This document is from a Calculus I class at New York University and covers antiderivatives. It begins with announcements about an upcoming quiz. The objectives are to find antiderivatives of simple functions, remember that a function whose derivative is zero must be constant, and solve rectilinear motion problems. It then outlines finding antiderivatives through tabulation, graphically, and with rectilinear motion examples. Examples are provided of finding the antiderivative of power functions like x^3 through identifying the power rule relationship between a function and its derivative.

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

53. Exponential Functions
Fact
If f(x) = ax , f′ (x) = (ln a)ax .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33

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

91. 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

92. 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

93. 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

94. 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

95. 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
. .F
6
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

96. 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
. .F
6
IP
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

97. 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
. .F
6
IP IP
.
. . . . . ..
F
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

98. 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
. .F
6
IP IP
.
. . . . . ..
F
.
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

99. 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
. .F
6
IP IP
.
. . . . . ..
F
. .
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

100. 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
. .F
6
IP IP
.
. . . . . ..
F
. . .
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

101. 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
. .F
6
IP IP
.
. . . . . ..
F
. . . .
1
. 2
. 3
. 4
. 5
. 6s
. . hape
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

102. 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
. .F
6
IP IP
.
. . . . . ..F
. . . . . . hape
1
. 2
. 3
. 4
. 5
. .s
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

103. 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
. .F
6
IP IP
.
?
.. ?
.. ?
.. ?
.. ?
.. ?F
.. .
. . . . . . hape
1
. 2
. 3
. 4
. 5
. .s
6
The only question left is: What are the function values?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

104. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
.
Solution
.
. ..
f
. . . . . . .
x
.
1 2 3 4 5 6
. . . . . .
.
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

105. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
.
Solution
.
We start with F(1) = 0. . ..
f
. . . . . . .
x
.
1 2 3 4 5 6
. . . . . .
.
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

106. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
.
Solution
.
We start with F(1) = 0. . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

107. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
.
Solution
.
.
We start with F(1) = 0. . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

108. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
.
Solution
.
.
We start with F(1) = 0. . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

109. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

110. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

111. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

112. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the 1 2 3 4 5 6
. . . . . .
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

113. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the . . . . .. .
1 2 3 4 5 6
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

114. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the . . . . .. .
1 2 3 4 5 6
specified monotonicity and
concavity .
. . . . . ..
F
. . . . .
1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

115. Could you repeat the question?
Problem
Below is the graph of a function f. Draw the graph of the antiderivative
for f with F(1) = 0.
y
. .
Solution
.
.
We start with F(1) = 0. . . ..
f
. . . . . . .
Using the sign chart, we x
.
draw arcs with the . . . . .. .
1 2 3 4 5 6
specified monotonicity and
concavity .
. . . . . ..
F
It’s harder to tell if/when F . . . . .
crosses the axis; more 1 2 3 4 5
. . . . . 6s
. . hape
IP
.
max
.
IP
.
min
.
about that later.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33

116. 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 25 / 33

117. Say what?
“Rectilinear motion” just means motion along a line.
Often we are given information about the velocity or acceleration
of a moving particle and we want to know the equations of motion.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 26 / 33

118. Application: Dead Reckoning
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 27 / 33

119. Application: Dead Reckoning
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 27 / 33

120. Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33

121. Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
F
constant acceleration. So a(t) = a = .
m
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33

122. Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
F
constant acceleration. So a(t) = a = .
m
Since v′ (t) = a(t), v(t) must be an antiderivative of the constant
function a. So
v(t) = at + C = at + v0
where v0 is the initial velocity.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33

123. Problem
Suppose a particle of mass m is acted upon by a constant force F.
Find the position function s(t), the velocity function v(t), and the
acceleration function a(t).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
F
constant acceleration. So a(t) = a = .
m
Since v′ (t) = a(t), v(t) must be an antiderivative of the constant
function a. So
v(t) = at + C = at + v0
where v0 is the initial velocity.
Since s′ (t) = v(t), s(t) must be an antiderivative of v(t), meaning
1 2 1
s(t) = at + v0 t + C = at2 + v0 t + s0
2 2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33

124. An earlier Hatsumon
Example
Drop a ball off the roof of the Silver Center. What is its velocity when it
hits the ground?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 29 / 33

125. An earlier Hatsumon
Example
Drop a ball off the roof of the Silver Center. What is its velocity when it
hits the ground?
Solution
Assume s0 = 100 m, and v0 = 0. Approximate a = g ≈ −10. Then
s(t) = 100 − 5t2
√ √
So s(t) = 0 when t = 20 = 2 5. Then
v(t) = −10t,
√ √
so the velocity at impact is v(2 5) = −20 5 m/s.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 29 / 33

126. Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33

127. Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
Solution (Setup)
While breaking, the car has acceleration a(t) = −20
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33

128. Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
Solution (Setup)
While breaking, the car has acceleration a(t) = −20
Measure time 0 and position 0 when the car starts braking. So
s(0) = 0.
The car stops at time some t1 , when v(t1 ) = 0.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33

129. Finding initial velocity from stopping distance
Example
The skid marks made by an automobile indicate that its brakes were
fully applied for a distance of 160 ft before it came to a stop. Suppose
that the car in question has a constant deceleration of 20 ft/s2 under the
conditions of the skid. How fast was the car traveling when its brakes
were first applied?
Solution (Setup)
While breaking, the car has acceleration a(t) = −20
Measure time 0 and position 0 when the car starts braking. So
s(0) = 0.
The car stops at time some t1 , when v(t1 ) = 0.
We know that when s(t1 ) = 160.
We want to know v(0), or v0 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33

130. Implementing the Solution
In general,
1
s(t) = s0 + v0 t + at2
2
Since s0 = 0 and a = −20, we have
s(t) = v0 t − 10t2
v(t) = v0 − 20t
for all t.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 31 / 33

131. Implementing the Solution
In general,
1
s(t) = s0 + v0 t + at2
2
Since s0 = 0 and a = −20, we have
s(t) = v0 t − 10t2
v(t) = v0 − 20t
for all t. Plugging in t = t1 ,
160 = v0 t1 − 10t2
1
0 = v0 − 20t1
We need to solve these two equations.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 31 / 33

132. Solving
We have
v0 t1 − 10t2 = 160
1 v0 − 20t1 = 0
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 32 / 33

133. Solving
We have
v0 t1 − 10t2 = 160
1 v0 − 20t1 = 0
The second gives t1 = v0 /20, so substitute into the first:
v0 ( v )2
v0 · − 10 0 = 160
20 20
or
v2 10v2
0
− 0
= 160
20 400
2v2 − v2 = 160 · 40 = 6400
0 0
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 32 / 33

134. Solving
We have
v0 t1 − 10t2 = 160
1 v0 − 20t1 = 0
The second gives t1 = v0 /20, so substitute into the first:
v0 ( v )2
v0 · − 10 0 = 160
20 20
or
v2 10v2
0
− 0
= 160
20 400
2v2 − v2 = 160 · 40 = 6400
0 0
So v0 = 80 ft/s ≈ 55 mi/hr
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 32 / 33

135. Summary
Antiderivatives are a useful
concept, especially in
motion y
. .
We can graph an .
.
. . ..
f
antiderivative from the . . . . . . .
graph of a function x
.
. . . . .. . F
1 2 3 4 5 6
We can compute
antiderivatives, but not .
always
f(x) = e−x
2
f′ (x) = ???
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 33 / 33