
cos x
x
1
sin x
x cos x
=
lim
x→ 0
sin x
x = 1
sin x
x
1
cos x
x
1
cos 0
=1 x=
= 1 x 1 = 1

8
Example4:
– Solution –
Therefore take the derivative of the numerator of denominator.
1
1
as x 5, 5-4 =1
lim
x→ 5
x – 5
ln(x -4)
5 – 5
ln(5-4)
as x 5,
0
0
=
=
x-4
= x -4

9
Example5: Find lim
X ∞
– Solution –
but x ∞,
x4
(2) = 2x4
sin x ∞, 2x4
∞
Example6: lim
X ∞
Divide the numerator and denominator by x4
as x ∞
2x4-x2-8x
X2
X4
x4
(2-
8x
X4
)
x4(2-
1
X2
8
X3
)
1
X2
8
X3
and 0
2x4
-x2
+8x
-5x4+7
X2
X4
2- +
8x
X4
-5+
7
X4
=
1
X2
2- +
8
X3
-5+
7
X4
=
1
X2
8
X3
7
X4Answer is=
−2
5
, , 0

10
Example7: lim
x ∞
Factor out the x from the numerator of denominator
x2
(3+
|x|=x if x ≥ 0
|x|=-x if x < 0
|x| 3+
6
x
2
X (
5
x
- 2)
sin x +∞ , |x|=x
x 3+
6
x
2
x (
5
x
- 2)
∞,
6
x
2 ,
5
x
0
3x2
+6
x -∞
3x2+6
=5-2x
– Solution –
66
x2
)
x (
5
x
−2)
=
√x2=|x|

=
= =
3+
6
x
2
(
5
x
- 2)
as x
, |x|=-x
Answer =-
√3
(−2)
=
√3
2
Answer is : -
√3
2
If lim
5-2x

11
-Continuity-
Definition: f(x) is continuous at x=a
if lim f(x) = f(a)
x a
Example1: Given the graph of f(x), shown below, determine if f(x)
is continuous at x = -2, x=0, x=3.
– Solution –
When x =-2, there are 2 values for y either 2 or -1, therefore it is
not continuous at x = -2.
if x = 0, there is only 1 value for y which is 1.therefore, it is
continuous at x = 0
if x=3,it is continuous because y is very close to 0 whether we go to
the right of 3 or to the left of 3.

12
Mean value theorem
If y = f(x) is continuous at every point of the closed interval [a, b]
and differentiable at every point of its interior (a, b), then there is
at least one point c in (a, b) at which
fˊ(x) =
f(b)−f(a)
b−a
Example1: show that f(x) = x2 satisfy the mean value theorem on
[0, 2] then find c.
– Solution –
fx is continuous on [0, 2] and differentiable on [0, 2].
Since f(0)=0, f(2)=4, the mean value theorem guarantees a point c
in (0, 2) for which:
fˊ(x) =2x fˊ(c) =2c
2c =
f(2)−f(0)
2−0
2c=
4−0
2−0
2c = 2 c=1
Which means the tangent line to f(x) = x2
at x= 1 has a slope of 2

13
Example2: Determine all the numbers c which satisfy the
conclusions of the mean value theorem for:
f(x)=x3
+2x2
-x on [-1,2]
– Solution –
fˊ(x) = 3x2+4x-1
fˊ(c) = 3c2
+4c-1
3c2+4c-1 =
f(2)−f(−1)
2−1
f(2) = (2)3+(2)2-2 = 14
f(-1) = (-1)3
+2(-1)2
-(-1) = 2
3c2+4c-1 =
14−2
3
=4
3c2+4c-1 =4
3c2+4c-1-4=0
3c2
+4c-5=0 use the quad formula
C=0.7863 Ignore the (-) answer
Since it is not in [-1, 2]

14
Derivatives
Example1: If y=(x3
+1)2
then
dy
dx
=?
– Solution –
Use the chain rule,
yˊ=
𝑑𝑦
𝑑𝑥
=2(x3
+1).3x2
=6x2
(x3
+1)
Example2: if y =
2x+3
3x+2
,
dy
dx
=?
– Solution –
Use the quotient rule,
yˊ=
yˊ=
−5
(3x+2)
2
Example3: if f(x) =ln(x+4+e-3x
), then fˊ(0) =?
– Solution –
If y=In u yˊ=
uˊ
u
fˊ(x)=
fˊ (0)=
2(3x+2)-3(2x+3) 6x+4-6x-9
(3x+2)2
(3x+2)2
=
1-3e-3x
X+4+e-3x
1-3e0
0+4+e0
=
1-3
4+1
=
-2
5

15
Example4: If y=x2sin 2x, then
dy
dx
=?
– Solution –
Use the product rule,
y= u times v yˊ=uˊv+vˊu
yˊ=2x sin 2x +2cos 2x (x2)
=2x sin 2x +2x2cos 2x
Take 2x as a common factor
2x ( sin 2x +x cos 2x)
Example5: what is the slope of the tangent line to the curve?
3y2
-2x2
=6-2xy at (3, 2)
– Solution –
Slope of the tangent line is the derivative. Use implicit
derivation.
6yyˊ-4x=6-(2y+yˊ.2x)
6yyˊ-4x=6-2y-2xyˊ
6yyˊ+2xyˊ=6-2y+4x
yˊ (6y+2x) = 6-2y+4x
yˊ=
6−2y+4x
6y+2x
but x=3, y=2
=
6−2(2)+4(3)
6(2)+2(3)
=
6−4+12
12+6
=
14
18
=
7
9

