1. .
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
. . . . . .
2. 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
3. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 3 / 41
4. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 4 / 41
5. 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(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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 5 / 41
6. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 6 / 41
7. Example
∫ 1
4
Estimate dx using the midpoint rule and four divisions.
0 1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
8. Example
∫ 1
4
Estimate dx using the midpoint rule and four divisions.
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
11. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 8 / 41
12. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 9 / 41
13. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
14. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b
f(x) dx
a
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
15. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b ∫ c
f(x) dx f(x) dx
a b
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
16. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b ∫ c ∫ c
f(x) dxf(x) dxf(x) dx
a a b
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
17. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
18. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b
f(x) dx
a
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
19. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ c
f(x) dx =
b∫
b
− f(x) dx
c
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
20. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ c ∫ c
f(x) dx f(x) dx =
a b∫
b
− f(x) dx
c
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
21. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 11 / 41
22. Comparison Properties of the Integral
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
23. Comparison Properties of the Integral
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
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b ∫ b
f(x) dx ≥ g(x) dx
a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
24. Comparison Properties of the Integral
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
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
25. Estimating an integral with inequalities
Example
∫ 2
1
Estimate dx using Property 8.
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
26. Estimating an integral with inequalities
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
27. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 14 / 41
28. 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 . . .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 15 / 41
29. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
30. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
31. Proving the Second FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
n
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
32. Proving the Second FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
33. Proving the Second FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
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 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
34. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
35. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
36. Proof continued
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
37. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
38. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
39. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
40. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
41. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
42. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
43. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
44. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
45. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
46. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
47. Proof continued
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 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
48. Proof continued
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)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
49. Proof Completed
We have shown for each n,
Sn = F(b) − F(a)
Which does not depend on n.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
50. 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→∞
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
51. Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
52. Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
53. Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
54. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
55. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
1
.
−1 1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
56. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
1
.
−1 1
Solution
∫ 1 [ ]1 ( [ )]
x3 1 1 4
A=2− x dx = 2 − 2
=2− − − =
−1 3 −1 3 3 3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
57. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 22 / 41
59. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
60. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
61. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
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)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
62. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
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)
(π )
=4 −0
4
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
63. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
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)
(π )
=4 −0 =π
4
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
64. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 25 / 41
65. Estimating an integral with inequalities
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 26 / 41
66. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
67. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx = ln x|2
1
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
68. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
69. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
70. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 28 / 41
71. The Integral as Total Change
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:
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
72. The Integral as Total Change
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:
Theorem
If v(t) represents the velocity of a particle moving rectilinearly, then
∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
73. The Integral as Total Change
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:
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
74. The Integral as Total Change
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:
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
75. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 30 / 41
76. 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).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
77. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
78. My first table of integrals
.
∫ ∫ ∫
[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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 32 / 41
79. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 33 / 41
80. Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
81. Computing Area with integrals
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
82. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
83. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
.
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
84. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
Instead compute the area as
∫ π/2
π
− sin y dy .
2 0
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
85. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
Instead compute the area as
∫ π/2
π π π/2
− sin y dy = −[− cos x]0 .
2 0 2
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
86. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
Instead compute the area as
∫ π/2
π π π/2 π
− sin y dy = −[− cos x]0 = −1 .
2 0 2 2
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
87. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
88. 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.
Solution
∫ 3
Consider (x − 1)(x − 2) dx.
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
89. Graph
y
. x
1 2 3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 37 / 41
90. 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.
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).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
91. 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.
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
92. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 39 / 41
93. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 40 / 41
94. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 41 / 41