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Lesson 23: Antiderivatives
1. Section 4.7
Antiderivatives
V63.0121.002.2010Su, Calculus I
New York University
June 16, 2010
Announcements
Quiz 4 Thursday on 4.1–4.4
. . . . . .
2. Announcements
Quiz 4 Thursday on
4.1–4.4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 2 / 33
3. Objectives
Given a ”simple“
elementary function, find a
function whose derivative
is that function.
Remember that a function
whose derivative is zero
along an interval must be
zero along that interval.
Solve problems involving
rectilinear motion.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 3 / 33
4. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 4 / 33
5. What is an antiderivative?
Definition
Let f be a function. An antiderivative for f is a function F such that
F′ = f.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 5 / 33
6. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
7. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
8. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
9. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d
(x ln x − x)
dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
10. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d 1
(x ln x − x) = 1 · ln x + x · − 1
dx x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
11. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d 1
(x ln x − x) = 1 · ln x + x · − 1 = ln x
dx x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
12. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?
Solution
d
dx
1
(x ln x − x) = 1 · ln x + x · − 1 = ln x
x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
13. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x y. Then f is continuous on
[x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y)
such that
f(y) − f(x)
= f′ (z) =⇒ f(y) = f(x) + f′ (z)(y − x)
y−x
But f′ (z) = 0, so f(y) = f(x). Since this is true for all x and y in (a, b),
then f is constant.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 7 / 33
14. When two functions have the same derivative
Theorem
Suppose f and g are two differentiable functions on (a, b) with f′ = g′ .
Then f and g differ by a constant. That is, there exists a constant C
such that f(x) = g(x) + C.
Proof.
Let h(x) = f(x) − g(x)
Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
So h(x) = C, a constant
This means f(x) − g(x) = C on (a, b)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 8 / 33
15. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 9 / 33
16. Antiderivatives of power functions
y
.
.(x) = x2
f
Recall that the derivative of a
power function is a power
function.
Fact (The Power Rule)
If f(x) = xr , then f′ (x) = rxr−1 .
.
x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
17. Antiderivatives of power functions
′
y f
. . (x) = 2x
.(x) = x2
f
Recall that the derivative of a
power function is a power
function.
Fact (The Power Rule)
If f(x) = xr , then f′ (x) = rxr−1 .
.
x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
18. Antiderivatives of power functions
′
y f
. . (x) = 2x
.(x) = x2
f
Recall that the derivative of a
power function is a power
function. F
. (x) = ?
Fact (The Power Rule)
If f(x) = xr , then f′ (x) = rxr−1 .
.
x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
19. Antiderivatives of power functions
′
y f
. . (x) = 2x
.(x) = x2
f
Recall that the derivative of a
power function is a power
function. F
. (x) = ?
Fact (The Power Rule)
If f(x) = xr , then f′ (x) = rxr−1 .
So in looking for antiderivatives
.
of power functions, try power x
.
functions!
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
20. Example
Find an antiderivative for the function f(x) = x3 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
21. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
22. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
23. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
r − 1 = 3 =⇒ r = 4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
24. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
25. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
So F(x) = x is an antiderivative.
4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
26. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
So F(x) = x is an antiderivative.
4
Check: ( )
d 1 4 1
x = 4 · x4−1 = x3
dx 4 4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
27. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
So F(x) = x is an antiderivative.
4
Check: ( )
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
28. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
So F(x) = x is an antiderivative.
4
Check: ( )
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4
Any others?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
29. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
So F(x) = x is an antiderivative.
4
Check: ( )
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4
1 4
Any others? Yes, F(x) = x + C is the most general form.
4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
30. Fact (The Power Rule for antiderivatives)
If f(x) = xr , then
1 r+1
F(x) = x
r+1
is an antiderivative for f…
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 12 / 33
31. Fact (The Power Rule for antiderivatives)
If f(x) = xr , then
1 r+1
F(x) =
x
r+1
is an antiderivative for f as long as r ̸= −1.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 12 / 33
32. Fact (The Power Rule for antiderivatives)
If f(x) = xr , then
1 r+1
F(x) =
x
r+1
is an antiderivative for f as long as r ̸= −1.