16
Example6: If y =
lnx
x
, then
dy
dx
=?
– Solution –
Use the quotient rule,
If y =
u
v
yˊ=
yˊ=
UˊV-VˊU
V2
1
x
(x)-1(lnx) 1-lnx
x2
x2

17
Example7: An equation of the line tangent to the graph of
f(x) =x(1-2x)3
at(1, -1) is:
– Solution –
Slope=fˊ(1)
fˊ(x) =1(1-2x)3 +3(1-2x)2 (-2)(x)
=(1-2x)3 -6x(1-2x)2
fˊ(x) = (1-2)3 -6(1-2)2
=(-1)3-6(-1)2
=-1-6 =-7
Equation of the line is:
y= mx+b
y=-7x+b x=1, y=-1
-1=-7+b b = 6
y=-7x+6
Example8: f(x) =√2𝑥 , then fˊ(x) =?
– Solution –
f(x) =√2𝑥=(2x)1/2
use the chain rule
fˊ(x) =
1
2
(2x)−1/2
. (2) =(2x)−1/2
=
x with 2, fˊ(2)=
1
(2x)1/2
1
√2x
=
1
√2.2
=
1
2
Replace

18
Example9: If y=2cos (
x
2
) then
y=2cos
x
2
yˊ=-2sin
x
2
times derivative of
x
2
which is
1
2
yˊ=-2.
1
2
Sin
x
2
= - Sin
x
2
yˊˊ =
Example10:
d
dx
(xlnx
) =?
– Solution –
y= xlnx
Place ln in front
Iny= Inxlnx
Iny= Inx.lnx =In2x Take derivatives
yˊ
y
=2lnx.
1
x
Cross multiply
yˊx= y (2lnx)
yˊ=
2ylnx
x
but y = xlnx
d2
y
dx2
= ?
d2
y
dx2
= -
1
2
Cos
x
2
– Solution –
2xlnx
.lnx
x
yˊ=

19
Example11: An equation for a tangent to the graph
y= arc sin
x
2
at the origin is:
– Solution –
Slope = fˊ(0)
If y= arc sin u
yˊ=
yˊ=derivative of
x
2
. derivative of arc sin
𝑥
2
yˊ=
1
2
yˊ=slope=
1
2
.1 =
1
2
y=
1
2
.x
1
1-u2
1
1-x2/4
at x=0

20
Example12: Find yˊ if y=√sin x
– Solution –
Y= (sinx)1/2 use the chain rule
yˊ=
1
2
(sinx)-1/2
. cos x
yˊ=
1
2
cos x .
1
√sin x
=
cos x
2√sin x
Example13: y=sin3
(3x2
-2x+1)
– Solution –
Use the power rule and chain rule
yˊ=3sin2
(3x2
-2x+1).cos (3x2
-2x+1) (6x-2)
yˊ=3(6x-2) sin2
(3x2
-2x+1).cos (3x2
-2x+1)

21
Example14:
d
dx
(lne2x)
– Solution –
y=lne2x
If y=lnu yˊ=
uˊ
u
yˊ=
Example15: If sin x= ey
what is
dy
dx
, 0< x < π
– Solution –
sin x = ey
place ln in the front
ln sin x= lney
ln sin x= ylne but lne=1
ln sin x= y if y = ln u yˊ=
uˊ
u
yˊ=
cos x
sin x
=cot x
Example16:If y=2cos(
x
2
), then
y=2cos
x
2
use the chain rule
yˊ=derivative of cos
x
2
times the derivative of
x
2
=2(-sin
x
2
)(
1
2
)=−sin
x
2
yˊˊ=
2e2x
e2x
=2
d2
y
dx2
?
– Solution –
d2
y
dx2
1
2
cos
x
2=

22
Example17: h(x) = (arctan(x3
+1) +2x)4
find hˊ(x)
– Solution –
Use the chain rule
hˊ(x)=4(arctan(x3
+1) +2x)3
× derivative arctan(x3
+1) +2x
Which is
hˊ(x)=4(arctan(x3
+1) +2x)3
.
hˊ(x)=4(arctan(x3
+1) +2x)3
.
1
(1+x3
)
× 3x2
+2
3x2
(1+x3
)
+2=
3x2
1+x3
+2
3x2
+2+x3
1+x3

23
-Rates-
Background information: Rate of change means derivative '' yˊ ''
or
dy
dx
.
Example1: A rectangular water tank is being filled at the
constant rate of 30 liter/sec .the base of the tank has
dimensions w=2m, L=30m.what is the rate of change of the
height of the water in the tank in cm/sec.
– Solution –
Constant rate of 30 lit/sec
lit is volume ; sec is time
dv
dt
= 30 lit/sec = 30000 cm3
/sec
V=L× W × H
Since L and W are constant, height is the only one that is
changing. Take the derivative.
dv
dt
=L× W ×
dH
dt
substitute.
30000= 300× 200 ×
dH
dt
30000= 60000×
dH
dt
dH
dt
= 0.5 cm/sec.

