At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.

1. Section 4.7
Antiderivatives
V63.0121.041, Calculus I
New York University
November 29, 2010
Announcements
Quiz 5 in recitation this week on §§4.1–4.4
. . . . . .

2. Announcements
Quiz 5 in recitation this
week on §§4.1–4.4
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 2 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 3 / 35

4. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 4 / 35

5. What is an antiderivative?
Definition
Let f be a function. An antiderivative for f is a function F such that
F′ = f.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 5 / 35

6. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

7. Hard problem, easy check
Example
Find an antiderivative for f(x) = ln x.
Solution
???
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

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?
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 7 / 35

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)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 8 / 35

15. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 9 / 35

16. Antiderivatives of power functions
y
f(x) = x2
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35

17. Antiderivatives of power functions
′
y f (x) = 2x
f(x) = x2
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35

18. Antiderivatives of power functions
′
y f (x) = 2x
f(x) = x2
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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35

19. Antiderivatives of power functions
′
y f (x) = 2x
f(x) = x2
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!
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35

20. Example
Find an antiderivative for the function f(x) = x3 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

21. Example
Find an antiderivative for the function f(x) = x3 .
Solution
Try a power function F(x) = axr
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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?
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35

30. Extrapolating to general power functions
Fact (The Power Rule for antiderivatives)
If f(x) = xr , then
1 r+1
F(x) = x
r+1
is an antiderivative for f…
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35

31. Extrapolating to general power functions
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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35

32. Extrapolating to general power functions
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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35

44. Graph of ln |x|
y
. f(x) = 1/x
x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35

45. Graph of ln |x|
y
F(x) = ln(x)
. f(x) = 1/x
x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35

46. Graph of ln |x|
y
F(x) = ln |x|
. f(x) = 1/x
x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 35

49. Antiderivatives of Polynomials
.
Example
Find an antiderivative for f(x) = 16x + 5.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35

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 for 1.
2
So ( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the antiderivative of f.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35

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 for 1.
2
So ( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the antiderivative of f.
Question
Do we need two C’s or just one?
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35

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 for 1.
2
So ( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the antiderivative of f.
Question
Do we need two C’s or just one?
Answer
Just one. A combination of two arbitrary constants is still an arbitrary constant.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35

53. Exponential Functions
Fact
If f(x) = ax , f′ (x) = (ln a)ax .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35

57. Logarithmic functions?
Remember we found
F(x) = x ln x − x
is an antiderivative of f(x) = ln x.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35

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).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35

60. Trigonometric functions
Fact
d d
sin x = cos x cos x = − sin x
dx dx
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35

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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35

63. More Trig
Example
Find an antiderivative of f(x) = tan x.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

64. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

65. More Trig
Example
Find an antiderivative of f(x) = tan x.
Solution
???
Answer
F(x) = ln(sec x).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

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. . . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35

71. Antiderivatives of piecewise functions
Example
{
x if 0 ≤ x ≤ 1;
Let f(x) = Find the antiderivative of f with
1 − x2 if 1 x.
F(0) = 1.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 35

72. Antiderivatives of piecewise functions
Example
{
x if 0 ≤ x ≤ 1;
Let f(x) = Find the antiderivative of f with
1 − x2 if 1 x.
F(0) = 1.
Solution
We can antidifferentiate each piece:

1 2
 x + C1 if 0 ≤ x ≤ 1;
F(x) = 2 1
x − x3 + C
 if 1 x.
2
3
The constants need to be chosen so that F(0) = 1 and F is continuous
(at 1).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 35

73. 
1 2
 x + C1 if 0 ≤ x ≤ 1;
F(x) = 2 1
x − x3 + C
 if 1 x.
2
3
Note
1 2
0 + C1 = C1 =⇒ C1 = 1
F(0) =
2
1 3
This means lim F(x) = 12 + 1 = . Now
x→1− 2 2
1 2
lim+ F(x) = 1 − + C2 = + C2
x→1 3 3
So for F to be continuous we need
3 2 5
= + C2 =⇒ C2 =
2 3 6
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 22 / 35

74. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 23 / 35

75. Finding Antiderivatives Graphically
Problem
Below is the graph of a function f. Draw the graph of an antiderivative
for f.
y
y = f(x)
.
x
1 2 3 4 5 6
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 24 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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
f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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
f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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
f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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
f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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 6F
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

104. 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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

105. 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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
F
1 2 3 4 5 6 shape
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

106. 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′′
. ⌣ ⌢ ⌢ ⌣ ⌣ F
x 1 2 3 4 5 6
1 2 3 4 5 6 IP IP
? ? ? ? ? ?F
1 2 3 4 5 6 shape
The only question left is: What are the function values?
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35

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
f
.
x
1 2 3 4 5 6
F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
x
1 2 3 4 5 6
F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

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
f
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

116. 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
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

117. 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
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
1 2 3 4 5 6 shape
IP
max
IP
min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

118. 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
We start with F(1) = 0. .
Using the sign chart, we x
1 2 3 4 5 6
draw arcs with the
specified monotonicity and
concavity F
It’s harder to tell if/when F 1 2 3 4 5 6 shape
crosses the axis; more
IP
max
IP
min
about that later.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35

119. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions
Finding Antiderivatives Graphically
Rectilinear motion
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 27 / 35

120. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 28 / 35

121. Application: Dead Reckoning
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 35

122. Application: Dead Reckoning
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 35

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).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35

124. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35

125. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35