Lesson 27: Evaluating Definite Integrals

Section 5.3
Evaluating Deﬁnite Integrals

V63.0121.027, Calculus I

December 3, 2009

Announcements
Final Exam is Friday, December 18, 2:00–3:50pm
Final is cumulative; topics will be represented roughly
according to time spent on them

. .
Image credit: docman
. . . . . .

Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral

Estimating the Definite Integral
The Midpoint Rule
Comparison Properties of the Integral

Evaluating Definite Integrals
Examples

The Integral as Total Change

Indefinite Integrals
My first table of integrals

Computing Area with integrals

. . . . . .

The definite integral as a limit

Definition
If f is a function defined on [a, b], the definite integral of f from a
to b is the number
∫ b ∑ n
f(x) dx = lim f (c i ) ∆x
a n→∞
i=1

b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point
n
in [xi−1 , xi ].

Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists and is the same for any choice of ci .
a

. . . . . .

Notation/Terminology

∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integration (a is the lower limit and b
the upper limit)
dx — ??? (a parenthesis? an inﬁnitesimal? a variable?)
The process of computing an integral is called integration

. . . . . .

Properties of the integral

Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant.
Then
∫ b
1. c dx = c(b − a)
a
∫ b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
a a a
∫ b ∫ b
3. cf(x) dx = c f(x) dx.
a a
∫ b ∫ b ∫ b
4. [f(x) − g(x)] dx = f(x) dx − g(x) dx.
a a a

. . . . . .

More Properties of the Integral

Conventions: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
This allows us to have
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b

. . . . . .

Deﬁnite Integrals We Know So Far

If the integral computes
an area and we know
the area, we can use
that. For instance,
y
.
∫ 1√
π
1 − x2 dx =
0 2

By brute force we
.
computed x
.
∫ 1 ∫ 1
1 1
x2 dx = x3 dx =
0 3 0 4

. . . . . .

The Midpoint Rule
Given a partition of [a, b] into n pieces, let ¯i be the midpoint of
x
[xi−1 , xi ]. Deﬁne
∑n
Mn = f(¯i ) ∆x.
x
i =1

. . . . . .

The Midpoint Rule
Given a partition of [a, b] into n pieces, let ¯i be the midpoint of
x
[xi−1 , xi ]. Deﬁne
∑n
Mn = f(¯i ) ∆x.
x
i =1

y
.
. = x2
y
Example
∫ 1
Cmpute M2 for x2 dx.
0 .
Solution

( ) ( )
1 1 2 1 3 2 5 .
M2 = · + · = . .
2 4 2 4 16 x
.
1
.
2
. . . . . .

Why are midpoints often better?
∫ 1
1
Compare L2 , R2 , and M2 for x2 dx = :
0 3
y
.
. = x2
y

. .
x
.
1
.
2

. . . . . .

∫ 1
1
0 3
y
.
( )2 . = x2
y
1 1 1 1
L2 = · (0)2 + · = = 0.125
2 2 2 8

.
. .
x
.
1
.
2

. . . . . .

∫ 1
1
0 3
y
.
( )2 . = x2
y
1 2 1 1 1
L2 = · (0) + · = = 0.125
2 2 2 8
( )2
1 1 1 5
R2 = · + · (1)2 = = 0.625
2 2 2 8 .
. .
x
.
1
.
2

. . . . . .

∫ 1
1
0 3
y
.
( )2 . = x2
y
1 1 1 1
L2 = · (0)2 + · = = 0.125
2 2 2 8
( )2
1 1 1 5
R2 = · + · (1)2 = = 0.625
2 2 2 8 .
( )2 ( )2
1 1 1 3 5
M2 = · + · = = 0.3125 . .
2 4 2 4 16 x
.
1
.
2

. . . . . .

∫ 1
1
0 3
y
.
( )2 . = x2
y
1 1 1 1
L2 = · (0)2 + · = = 0.125
2 2 2 8
( )2
1 1 1 5
R2 = · + · (1)2 = = 0.625
2 2 2 8 .
( )2 ( )2
1 1 1 3 5
M2 = · + · = = 0.3125 ..
2 4 2 4 16 x
.
1
.
2
Where f is monotone, one of Ln and Rn will be too much, and the
other two little. But Mn allows overestimates and underestimates
to counteract.

. . . . . .