24
Example2: when a circular shield of bronze is heated over a fire,
its radius increases at the rate of 0.2 cm/sec.at what rate is the
shield’s area increasing when the radius is 40 cm?
– Solution –
Radius increasing at the rate of 0.2 cm/sec means
𝑑𝑟
𝑑𝑡
=0.2
A circle=𝜋 r2
Take the derivative with respect to time
𝑑𝐴
𝑑𝑡
= 𝜋 (2r)
𝑑𝑟
𝑑𝑡
. Substitute
𝑑𝐴
𝑑𝑡
= 3.14× 2 ×40×0.2= 50.40 cm3
/sec

25
Example3: A right cylindrical tank is filled with water. The tank
stands upright has radius =30cm.how fast does the height of
water in the tank drop when the when the water is being
drained at the rate of 20cm3
/sec.
– Solution –
Cm3 is the unit of volume
Sec = = = = time
Rate means derivative
Given:
𝑑𝑣
𝑑𝑡
=20
Vcyl =𝜋r2h
Radius =30=constant
The only variable is the height
𝑑𝑉
𝑑𝑡
= 𝜋r2 𝑑ℎ
𝑑𝑡
substitute
-20=3.14×302×
𝑑ℎ
𝑑𝑡
-20=2826×
𝑑ℎ
𝑑𝑡
𝑑ℎ
𝑑𝑡
= - 0.00707cm/sec

26
Sketching a graph of f(x)
Given the graph fˊ(x)
Example1: sketch a graph of f(x),given the graph of fˊ(x)
5
-1 1
– Solution –
fˊ(x) graph from (-∞, −1) is under the x-ax is fˊ(x) is < 0
(decreasing) fˊ(x)=0 at x=-1 it’s a vertex
fˊ(x)=0 at x=1
from -1< x < 0 fˊ(x)> 0 f(x) is increasing
from 0< x < 1, fˊ(x)> 0 f(x) is increasing
-1
-1

27
0 1
for x>1 fˊ(x) < 0 (under the x-ax is)
the function is decreasing.
1
Put all the mini graph together.
y
-1 1 x

28
Sketching a graph of fˊ(x)
Given the graph f(x)
Background information: At the vertices fˊ(x)=0 (that is the point
is on the x-axis) if the function is increasing which means fˊ(x)> 0
The graph is on top if the x-axis. If fˊ(x)< 0 graph is the below
the x-axis.
Example1: Sketch the graph of fˊ(x) for the graph below.
– Solution –
fˊ(x)> 0 from x=-5 to x=-3; (-5,-3) the vertex is at x=-3
y
x
-5 -3
(5, 0)
f(x)
(-3, 2)
(-5, 2)
(2,-5)

29
Now from x=-3 to x=2, fˊ(x)< 0, the graph is under the x-axis
y
x
At x=2, it is a vertex
fˊ(x) is on the x-axis.
from x>2 fˊ(x)> 0
f(x) is on top of the x-axis
1 2
Let’s combine all the graph of fˊ(x)
y
x
.
2-3

30
Example2: Given fˊ(x) graph, graph f(x)
graph f(x)
– Solution –
When fˊ(x) is under the x-axis
It is negative the function is decreasing
When fˊ(x) is over the x-axis it is +
The function is increasing.
At x = 0 f(x) is constant.

31
-Modeling and optimization-
Example1: Find 2numbers whose sum is 20 and whose product is
as large as possible.
– Solution –
If one number is x
2nd
number is 20-x
their product: x (20-x) is to be maximized.
Let y = x (20-x) = 20x-x2
Maximum yˊ= 0
yˊ= 20-2x=0 x=10
Numbers are 10, 10

32
Example2: An open box is to be made by cutting congruent
squares of side length x from the corners of a 20 by 25 inch
sheet of tin and bending up the sides. How large should the
squares be to make the box hold as much as possible?
– Solution –
x
Since x is cut from each corner.
L=20-2x
W=25-2x
V(x)=l×w×h=x(20-2x)(25-2x)
=x(500-90x+4x2
)
=500x-90x2
+4x3
Maximum vˊ=0
500-180x+12x2
=0
X1=3.68 and x2=11.32. x1 is in the domain. Since l and w
cannot be (-)
20ˊˊ
25ˊˊ

33
Example3: suppose revenue r(x) =9x and the cost C(x) =x3
-6x2
+15x,
where x represents 1000ˊs of units.is there a production level that
maximizes the profit? What is it?
– Solution –
Profit = cost - revenue
=(x3-6x2+15x)-(9x)
= x3-6x2+6x
Max derivative = 0
3x2
-12x+6=0. Divide by 3
X2-4x+2=0. Use the quad formula
X1=0.586 thousand units or
X=3.41 thousand units. If you can draw the graph.
Max profit is when f(x) is on top of C(x)
at x=3.41
X=3.41
X=.586
Cost
c(x)

34
Maximum, Minimum and points of Inflection
Example1: The graph of y= 5 𝑥4
– 𝑥5
has a point of inflection at
-Solution-
Y= 5 𝑥4
- 𝑥5
𝑦′
= 20𝑥3
- 5 𝑥4
𝑦′
′= 60𝑥2
- 20𝑥3
𝑦′
′=0 60𝑥2
- 20𝑥3
= 0
= 20𝑥2(3 − 𝑥) = 0
X = 0 and x = 3
at x = 0 y = 5(0)4
- (0)5
= 0
at x = 3 y =5(3)4
- (3)5
= 162
(0, 0) (3, 162)

35
Example2: The function defined by f(x) = √3 cosx + 3 sinx has an
amplitude of:
-Solution-
Amplitude = Maximum 𝑓′
(x) = 0
𝑓′
(x) = −√3 sinx + 3 cosx = 0
3 cosx =√3 sinx Divide by cosx
3 =√3
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
tanx =
3
√3
= √3
Example3:If f(x) = x +
1
𝑥
,then the set of values for which fˊ
increases is:
-Solution-
f increases when f ‘(x) > 0
f‘(x) = 1 -
1
𝑥2 =
𝑥2−1
𝑥2
𝑥2
– 1 = 0 when x = ± 1
-∞ -1 1 ∞
+ - +
(-∞,-1)U (1, ∞)

