Lesson 23: Antiderivatives (Section 021 slides)

Section 4.7
Antiderivatives

V63.0121.021, Calculus I

New York University

November 30, 2010

Announcements

Quiz 5 in recitation this week on §§4.1–4.4
Final Exam: Monday, December 20, 12:00–1:50pm

. . . . . .

Announcements

Quiz 5 in recitation this
week on §§4.1–4.4
Final Exam: Monday,
December 20,
12:00–1:50pm

. . . . . .

V63.0121.021, Calculus I (NYU) Section 4.7 Antiderivatives November 30, 2010 2 / 37

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.

. . . . . .


Outline

What is an antiderivative?

Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Antiderivatives of piecewise functions

Finding Antiderivatives Graphically

Rectilinear motion

. . . . . .



Definition
Let f be a function. An antiderivative for f is a function F such that
F′ = f.

. . . . . .


Who cares?

Question
Why would we want the antiderivative of a function?

Answers

For the challenge of it
For applications when the derivative of a function is known but the
original function is not
Biggest application will be after the Fundamental Theorem of
Calculus (Chapter 5)

. . . . . .


Hard problem, easy check

Example
Find an antiderivative for f(x) = ln x.

. . . . . .



Example

Solution
???

. . . . . .



Example

Solution
???

Example
is F(x) = x ln x − x an antiderivative for f(x) = ln x?

. . . . . .



Example

Solution
???

Example

Solution

d
(x ln x − x)
dx

. . . . . .



Example

Solution
???

Example

Solution

d 1
(x ln x − x) = 1 · ln x + x · − 1
dx x

. . . . . .



Example

Solution
???

Example

Solution

d 1
(x ln x − x) = 1 · ln x + x · − 1 = ln x
dx x

. . . . . .



Example

Solution
???

Example

Solution

d
dx
1
(x ln x − x) = 1 · ln x + x · − 1 = ln x
x

. . . . . .


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.

. . . . . .


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)

. . . . . .


Outline


Power functions
Combinations


Rectilinear motion

. . . . . .


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

. . . . . .



′
y f (x) = 2x
f(x) = x2
function.

.
x

. . . . . .



′
y f (x) = 2x
f(x) = x2
function. F(x) = ?


.
x

. . . . . .



′
y f (x) = 2x
f(x) = x2
function. F(x) = ?


So in looking for antiderivatives
.
of power functions, try power x
functions!

. . . . . .


Example
Find an antiderivative for the function f(x) = x3 .

. . . . . .


Example

Solution

Try a power function F(x) = axr

. . . . . .


Example

Solution

Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .

. . . . . .


Example

Solution


r − 1 = 3 =⇒ r = 4

. . . . . .


Example

Solution

1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4

. . . . . .


Example

Solution

1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
So F(x) = x is an antiderivative.
4

. . . . . .


Example

Solution

1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
4
Check: ( )
d 1 4 1
x = 4 · x4−1 = x3
dx 4 4

. . . . . .


Example

Solution

1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
4
Check: ( )
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4

. . . . . .


Example

Solution

1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
4
Check: ( )
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4

Any others?

. . . . . .


Example

Solution

1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4
1 4
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
. . . . . .


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…

. . . . . .



If f(x) = xr , then
1 r+1
F(x) = x
r+1
is an antiderivative for f as long as r ̸= −1.

. . . . . .



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.

. . . . . .


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.

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

If x 0,
d
ln |x|
dx

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

If x 0,
d d
ln |x| = ln(x)
dx dx

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

If x 0,
d d 1
ln |x| = ln(x) =
dx dx x

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x

If x 0,
d
ln |x|
dx

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x

If x 0,
d d
ln |x| = ln(−x)
dx dx

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

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

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

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

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

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

. . . . . .


{
ln(x) if x 0;
F(x) = ln |x| =
ln(−x) if x 0.

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.
. . . . . .


Graph of ln |x|

y

. f(x) = 1/x
x

. . . . . .


Graph of ln |x|

y

F(x) = ln(x)

. f(x) = 1/x
x

. . . . . .


Graph of ln |x|

y

F(x) = ln |x|

. f(x) = 1/x
x

. . . . . .


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.

. . . . . .


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
. . . . . .


Antiderivatives of Polynomials
.
Example
Find an antiderivative for f(x) = 16x + 5.