Example
∫ 1
4
Estimate dx using the midpoint rule and four divisions.
0 1 + x2

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution
Dividing up [0, 1] into 4 pieces gives

1 2 3 4
x 0 = 0 , x1 = , x 2 = , x3 = , x4 =
4 4 4 4
So the midpoint rule gives
( )
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

1 2 3 4
x 0 = 0 , x1 = , x 2 = , x3 = , x4 =
4 4 4 4
( )
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
( )
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

1 2 3 4
x 0 = 0 , x1 = , x 2 = , x3 = , x4 =
4 4 4 4
( )
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
( )
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
150, 166, 784
= ≈ 3.1468
47, 720, 465

. . . . . .

Theorem
Let f and g be integrable functions on [a, b].
6. If f(x) ≥ 0 for all x in [a, b], then
∫ b
f(x) dx ≥ 0
a

. . . . . .

The integral of a nonnegative function is nonnegative

Proof.
If f(x) ≥ 0 for all x in [a, b], then for any number of divisions n
and choice of sample points {ci }:
n
∑ n
∑
Sn = f(ci ) ∆x ≥ 0 · ∆x = 0
i=1 ≥0 i =1

Since Sn ≥ 0 for all n, the limit of {Sn } is nonnegative, too:
∫ b
f(x) dx = lim Sn ≥ 0
a n→∞
≥0

. . . . . .

Theorem
∫ b
f(x) dx ≥ 0
a

7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b ∫ b
f(x) dx ≥ g(x) dx
a a

. . . . . .

The deﬁnite integral is “increasing”

Proof.
Let h(x) = f(x) − g(x). If f(x) ≥ g(x) for all x in [a, b], then
h(x) ≥ 0 for all x in [a, b]. So by the previous property
∫ b
h(x) dx ≥ 0
a

This means that
∫ b ∫ b ∫ b ∫ b
f(x) dx − g(x) dx = (f(x) − g(x)) dx = h(x) dx ≥ 0
a a a a

So ∫ ∫
b b
f(x) dx ≥ g(x) dx
a a

. . . . . .

Theorem
∫ b
f(x) dx ≥ 0
a

7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b ∫ b
f(x) dx ≥ g(x) dx
a a

8. If m ≤ f(x) ≤ M for all x in [a, b], then
∫ b
m(b − a) ≤ f(x) dx ≤ M(b − a)
a
. . . . . .

Bounding the integral using bounds of the function

Proof.
If m ≤ f(x) ≤ M on for all x in [a, b], then by the previous property
∫ b ∫ b ∫ b
m dx ≤ f(x) dx ≤ M dx
a a a

By Property 1, the integral of a constant function is the product of
the constant and the width of the interval. So:
∫ b
m(b − a) ≤ f(x) dx ≤ M(b − a)
a

. . . . . .

Example
∫ 2
1
Estimate dx using Property 8.
1 x

. . . . . .

Example
∫ 2
1
Estimate dx using Property 8.
1 x
Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
we have ∫ 2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or ∫ 2
1 1
≤ dx ≤ 1
2 1 x

. . . . . .

Socratic proof

The deﬁnite integral of
velocity measures
displacement (net
distance)
The derivative of
displacement is velocity
So we can compute
displacement with the
deﬁnite integral or the
antiderivative of velocity
But any function can be
a velocity function, so
...

. . . . . .

Theorem of the Day

Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another function F,
then ∫ b
f(x) dx = F(b) − F(a).
a

. . . . . .

Theorem of the Day

Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another function F,
then ∫ b
f(x) dx = F(b) − F(a).
a

Note
In Section 5.3, this theorem is called “The Evaluation Theorem”.
Nobody else in the world calls it that.

. . . . . .

Proving the Second FTC

b−a
Divide up [a, b] into n pieces of equal width ∆x = as
n
usual. For each i, F is continuous on [xi−1 , xi ] and differentiable
on (xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with

F(xi ) − F(xi−1 )
= F′ (ci ) = f(ci )
x i − x i −1

Or
f(ci )∆x = F(xi ) − F(xi−1 )

. . . . . .

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

Form the Riemann Sum:
n
∑ n
∑
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i =1 i =1

= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
= F(xn ) − F(x0 ) = F(b) − F(a)

. . . . . .

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

Form the Riemann Sum:
n
∑ n
∑
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i =1 i =1

= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
= F(xn ) − F(x0 ) = F(b) − F(a)

See if you can spot the invocation of the Mean Value Theorem!

. . . . . .

We have shown for each n,

Sn = F(b) − F(a)

so in the limit
∫ b
f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a)
a n→∞ n→∞

. . . . . .

Example
Find the area between y = x3 and the x-axis, between x = 0 and
x = 1.

.

. . . . . .

Example
x = 1.

Solution

∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .

. . . . . .

Example
x = 1.

Solution

∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .

Here we use the notation F(x)|b or [F(x)]b to mean F(b) − F(a).
a a

. . . . . .

