Lesson 25: Evaluating Definite Integrals (Section 041 slides)

.

Section 5.3
Evaluating Definite Integrals

V63.0121.041, Calculus I

New York University

December 6, 2010

Announcements

Today: Section 5.3
Wednesday: Section 5.4
Monday, December 13: Section 5.5
”Monday,” December 15: Review and Movie Day!
Monday, December 20, 12:00–1:50pm: Final Exam
. . . . . .

Announcements

Today: Section 5.3
Wednesday: Section 5.4
Monday, December 13:
Section 5.5
”Monday,” December 15:
Review and Movie Day!
Monday, December 20,
12:00–1:50pm: Final Exam

. . . . . .

V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41

Objectives

Use the Evaluation
Theorem to evaluate
definite integrals.
Write antiderivatives as
indefinite integrals.
Interpret definite integrals
as “net change” of a
function over an interval.

. . . . . .


Outline

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

Examples

The Integral as Total Change

Indefinite Integrals
My first table of integrals

Computing Area with integrals

. . . . . .



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(ci ) ∆x
a n→∞
i=1

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

. . . . . .


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
x0 = 0, x1 = , x2 = , 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
x0 = 0, x1 = , x2 = , 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
x0 = 0, x1 = , x2 = , 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 (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
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b

. . . . . .


Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
a a b

y

.
a b c x
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

∫ b
f(x) dx
a

.
a b c x
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

∫ b ∫ c
f(x) dx f(x) dx
a b

.
a b c x
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

∫ b ∫ c ∫ c
f(x) dxf(x) dxf(x) dx
a a b

.
a b c x
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

.
a c x
b
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

∫ b
f(x) dx
a

.
a c x
b
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

∫ c
f(x) dx =
b∫
b
− f(x) dx
c
.
a c x
b
. . . . . .


Theorem
∫ c ∫ b ∫ c
a a b

y

∫ c ∫ c
f(x) dx f(x) dx =
a b∫
b
− f(x) dx
c
.
a c x
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 4

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

. . . . . .



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

. . . . . .



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

. . . . . .



Theorem
∫ b
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

. . . . . .


Estimating an integral with inequalities

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

. . . . . .



Example
∫ 2
1
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

. . . . . .


Outline


Examples




. . . . . .


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 usual.
n

. . . . . .



b−a
n
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 )
xi − xi−1

. . . . . .



b−a
n
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 )
xi − xi−1

Or
f(ci )∆x = F(xi ) − F(xi−1 )

. . . . . .


Proof continued

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

. . . . . .


Proof continued

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

Form the Riemann Sum:

. . . . . .


Proof continued

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1

. . . . . .


Proof continued

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

∑
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 ))

. . . . . .


Proof continued

We have for each i

f(ci )∆x = F(xi ) − F(xi−1 )

∑
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)

. . . . . .


Proof Completed

We have shown for each n,

Sn = F(b) − F(a)

Which does not depend on n.

. . . . . .


Proof Completed

We have shown for each n,

Sn = F(b) − F(a)

Which does not depend on n.
So in the limit
∫ b
f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a)
a n→∞ n→∞

. . . . . .


Computing area with the Second FTC

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

.

. . . . . .



Example

Solution

∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .

. . . . . .



Example

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 the line y = 1.

. . . . . .


Example

1

.
−1 1

. . . . . .


Example

1

.
−1 1

Solution

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


Computing an integral we estimated before

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

. . . . . .


Example
∫ 1
4
0 1 + x2

Solution

1 2 3 4
x0 = 0, x1 = , x2 = , 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
. . . . . .



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
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

. . . . . .



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

. . . . . .


Outline


Examples




. . . . . .



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
. . . . . .


Outline


Examples




. . . . . .


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

. . . . . .


.

∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
xn+1
n
x 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

. . . . . .


Outline


Examples




. . . . . .



Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex .

. . . . . .



Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex .

Solution
The answer is ∫ 4
4
ex dx = ex |1 = e4 − e.
1

. . . . . .



Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.

. . . . . .



Example

Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.

.
x
1
. . . . . .



Example

Solution
∫ 1
0
arcsin.
Instead compute the area as
∫ π/2
π
− sin y dy .
2 0
x
1
. . . . . .



Example

Solution
∫ 1
0
arcsin.
∫ π/2
π π π/2
− sin y dy = −[− cos x]0 .
2 0 2
x
1
. . . . . .



Example

Solution
∫ 1
0
arcsin.
∫ π/2
π π π/2 π
− sin y dy = −[− cos x]0 = −1 .
2 0 2 2
x
1
. . . . . .


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, 1)
0
and (2, 3], and negative on (1, 2).

. . . . . .


Example

Solution
∫ 3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
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 [ ]2 [ ]3
= − 1 3
3x
3 2
2x + 2x − 1 x3 − 3 x2 + 2x + 1 x3 − 3 x2 + 2x
3 2 3 2
( ) 0 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
x4 1 1
3
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.

. . . . . .


Summary

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 25: Evaluating Definite Integrals (Section 041 slides)

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Lesson 25: Evaluating Definite Integrals (Section 041 slides)