Fact
1
If f(x) = x−1 = , then
x
F(x) = ln |x| + C
is an antiderivative for f.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 12 / 33
33. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
34. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
ln |x|
dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
35. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d d
ln |x| = ln(x)
dx dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
36. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d d 1
ln |x| = ln(x) =
dx dx x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
37. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
38. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
ln |x|
dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
39. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d d
ln |x| = ln(−x)
dx dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
40. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d d 1
ln |x| = ln(−x) = · (−1)
dx dx −x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
41. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d d 1 1
ln |x| = ln(−x) = · (−1) =
dx dx −x x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
42. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x) =
1
−x
· (−1) =
1
x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
43. What's with the absolute value?
{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x
If x 0,
d
dx
ln |x| =
d
dx
ln(−x) =
1
−x
· (−1) =
1
x
We prefer the antiderivative with the larger domain.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
44. Graph of ln |x|
y
.
. f
.(x) = 1/x
x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 14 / 33
45. Graph of ln |x|
y
.
F
. (x) = ln(x)
. f
.(x) = 1/x
x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 14 / 33
46. Graph of ln |x|
y
.
. (x) = ln |x|
F
. f
.(x) = 1/x
x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 14 / 33
47. Combinations of antiderivatives
Fact (Sum and Constant Multiple Rule for Antiderivatives)
If F is an antiderivative of f and G is an antiderivative of g, then
F + G is an antiderivative of f + g.
If F is an antiderivative of f and c is a constant, then cF is an
antiderivative of cf.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 15 / 33
48. Combinations of antiderivatives
Fact (Sum and Constant Multiple Rule for Antiderivatives)
If F is an antiderivative of f and G is an antiderivative of g, then
F + G is an antiderivative of f + g.
If F is an antiderivative of f and c is a constant, then cF is an
antiderivative of cf.
Proof.
These follow from the sum and constant multiple rule for derivatives:
If F′ = f and G′ = g, then
(F + G)′ = F′ + G′ = f + g
Or, if F′ = f,
(cF)′ = cF′ = cf
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 15 / 33
49. Antiderivatives of Polynomials
Example
Find an antiderivative for f(x) = 16x + 5.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 16 / 33
50. Antiderivatives of Polynomials
Example
Find an antiderivative for f(x) = 16x + 5.
Solution
1 2
The expression x is an antiderivative for x, and x is an antiderivative
2
for 1. So
( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the antiderivative of f.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 16 / 33
51. Antiderivatives of Polynomials
Example
Find an antiderivative for f(x) = 16x + 5.
Solution
1 2
The expression x is an antiderivative for x, and x is an antiderivative
2
for 1. So
( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the antiderivative of f.
Question
Why do we not need two C’s?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 16 / 33
52. Antiderivatives of Polynomials
Example
Find an antiderivative for f(x) = 16x + 5.
Solution
1 2
The expression x is an antiderivative for x, and x is an antiderivative
2
for 1. So
( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the antiderivative of f.
Question
Why do we not need two C’s?
Answer . . . . . .
A combination of two arbitrary constants is still an arbitrary constant.16 / 33
V63.0121.002.2010Su, Calculus I (NYU)
Section 4.7 Antiderivatives June 16, 2010
54. Exponential Functions
Fact
If f(x) = ax , f′ (x) = (ln a)ax .
Accordingly,
Fact
1 x
If f(x) = ax , then F(x) = a + C is the antiderivative of f.
ln a
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
55. Exponential Functions
Fact
If f(x) = ax , f′ (x) = (ln a)ax .
Accordingly,
Fact
1 x
If f(x) = ax , then F(x) = a + C is the antiderivative of f.
ln a
Proof.
Check it yourself.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
56. Exponential Functions
Fact
If f(x) = ax , f′ (x) = (ln a)ax .