Example
Find the area enclosed by the parabola y = x2 and y = 1.

. . . . . .

Example

.
1
.

. . .
−
. 1 1
.

. . . . . .

Example

.
1
.

. . .
−
. 1 1
.

Solution

∫ 1 [ ]1 [ ( )]
2 x3 1 1 4
A=2− x dx = 2 − =2− − − =
−1 3 −1 3 3 3

. . . . . .

Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
(π )
=4 −0
4

. . . . . .

Example
∫ 1
4
0 1 + x2
Solution

∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
(π )
=4 −0 =π
4

. . . . . .

Example
∫ 2
1
Evaluate dx.
1 x

. . . . . .

Example
∫ 2
1
Evaluate dx.
1 x
Solution

∫ 2
1
dx
1 x

. . . . . .

Example
∫ 2
1
Evaluate dx.
1 x
Solution

∫ 2
1
dx = ln x|2
1
1 x

. . . . . .

Example
∫ 2
1
Evaluate dx.
1 x
Solution

∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1

. . . . . .

Example
∫ 2
1
Evaluate dx.
1 x
Solution

∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2

. . . . . .


Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a

or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramiﬁcations:

. . . . . .


∫ b
F′ (x) dx = F(b) − F(a),
a


Theorem
If v(t) represents the velocity of a particle moving rectilinearly,
then ∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0

. . . . . .


∫ b
F′ (x) dx = F(b) − F(a),
a


Theorem
If MC(x) represents the marginal cost of making x units of a
product, then
∫ x
C(x) = C(0) + MC(q) dq.
0

. . . . . .


∫ b
F′ (x) dx = F(b) − F(a),
a


Theorem
If ρ(x) represents the density of a thin rod at a distance of x from
its end, then the mass of the rod up to x is
∫ x
m(x) = ρ(s) ds.
0

. . . . . .

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation and
integration, we use the indeﬁnite integral notation
∫
f(x) dx

for any function whose derivative is f(x).

. . . . . .

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation and
integration, we use the indeﬁnite integral notation
∫
f(x) dx

for any function whose derivative is f(x). Thus
∫
x2 dx = 1 x3 + C.
3

. . . . . .

My ﬁrst table of integrals
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
x n +1
xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx
n+1 ∫
∫
1
ex dx = ex + C dx = ln |x| + C
x
∫ ∫
ax
sin x dx = − cos x + C ax dx = +C
ln a
∫ ∫
cos x dx = sin x + C csc2 x dx = − cot x + C
∫ ∫
sec2 x dx = tan x + C csc x cot x dx = − csc x + C
∫ ∫
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
1 − x2
∫
1
dx = arctan x + C
1 + x2
. . . . . .

Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis,
and the vertical lines x = 0 and x = 3.

. . . . . .

Example

Solution ∫
3
Consider (x − 1)(x − 2) dx.
0

. . . . . .

Graph
y
.

. . . . x
.
1
. 2
. 3
.

. . . . . .

Example

Solution ∫
3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on
0
[0, 1) and (2, 3], and negative on (1, 2).

. . . . . .

Example

Solution ∫
3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on
0
[0, 1) and (2, 3], and negative on (1, 2). If we want the area of
the region, we have to do
∫ 1 ∫ 2 ∫ 3
A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
0 1 2
[1 ]1 [1 3 ]2 [1 ]3
= 3 x3 − 3 x2 + 2x 0 − 3x − 3 x2 + 2x + 3 3 2
3 x − 2 x + 2x
( 2 ) 2 1 2
5 1 5 11
= − − + = .
6 6 6 6

. . . . . .

Interpretation of “negative area” in motion

There is an analog in rectlinear motion:
∫ t1
v(t) dt is net distance traveled.
t0
∫ t1
|v(t)| dt is total distance traveled.
t0

. . . . . .

What about the constant?

It seems we forgot about the +C when we say for instance
∫ 1 1
3 x4 1 1
x dx = = −0=
0 4 0 4 4

But notice
[ 4 ]1 ( )
x 1 1 1
+C = + C − (0 + C) = + C − C =
4 0 4 4 4

no matter what C is.
So in antidifferentiation for deﬁnite integrals, the constant is
immaterial.

. . . . . .

What have we learned today?

The second Fundamental Theorem of Calculus:
∫ b
f(x) dx = F(b) − F(a)
a

where F′ = f.
Deﬁnite integrals represent net change of a function over an
interval. ∫
We write antiderivatives as indeﬁnite integrals f(x) dx

. . . . . .

Lesson 27: Evaluating Definite Integrals

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