36
Example4: What are the values of x for which the function f
defined by f (x) = ( 𝑥2
− 3 ) 𝑒−𝑥
is increasing?
-Solution-
f(x) is increasing if f’ ( x ) > 0
f’(x) = 2 × 𝑒−𝑥
- 𝑒−𝑥
(𝑥2
− 3)
= 𝑒−𝑥
(2𝑥 - 𝑥2
+ 3)
f’( x )=0 only if 𝑥2
+2 𝑥 + 3 = 0
Since 𝑒−2𝑥
is always > 0
𝑥2
- 2𝑥 - 3 = 0
(X-3) (x + 1) = 0 x = 3, x = - 1
-∞ -1 3 ∞
- + -
f’( x ) > 0 when - 1 < x < 3

37
Example5: Find all points of inflections
-Solution-
y=e-x2
yˊ=-2 x e-x2
yˊˊ use the product rule;
yˊˊ=-2 e-x2
+(-2x e-x2
(-2x))
=-2 e-x2
+4x
2
e-x2
Set yˊˊ to 0
-2 e-x2
+4x
2
e-x2=0
e-x2(-2+4x2
)=0
e-x2 is always >0
-2+4x2
=0
4x2=2 x2=
1
2
, x=
√2
2
When x=
√2
2
y=e-1/2 or
1
√𝑒
x=
−√2
2
y=e1/2 =
1
√𝑒
points of inflection (
−√2
2
,
1
√𝑒
) , (
√2
2
,
1
√𝑒
)
y=e-x2

38
Example6: Determine the concavity of
y = 3 + sin x on (0, 2𝜋 )
-Solution-
Y’ = cosx
Y’’ = - sinx
0
𝜋
2
𝜋 3𝜋 2𝜋
Sign of y₺ - - + +
It is concave downward (o , 𝜋 )
It is concave upward (𝜋 , 2𝜋 )

39
Example7: You have been asked to design a one - liter oil can
shaped like a right circular cylinder. What dimensions will use the
least material?
-Solution-
1 liter = 1000𝑐𝑚3
πr2h = 1000
Surface area of a cylinder = 2 π𝑟2
+ 2 πrh
h =
1000
π𝑟2
S.A = 2π𝑟2
+ 2πr
1000
π𝑟2
S.A = 2π𝑟2
+
2000
𝑟
Minimum = derivative of S.A = 0
𝑑(𝑆,𝐴)
𝑑𝑟
=
4𝜋𝑟
1
+
−2000
𝑟2 = 0
=
4𝜋𝑟3−2000
𝑟2 = 0
= 4𝜋𝑟3
− 2000 = 0 𝑟3
=
2000
4𝜋
=
2000
4 𝑥 3.14
𝑟3
= 159.23 r = ∛159.23 =5.42 ‘’

40
Example8: Find the extreme value
f (x) = ln
𝑥
1+𝑥2
-Solution-
f ‘ (x) =
1(1+𝑥2)−2𝑥.𝑥
𝑥
1+𝑥2
=
1+𝑥2− 2𝑥2
𝑥
1+𝑥2
f ‘ (x) =
− 𝑥2+ 1
𝑥
1+𝑥2
=
− 𝑥2+ 1
𝑥(1+𝑥2)
f ‘ (x) = 0 when the numerator = 0
- 𝑥2
+ 1 = 0 x = ± 1
-∞ -1 1 ∞
- + -
fˊ(x)
Minimum at (-1, ln
1
2
), Maximum (1 ,ln
1
2
)

41
Riemann Sum
Example1: Approximate the area under the curve
f (x) = 𝑥2
+ 2 - 2 ≤ 𝑥 ≤ 1
Using 6 sub- intervals and right end points
-Solution-
a = -2, b = 1, ∆ x =
𝑏−𝑎
𝑥
=
1−(−2)
6
= 0.5
f (x ) = 𝑥2
+ 2
Area = (Σ f (xi) ∆ x) = ∆ x (f (-1.5) + f (-1) + f (-0.5) +f (0) + f (0.5) +f (1)
f (-1.5) = (-1.5)2 +2= 2.25 + 2 = 4.25
f (-1) = (-1)2 + 2 = 3
f (-0.5) = (-0.5)2 + 2 = 2.25
f (0) = (0)2 + 2 = 2
f (0.5) = (0.5)2 + 2 =2.25
f (1) = (1)2 + 2 = 3
Area = 0.5 (4.25 + 3+2.25+2+2.25+3)
= 0.5 (1675) = 8.375

42
Example2: Approximate the area under the curve
f (x) = √ 𝑥 + 1 -1 ≤ x ≤ 0
Using 4 sub-intervals and left end points
a =- 1, b= 0 ∆ x =
𝑏−𝑎
𝑥
=
0−(−1)
4
= 0.25
Area = = ( Σ f (xi) ∆ x)
= ∆ x (f (-1) + f (-0.75) + f (-0.5) + f(-0.25)
f (-1) =√-1+1 = 0
f (-0.75) =√-0.75+1 = √0.25 = 0.5
f (-0.5) = √ -0.5+1= √0.5 = 0.707
f (-0.25) = √ -0.25+1= √0.75 = 0.866
Area = 0.25 (0+0.5+0.707+0.866)
= 0.25(2.073) = 0.518
Example3: A rectangular canal, 5m wide and 100 m long has an
even bottom. Depth measurements are taken at every 20m along
the length of the canal. Use these depth measurements to
construct a Riemann sum using right end points to estimate the
volume of water in the canal.
Distance 0m 2m 40m 60m 80m 100m
Depth 2m 1.6m 1.8m 2.1m 2.1m 1.9m

43
-Solution-
Volume = width x Area
Width = 5m
Area = ∑ 𝑓 ( 𝑥) ∆x
Area = 20 (f (20) + f (40) +f (60) +f (80) +f (100))
= 20 (1.6+1.8+2.1+2.1+1.9)
=20(9.5) = 190
Volume = 5x190 = 950 m3