. . . . . .


.
Example

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.

. . . . . .


.
Example

Solution
1 2
2
So ( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2

Question
Do we need two C’s or just one?

. . . . . .


.
Example

Solution
1 2
2
So ( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2

Question
Do we need two C’s or just one?

Answer
Just one. A combination of two arbitrary constants is still an arbitrary constant.
. . . . . .


Exponential Functions

Fact
If f(x) = ax , f′ (x) = (ln a)ax .

. . . . . .



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

. . . . . .



Fact
If f(x) = ax , f′ (x) = (ln a)ax .

Accordingly,
Fact
1 x
ln a

Proof.
Check it yourself.

. . . . . .



Fact
If f(x) = ax , f′ (x) = (ln a)ax .

Accordingly,
Fact
1 x
ln a

Proof.
Check it yourself.

In particular,
Fact
If f(x) = ex , then F(x) = ex + C is the antiderivative of f.
. . . . . .


Logarithmic functions?

Remember we found

F(x) = x ln x − x

is an antiderivative of f(x) = ln x.

. . . . . .



Remember we found

F(x) = x ln x − x

This is not obvious. See Calc II for the full story.

. . . . . .



Remember we found

F(x) = x ln x − 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).
. . . . . .



Fact

d d
sin x = cos x cos x = − sin x
dx dx

. . . . . .



Fact

d d
dx dx

So to turn these around,
Fact

The function F(x) = − cos x + C is the antiderivative of f(x) = sin x.

. . . . . .



Fact

d d
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.

. . . . . .


More Trig
Example
Find an antiderivative of f(x) = tan x.

. . . . . .


More Trig
Example

Solution
???

. . . . . .


More Trig
Example

Solution
???

Answer
F(x) = ln | sec x|.

. . . . . .


More Trig
Example

Solution
???

Answer
F(x) = ln | sec x|.

Check

d 1 d
= · sec x
dx sec x dx

. . . . . .


More Trig
Example

Solution
???

Answer
F(x) = ln | sec x|.

Check

d 1 d 1
= · sec x = · sec x tan x
dx sec x dx sec x

. . . . . .


More Trig
Example

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

. . . . . .


More Trig
Example

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 Trig
Example

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. . . . . . .


Example
Let {
x if 0 ≤ x ≤ 1;
f(x) =
1 − x2 if 1 x.
Find the antiderivative of f with F(0) = 1.

. . . . . .


Example
Let {
x if 0 ≤ x ≤ 1;
f(x) =
1 − x2 if 1 x.
Find the antiderivative of f with 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).
. . . . . .



1 2
 x + C1 if 0 ≤ x ≤ 1;
F(x) = 2 1
x − x3 + C
 if 1 x.
2
3
1 2
Note F(0) = 0 + C1 = C1 , so if F(0) is to be 1, C1 = 1.
2

. . . . . .



1 2
 x + C1 if 0 ≤ x ≤ 1;
F(x) = 2 1
x − x3 + C
 if 1 x.
2
3
1 2
2
1 3
This means lim F(x) = 12 + 1 = .
x→1 − 2 2

. . . . . .



1 2
 x + C1 if 0 ≤ x ≤ 1;
F(x) = 2 1
x − x3 + C
 if 1 x.
2
3
1 2
2
1 3
This means lim F(x) = 12 + 1 = .
x→1 − 2 2
On the other hand,
1 2
lim+ F(x) = 1 − + C2 = + C2
x→1 3 3

3 2 5
So for F to be continuous we need = + C2 . Solving, C2 = .
2 3 6

. . . . . .


Outline


Power functions
Combinations


Rectilinear motion

. . . . . .



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

. . . . . .


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

. . . . . .




. + 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

. . . . . .




. + + 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

. . . . . .




. + + − 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

. . . . . .




. + + − − 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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

. . . . . .




. + + − − + 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?
. . . . . .


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
. . . . . .


Problem

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
. . . . . .


Problem

y
Solution
f
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
. . . . . .


Problem

y
Solution
f
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.
. . . . . .


Outline


Power functions
Combinations


Rectilinear motion

. . . . . .


Lesson 23: Antiderivatives (Section 021 slides)

Lesson 23: Antiderivatives (Section 021 slides)

More Related Content

Lesson 23: Antiderivatives (Section 021 slides)