Accordingly,
Fact
1 x
If f(x) = ax , then F(x) = a + C is the antiderivative of f.
ln a
Proof.
Check it yourself.
In particular,
Fact
If f(x) = ex , then F(x) = ex + C is the antiderivative of f.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
57. Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 18 / 33
58. Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
This is not obvious. See Calc II for the full story.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 18 / 33
59. Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
This is not obvious. See Calc II for the full story.
ln x
However, using the fact that loga x = , we get:
ln a
Fact
If f(x) = loga (x)
1 1
F(x) = (x ln x − x) + C = x loga x − x+C
ln a ln a
is the antiderivative of f(x).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 18 / 33
60. Trigonometric functions
Fact
d d
sin x = cos x cos x = − sin x
dx dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 19 / 33
61. Trigonometric functions
Fact
d d
sin x = cos x cos x = − sin x
dx dx
So to turn these around,
Fact
The function F(x) = − cos x + C is the antiderivative of f(x) = sin x.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 19 / 33
62. Trigonometric functions
Fact
d d
sin x = cos x cos x = − sin x
dx dx
So to turn these around,
Fact
The function F(x) = − cos x + C is the antiderivative of f(x) = sin x.
The function F(x) = sin x + C is the antiderivative of f(x) = cos x.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 19 / 33
63. More Trig
Example
Find an antiderivative of f(x) = tan x.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
64. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
65. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
66. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d 1 d
= · sec x
dx sec x dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
67. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d 1 d 1
= · sec x = · sec x tan x
dx sec x dx sec x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
68. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d 1 d 1
= · sec x = · sec x tan x = tan x
dx sec x dx sec x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
69. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
·
d
sec x dx
sec x =
1
sec x
· sec x tan x = tan x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
70. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
Check
d
dx
=
1
·
d
sec x dx
sec x =
1
sec x
· sec x tan x = tan x
More about this later. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
71. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 21 / 33
72. Finding Antiderivatives Graphically
Problem
Below is the graph of a function f. Draw the graph of an antiderivative
for f.
y
.
.
. . . = f(x)
y
. . . . . . .
x
.
1
. 2
. 3
. 4
. 5
. 6
.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 22 / 33
73. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
′
. . . . . . .. = F
f
y
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
74. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
′
. .. .
+ . . . .. = F
f
y
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
75. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
′
. .. .. .
+ + . . .. = F
f
y
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
76. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − ′
. .. .. .. . . .. = F
f
y
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
77. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − ′
. .. .. .. .. . .. = F
f
y
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
78. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
79. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1↗2
. . . 3
. 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
80. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1↗2↗3
. . . . . 4
. 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
81. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1↗2↗3↘4
. . . . . . . 5
. 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
82. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1↗2↗3↘4↘5
. . . . . . . . . 6F
..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
83. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . . . . . . . . . ..
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
84. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . . . . . . .
max
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
85. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . .. . . . . .
max min
.
. . ′ ′′
. . . . . . . . . . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
86. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . .. . . . . .
max min
.
. . ′ ′′
. . . . . . . .. + .
+ . . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
87. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . .. . . . . .
max min
.
. . ′ ′′
. . . . . . . .. + .. − .
+ − . . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
88. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . .. . . . . .
max min
.
. . ′ ′′
. . . . . . . .. + .. − .. − .
+ − − . .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
89. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . .. . . . . .
max min
.
. . ′ ′′
. . . . . . . .. + .. − .. − .. + .
+ − − + .. = F
f
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
90. Using f to make a sign chart for F
Assuming F′ = f, we can make a sign chart for f and f′ to find the
intervals of monotonicity and concavity for F:
+ + − − + f ′
. . . . . . . . . . . .. = F
y
. 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F
. . .. . . . .. . . . . .
max min
.
. . + − − + + f′ ′′
. . . . . . . .. + .. − .. − .. + .. + . . = F
1 2 3 4 5 6
. . . . . .
x
. 1
. 2
. 3
. 4
. 5
. 6F
..
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33