44
Trapezoidal Method
Of integration
Example1: Use the trapezoidal rule, with n= 4 to estimate
∫ 𝑥
2
1
2 dx
-Solution-
Break [1,2] into 4 equal lengths
For n = 4 h =
(2−1)
4
= 0.25
T =
ℎ
2
[y0 + 2y1 +2y2 + 2y3 + y4] y y=x2
T =
0.25
2
[ 1 1.25 1.50 1.75 2 x
Y0 = (1)2 = 1 y1 = x2 = (1.25)2 = 1.565
Y2 = (1.50)2 = 2.25
Y3 = (1.75)2 = 3.0625
Y4 = (2)2 = 4
T =
0.25
2
[1+2(1.565) +2+ (2.25) +2(3.0625) +4]
T = 0.125 [18.755] = 2.344

45
Example2: An observer measures the outside temperature every
hour from noon to midnight, recording the temperatures in the
following table:
Time N 1 2 3 4 5 6 7 8 9 10 11 M
Temp 63 65 66 68 70 69 68 68 65 64 62 58 55
What was the average temperature for the 12 hour period?
-Solution-
Average temp =
1
𝑏−𝑎
∫ 𝑓 ( 𝑥) 𝑑𝑥
𝑏
𝑎
We can approximate it by using the trapezoidal rule:
T =
ℎ
2
[y0 + 2y1 + 2y2 + 2y3 + …………..+y12]
Make h = 1
=
1
2
[63+2x65+2x66+2x68+2x70+2x64+2x68+2x68+2x65+2x64+2x62+2x58
+55]
= 782

46
Integrals
Example1: Find ∫
𝑥𝑑𝑥
√3𝑥2+5
-Solution-
let u = 3𝑥2
+5
du = 6 x d 𝑥 d 𝑥 =
𝑑𝑢
6𝑥
∫
𝑥 .𝑑𝑥
6 𝑥√ 𝑢
=
1
6
∫
𝑑𝑢
√𝑢
=
1
6
∫ 𝑢 -1/2
dx
=
1
6
(
1
−1
2
+1
𝑢−
1
2
+1
) =
1
6
(2𝑢1/2
)
=
1
3
𝑢1/2
=
1
3
(3𝑥2
+5 )
1
2 + c
Example2: ∫ (2𝑘𝑥 − 𝑥2
) 𝑑𝑥
𝑘
0
= 18, then k = ?
-Solution-
∫ (2𝑘𝑥 − 𝑥2)𝑑𝑥
𝑘
0
= ∫
𝑘
0
2𝑘𝑥 𝑑𝑥 - ∫ 𝑥2
𝑑𝑥
= 2k ∫
𝑘
0
𝑥 𝑑𝑥 - ∫ 𝑥2
𝑑𝑥
= 2k.
1
2
𝑥2
-
1
3
x3
= k𝑥2
-
1
3
x3
F (k) = k(k)2
-
1
3
(k)3
F (0) = 0
F(k) – F (0) = k3
-
1
3
k3
=
−2
3
k3
−2
3
k3
= 18 -2k3
= 54
K3 = - 27 𝑘 = √−27
3
= - 3

47
Example3: ∫
cos 𝑥
√1+ 𝑠𝑖𝑛𝑥
𝜋
2
0
d𝑥 =
-Solution-
Let u = 1+ sin𝑥
d𝑢 = cos 𝑥 d𝑥 d 𝑥 =
𝑑𝑢
cos 𝑥
∫
cos 𝑥
√ 𝑢
𝜋
2
0
.
𝑑𝑢
cos 𝑥
= ∫
𝑑𝑢
√𝑢
𝜋
2
0
= ∫ 𝑢 −1/2
𝜋
2
0
d𝑢 =
1
−1
2
+1
𝑢
−1
2 +1 = 2 𝑢
1
2
= 2(1+sin 𝑥 )
1
2
F(
𝜋
2
) = 2 ( 1+ sin
𝜋
2
)
1
2 = 2( 2)
1
2 = 2√2
F (0) = 2(1 + sin0)
1
2 = 2 ( 1)
1
2 = 2
F (
𝜋
2
) - F ( 0 ) = 2√2 - 2
Example4: If
𝑑𝑦
𝑑𝑡
= 𝑘𝑦 and K is a non-Zero constant, then y could be
-Solution-
𝑑𝑦
𝑑𝑡
=
𝑘𝑦
1
cross multiply
𝑑𝑦 = 𝑘𝑦 𝑑𝑡 Divide by y
𝑑𝑦
𝑦
= 𝑘𝑑𝑡 Integrate
∫
𝑑𝑦
𝑦
= ∫ 𝑘𝑑𝑡
lny = 𝑘𝑡
𝑒 𝑘𝑡
= 𝑦 or 𝑦 =𝑒 𝑘𝑡
+ c

48
Example5: What is the average value of y = 𝑥2
√𝑥3
+ 1
On the interval [0, 2]
-Solution-
Average value =
1
2−0
∫ ( 𝑥2
√𝑥
2
0
3 + 1 ) 𝑑𝑥
Let u = 𝑥3 + 1
𝑑𝑢 = 3𝑥2
d 𝑥 𝑑𝑥 =
𝑑𝑢
3𝑥2
= ∫ 𝑥
2
0
2
√ 𝑢.
𝑑𝑢
3𝑥2
=
1
3
∫ √ 𝑢. 𝑑𝑢
=
1
3
∫ 𝑢
1
2 =
1
3
1
1
2
+ 1
𝑢
1
2
+ 1.
=
1
3
.
2
3
𝑢
3
2
=
2
9
𝑢
3
2 =
2
9
(𝑥3
+ 1 )3/2
=
2
9
√(𝑥3 + 1 )3
F (2) =
2
9
√ 93
=
2
9
x 27 = 6
F (0)=
2
9
√1 =
2
9
F (2) – f (0) = 6 -
2
9
= 5
7
9
=
52
9
Average value =
1
2
.
52
9
=
26
9

49
Example6: Let F (x) anti derivative of
If f (1) = 0 then f (9) =?
dx let 𝑢 = 𝑙𝑛𝑥
𝑑𝑢
1
𝑥
𝑑𝑥 and 𝑑𝑥 = 𝑥𝑑𝑢
x 𝑑𝑢
= ∫ 𝑢3
𝑑𝑢 =
1
4
𝑢4
+ c
=
1
4
(𝑙𝑛𝑥)4
+ c
f (1 ) = 0 0 =
1
4
(ln1+ c , c = 0 )
f (𝑥) =
1
4
(lnx)4
f (9)= =
1
4
(ln9)4
=
1
4
(2.14)4
=
1
4
(23.30) = 5.82
Example7: ∫(sin(2𝑥) + cos(2𝑥) 𝑑𝑥 =
-Solution-
∫(𝑠𝑖𝑛 2𝑥 𝑑𝑥 + ∫ cos2𝑥 𝑑𝑥
=-
1
2
cos2𝑥 +
1
2
sin2𝑥 + 𝑐
(lnx)3
x
(lnx)3
x
-Solution-
u3
x

50
Example8: Compute ∫ .
1
2
0
-Solution-
It looks like ∫
1
1+𝑡2
= arc tanx + c
Let 2t = 𝑥 4𝑡2
=x2
𝑑𝑥 = 2𝑑𝑡
∫
4 𝑑𝑡
1+𝑥2
1
2
0
= ∫
𝑑𝑥 . 4
2 (1+𝑥2 )
1
2
0
= 2 ∫
𝑑𝑥
( 1+ 𝑥2 )
= 2 arc tan𝑥
= 2 arc tan 2 𝑡
F (
1
2
)= 2 arc tan2.
1
2
= 2 arc tan 1
= 2 x
𝜋
4
=
𝜋
2
F ( 0 ) = 2 arc tan 2 ( 0 ) = 0
F (
1
2
) – F ( 0 ) =
𝜋
2
4
1+4t2
dt

51
Example9:∫ √𝑥2 − 2𝑥 + 1
1
0
𝑑𝑥
-Solution-
∫ √( 𝑥 − 1 )
1
0
2
= ∫
1
0
x – 1 𝑑𝑥
Since x-1<0 when x is between 0 and 1
= ∫
1
0
(1 − x) 𝑑𝑥 =∫ 1 𝑑𝑥-∫
1
0
𝑥𝑑𝑥
= 𝑥 −
1
2
x2
F ( 1 ) = 1 −
1
2
(1)2
=
1
2
F ( 0 ) = 0 −
1
2
(0)2
=0
F ( 1 ) – F ( 0 ) =
1
2
- 0 =
1
2
Example10:∫
cos 𝑥
sin 𝑥
𝜋
2
𝜋
4
d𝑥
-Solution-
Lot sin𝑥 = u du = 𝑐𝑜𝑠𝑥 . 𝑑𝑥 𝑑𝑥 =
𝑑𝑢
cos 𝑥
∫
cos 𝑥
𝑢
𝜋
2
𝜋
4
.
𝑑𝑢
𝑐𝑜𝑠𝑥
= ∫
𝑑𝑢
𝑢
𝜋
2
𝜋
4
= lnu
= 𝑙𝑛𝑠𝑖𝑛𝑥
F (
𝜋
2
) = 𝑙𝑛𝑠𝑖𝑛
𝜋
2
= 0
F (
𝜋
4
) = 𝑙𝑛𝑠𝑖𝑛
𝜋
4
= 𝑙𝑛
√2
2
F (
𝜋
2
) – F (
𝜋
4
) = 0- 𝑙𝑛
√2
2
=-
ln √2
2

52
= - ln
2
1
2
2
= - (ln 2
1
2 - 𝑙𝑛2 ) = -
1
2
𝑙𝑛2 + 𝑙𝑛2
Example11:∫
𝑑𝑥
√1+𝑥
8
0
-Solution-
∫
𝑑𝑥
√1+𝑥
8
0
Let 1 + 𝑥= 𝑢
𝑑𝑢 = 𝑑𝑥
∫
𝑑𝑥
√𝑢
8
0
= ∫ 𝑢
−1
2
8
0
𝑑𝑢 =
1
−1
2
+ 1
𝑢-1/2+1
= 2𝑢
1
2 = 2 (1 + 𝑥 )1/2
F (8) = 2 (1 + 8)1/2
= 2 (9)1/2
=2 √9 = 6
F (0) = 2 (1 + 0)1/2
= 2 (1) = 2
F (8) – F (0) = 6 – 2= 4

53
Example12: ʃ0
1 𝑥2
𝑥2+1
𝑑𝑥
– Solution –
Use long division 1
X
2+1 X
2
− 𝑥2+1
−1
X
2+1
ʃ0
1
( 1 −
1
𝑥2 + 1
) 𝑑𝑥
ʃ0
1
1 dx- ʃ0
1 1
𝑥2+ 1
dx
= x- arc tanx
F (1) = 1 - arc tanx 1 = 1 -
𝜋
4
F (0) = 0 – arctan0 = 0
F (1) – F (0) = 1 -
𝜋
4
=1-

54
Example13:∫ 𝑙𝑛𝑥 𝑑𝑥
– Solution –
∫ 𝑙𝑛𝑥 𝑑𝑥
Use integration by parts
∫ 𝑢𝑑𝑣 = u v - ∫ vdu
Let u = 𝑙𝑛𝑥 and dv = d x v = x
∫ 𝑙𝑛𝑥𝑑𝑥 = lnx ( 𝑥 )- ∫( 𝑥.
1
𝑥
) 𝑑𝑥
= 𝑥 𝑙𝑛𝑥 - ∫ 1 𝑑𝑥 = x 𝑙𝑛𝑥 –𝑥 + c

55
Area Between 2 Graphs
Example1: Find the area bounded by f (x) = 𝑥2
+ 2
And g (x) = - x, x = 0 and x = 1
First you need to find which graph is on top.
X2+2
-x
Since x2 + 2 is on top, the area of x2 + 2 should be the first
Area ∫[ (𝑥2
+ 2 ]– ( −𝑥 ) ] 𝑑𝑥
ʃ( 𝑥2
+ 2 + 𝑥) 𝑑𝑥
ʃ0
1 ( 𝑥2
+ 𝑥 + 2 ) 𝑑𝑥 =
1
3
x3+
1
2
x2 + 2x
F (1) =
1
3
+
1
2
+ 2 = 2
5
6
F = (0) = 0, Area = F (1) – F (0) = 2
5
6

56
Example2: Find the area bounded by f (x) = 2 – x2 and g (x) = x
– Solution –
Graph the 2 function to see which graph is on top Y x
Since 2 - x2 is on top X
Area ∫ ( (2 − 𝑥2 )– 𝑥 )𝑑𝑥
= ∫ ( 2 − 𝑥2
− 𝑥 ) 𝑑𝑥 -X2+2
= 2 x-
1
3
𝑥 3-
1
2
x2
The lower and upper bound is where the 2 graphs intersect
2 – x2 = x
2 - x2 – x = 0 ( multiply by ( - )
X2 + x – 2 = 0 factor
(x + 2) (x – 1) = 0
X = -2 and x = 1
F (1) = 2 -
1
3
.
1
2
= 1.17
F (- 2) = 2 (- 2) -
1
3
(- 2)3 -
1
2
(- 2) 2 = - 2.34
Area F (1) – F (-2) = 1.17 – (- 2.34) = 3.51

57
Example3: Find the area between sin x and cos x
𝜋/2
Y = sin x
Y = cos x
cosx
The graph show that sin x is on the top
∫(sin 𝑥 − 𝑐𝑜𝑠𝑥 ) 𝑑𝑥 = −𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥
They intersect at 𝑠𝑖𝑛𝑥 = 𝑐𝑜𝑠𝑥
At
𝜋
4
and
5𝜋
4
sinx
5𝜋
4
𝜋
4
F (
5𝜋
4
) = − 𝑐𝑜𝑠
5𝜋
4
− 𝑠𝑖𝑛
5𝜋
4
=-
− √2
2
-
− √2
2
=
+ √2
2
+
√2
2
=
2√2
2
= √2 = 1.414
F (
𝜋
4
) = −𝑐𝑜𝑠
𝜋
4
− 𝑠𝑖𝑛
𝜋
4
= - 0.707 – 0.707 =- 1.414
Area = 1.414- (-1.414) = 2.828
– Solution –

58
Volume
The volume of a solid of known integrable cross section area A (x)
from x = a to x = b is the integral from a to b
v= ʃ𝑎
𝑏
A ( 𝑥 ) 𝑑𝑥
Example1: Slicing Method: A pyramid 3m high has congruent
triangular sides and a square base that is 3m on each side. Each cross
section of pyramid parallel to the base is a square. Find the volume of
the pyramid.
– Solution –
A (x) = x2
V= ʃ0
3
𝑥2 𝑑𝑥 =
1
3
x 3
= 9m3
3
x x
x
3

59
Example2: Circular cross section:
The region between the graph f (x) = 2 + 𝑥𝑐𝑜𝑠𝑥 and the x- axis over the
interval [- 2, 2] is revolved about the x – axis to generate a solid. Find
the volume of solid.
– Solution –
y
2 f(x)
x
-2
A = 𝜋 (𝑓(x))2
V= ʃ−2
2
𝐴( 𝑥) 𝑑𝑥
= ʃ𝜋(2 + 𝑥𝑐𝑜𝑠𝑥)−2
2 2
= ʃ−2
2
𝜋 ( 4 + 4 𝑥 𝑐𝑜𝑠𝑥 +x2cos2x) dx
= 52.42 units cubed.

60
Example3: washer cross section:
The region in the first quadrant enclosed by the x-axis and the graphs of
y = cosx and y = sinx is revolved about the x-axis to form a solid. Find
its volume.
– Solution –
Graph of y =𝑐𝑜𝑠𝑥
at
𝜋
4
, 𝑐𝑜𝑠𝑥 =
√2
2
Graph of y = 𝑠𝑖𝑛𝑥 x
y 𝜋/2 𝜋 3𝜋/2 2𝜋
1 -1
𝜋/2 𝜋 3𝜋/2 2𝜋 x
-1
The cosine function is on top of the sine function.
The cross section is an washer.
V= ʃ0
𝜋
4
𝜋(𝑐𝑜𝑠2x – 𝑠𝑖𝑛2x)𝑑𝑥
= 𝜋 ʃ0
𝜋
4
cos2x dx 𝜋/4
=
1
2
𝑠𝑖𝑛2𝑥 =
𝜋
2
.
1

61
Example4: Using cylindrical shells. The region bounded by the curve y
=√ 𝑥 , the x-axis, and the line x = 4 is revolved about the X- axis to
generate a solid. Find the volume of the solid.
– Solution –
Thickness= dy
Radius= y
1 2 3
X=4
Since it rotates around
The x- axis solve for x.
y = √ 𝑥 square both sides.
y2 =x
v= ʃ0
2
2𝜋 (times) (shell height) dy
= ʃ0
2
2𝜋(𝑦)(4 − 𝑦2)dy
=2𝜋 ʃ0
2
(4𝑦 − 𝑦3)dy
=2𝜋 × 4 ǀ0
2 𝑦2
2
-
1
4
y4
= 8𝜋
dy
x=y2
y = √ 𝑥

62
Exponential Growth And Decay
Example1: Use the fact that world population was 2560 million
people in 1950 and 3040 million in 1960.To model the population of the
world in the second half of the 20th century .What is the relative growth
rate K? Use the model to estimate the world population in 1993?
– Solution –
Growth rate is proportion to the population rise?
𝑑𝑝
𝑑𝑡
= K P
P (t) = p (0) ekt
3040 = 2560 ekt
From 1950 to 1960 t = 10 years
3040 = 2560 e10k Divided both side by 2560
1.1875 = e10k
Take natural log of both sides ln 1.1875 = 10 k lne.But lne = 1
K =
ln 1.1875
10
= 0.01718
Therefore P (t) = 2560e0.01718t
To estimate the world population in 1993.
There are 43 years from 1590 to 1993.
Replace t with 43

63
P(43)=2560e(0.01718 x 43)
use a calculator
= 5503 million
Example2: The half-life t
1
2
of radium 226 is 1590 years.
a, A sample of radium 226 has a mass of 100mg.Find the formula for
the mass of radium that remains after 10 years.
– Solution –
m (t) = m (0) ekt
Half-life means you lose
1
2
of the mass. Final amount =
1
2
of 100 = 50
50 = 100 ekt
But t = 1590 years
50 = 100e1590k Divide both side by 100
0.5 = e1590k Take natural log of both sides
ln 0.5= 1950 k lne , but lne = 1
K=
ln 0.5
1590
= -0.000435

64
Therefore m(t) = m(0)e (-0.000435t)
m(t) =100e(-0.000435t)
b, when will the mass be reduced to 30 mg?
– Solution –
30 = 100e - 0.000435t Divided by 100
0.3= e - 0.000435t
Take ln of both sides.
Ln 0.3 = -0.000435 t lne , lne = 1
t=
𝑙𝑛0.3
−0.000435
= 2767.75 years

65
Non-traditional Problems
Example1: Let p(x) = 2x3 + kx +1 Find K if the remainder of the
division of p(x) by x-2 is 10.
– Solution –
2 2 0 k 1
4 8 2k+16
4 k+8 2k+17
Remainder is 2k+ 17 = 10
-17 = -17
2k = - 7 k = -7/2
Example2: The average rate of change of the Function f defined by f(x)
= sin 𝑥 + 𝜋
On the closed interval [0,𝜋 ] is.
– Solution –
Average rate of change is the slole
𝑓 (𝜋)−𝑓 (0)
𝜋−0
=
(sin(𝜋)+𝜋)− 0
𝜋
But 𝑠𝑖𝑛𝜋 = 0 , =1

66
Example3: Functions f, g and h are defined as follows
g(x) = f (x2)
f(x) = h (x3 + 1)
hˊ(x) = 2x+1
g’(x)=?
– Solution –
h’(x)=2x+1 h (x) = ∫(2𝑥 + 1 )
= 𝑥2
+ 𝑥 + 𝑐
𝑓(𝑥) = ℎ (𝑥3
+ 1)
∫( 𝑥) = ( x3 + 1 )2 + (x3 + 1)
= x6 + 2x3+ 1+x3 +1= x6 + 3x3 + 2
g (x) = f (x2)
= (x2)6 + 3 x6 + 2
=x12 +3x6+2
g’ (x)= 12x11 +18x5

67
Example4:ln( 𝑥 − 2) < 0 if and only if.
– Solution –
𝑙𝑛𝑢 is < 0 when
0< 𝑢 < 1
0 < 𝑥 − 2 < 1
+2 +2 +2
2 < 𝑥 < 3
Example5: If the function f is defined by by x = x5 -1,Then 𝑓 -1 , the
inverse function of f, is defined by 𝑓 -1(x)=
– Solution –
x = x5 - 1 replace x with y and y with x
f(x) = y5 – 1 solve for y5
y5 = x - 1
y = f-1 (x) = √ 𝑥 + 1
5

68
Example6: If f(x)’ and g’ (x) exist and f’(x)>g’(x) for all real x, then the
graph of y = f (x) and the graph of y = g(x)
A, Intersect exactly once.
B, Intersect more than once.
C, Do not intersect.
– Solution –
Pick a function where f’(x) > g’ (x)
f(x) = x2 + 5x f’(x) = 2x + 5
g(x) = x2+2x g’(x) = 2x+2
The graphs intersect once.

69
Example7: If f is continuous and if f’(x) =f(x) for all real numbers x,
then,
∫ 𝑓 2( 𝑥) 𝑑𝑥 =
3
1
– Solution –
The function where f(x) = f’(x) is ex
∫ 𝑒2𝑥3
1
=
1
2
𝑒2𝑥
F(3) =
1
2
𝑒6
F(1) =
1
2
e2
F(3) – F (1) =
1
2
𝑒6
-
1
2
e2 =
1
2
(𝑒6
- e2)
=
1
2
(f (6)-f (2)

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