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
dx
(x ln x − x)

. . . . . .
Example
Solution
???
Example
Solution
d
dx
(x ln x − x) = 1 · ln x + x ·
1
x
− 1

. . . . . .
Example
Solution
???
Example
Solution
d
dx
(x ln x − x) = 1 · ln x + x ·
1
x
− 1 = ln 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)
y − x
= f′
(z) =⇒ f(y) = f(x) + f′
(z)(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
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) = x2

. . . . . .
function.
If f(x) = xr
, then f′
(x) = rxr−1
.
..
x
.
y
.
f(x) = x2
.
f′
(x) = 2x

. . . . . .
function.
If f(x) = xr
, then f′
(x) = rxr−1
.
..
x
.
y
.
f(x) = x2
.
f′
(x) = 2x
.
F(x) = ?

. . . . . .
function.
If f(x) = xr
, then f′
(x) = rxr−1
.
So in looking for antiderivatives
of power functions, try power
functions!
..
x
.
y
.
f(x) = x2
.
f′
(x) = 2x
.
F(x) = ?

. . . . . .
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
Then F′
(x) = arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4

. . . . . .
Example
.
Solution
Then F′
(x) = arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.

. . . . . .
Example
.
Solution
Then F′
(x) = arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.

. . . . . .
Example
.
Solution
Then F′
(x) = arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3

. . . . . .
Example
.
Solution
Then F′
(x) = arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3

Any others?

. . . . . .
Example
.
Solution
Then F′
(x) = arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
Check:
d
dx
(
1
4
x4
)
= 4 ·
1
4
x4−1
= x3

Any others? Yes, F(x) =
1
4
x4
+ C is the most general form.

. . . . . .
Extrapolating to general power functions
Fact (The Power Rule for antiderivatives)
If f(x) = xr
, then
F(x) =
1
r + 1
xr+1
is an antiderivative for f…

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

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

. . . . . .
What's with the absolute value?
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
The domain of F is all nonzero numbers, while ln x is only defined
on positive numbers.

. . . . . .
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
If x 0,
d
dx
ln |x|

. . . . . .
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
If x 0,
d
dx
ln |x| =
d
dx
ln(x)

. . . . . .
F(x) = ln |x| =
{
ln(x) if x 0;
ln(−x) if x 0.
If x 0,
d
dx
ln |x| =
d
dx
ln(x) =
1
x

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

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

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

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

. . . . . .
F(x) = ln |x| =
{
ln(x) if x 0;
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|
.. x.
y
. f(x) = 1/x

. . . . . .
Graph of ln |x|
.. x.
y
. f(x) = 1/x.
F(x) = ln(x)

. . . . . .
Graph of ln |x|
.. x.
y
. f(x) = 1/x.
F(x) = ln |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
The expression
1
2
x2
is an antiderivative for x, and x is an antiderivative for 1.
So
F(x) = 16 ·
(
1
2
x2
)
+ 5 · x + C = 8x2
+ 5x + C
is the antiderivative of f.

. . . . . .
..
Example
Solution
The expression
1
2
x2
So
F(x) = 16 ·
(
1
2
x2
)
+ 5 · x + C = 8x2
+ 5x + C
Question
Do we need two C’s or just one?

. . . . . .
..
Example
Solution
The expression
1
2
x2
So
F(x) = 16 ·
(
1
2
x2
)
+ 5 · x + C = 8x2
+ 5x + C
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
If f(x) = ax
, then F(x) =
1
ln a
ax
+ C is the antiderivative of f.

. . . . . .
Fact
If f(x) = ax
, f′
(x) = (ln a)ax
.
Accordingly,
Fact
If f(x) = ax
, then F(x) =
1
ln a
ax
Proof.
Check it yourself.

. . . . . .
Fact
If f(x) = ax
, f′
(x) = (ln a)ax
.
Accordingly,
Fact
If f(x) = ax
, then F(x) =
1
ln a
ax
Proof.
Check it yourself.
In particular,
Fact
If f(x) = ex
, then F(x) = ex

. . . . . .
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.
However, using the fact that loga x =
ln x
ln a
, we get:
Fact
If f(x) = loga(x)
F(x) =
1
ln a
(x ln x − x) + C = x loga x −
1
ln a
x + C
is the antiderivative of f(x).

. . . . . .
Fact
d
dx
sin x = cos x
d
dx
cos x = − sin x

. . . . . .
Fact
d
dx
sin x = cos x
d
dx
cos x = − sin x
So to turn these around,
Fact
The function F(x) = − cos x + C is the antiderivative of f(x) = sin x.

. . . . . .
Fact
d
dx
sin x = cos x
d
dx
cos x = − sin x
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
dx
=
1
sec x
·
d
dx
sec x

. . . . . .
More Trig
Example
Solution
???
Answer
F(x) = ln | sec x|.
Check
d
dx
=
1
sec x
·
d
dx
sec x =
1
sec x
· sec x tan x

. . . . . .
More Trig
Example
Solution
???
Answer
F(x) = ln | sec x|.
Check
d
dx
=
1
sec x
·
d
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
sec x
·
d
dx
sec x =
1
sec x
· sec x tan x = tan x
More about this later.

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

. . . . . .
Example
Let
f(x) =
{
x if 0 ≤ x ≤ 1;
1 − x2
if 1 x.
Find the antiderivative of f with F(0) = 1.
Solution
We can antidifferentiate each piece:
F(x) =



1
2
x2
+ C1 if 0 ≤ x ≤ 1;
x −
1
3
x3
+ C2 if 1 x.
The constants need to be chosen so that F(0) = 1 and F is continuous
(at 1).

. . . . . .
F(x) =



1
2
x2
+ C1 if 0 ≤ x ≤ 1;
x −
1
3
x3
+ C2 if 1 x.
Note F(0) =
1
2
02
+ C1 = C1, so if F(0) is to be 1, C1 = 1.

. . . . . .
F(x) =



1
2
x2
+ C1 if 0 ≤ x ≤ 1;
x −
1
3
x3
+ C2 if 1 x.
Note F(0) =
1
2
02
+ C1 = C1, so if F(0) is to be 1, C1 = 1.
This means lim
x→1−
F(x) =
1
2
12
+ 1 =
3
2
.

. . . . . .
F(x) =



1
2
x2
+ C1 if 0 ≤ x ≤ 1;
x −
1
3
x3
+ C2 if 1 x.
Note F(0) =
1
2
02
+ C1 = C1, so if F(0) is to be 1, C1 = 1.
This means lim
x→1−
F(x) =
1
2
12
+ 1 =
3
2
.
On the other hand,
lim
x→1+
F(x) = 1 −
1
3
+ C2 =
2
3
+ C2
So for F to be continuous we need
3
2
=
2
3
+ C2. Solving, C2 =
5
6
.

. . . . . .
Outline
Power functions
Combinations
Rectilinear motion

. . . . . .
Problem
Below is the graph of a function f. Draw the graph of an antiderivative
for f.
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
y = f(x)

. . . . . .
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:
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
.

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
..

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
...

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
....

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
.....

. . . . . .
Assuming F′
to find the
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
......
.. f = F′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
. +. +. −. −. +.
↗
.
↗
.
↘
.
↘
.
↗
.
max
.
min
.
f′
= F′′
.
F
..
1
..
2
..
3
..
4
..
5
..
6
.
++
.
−−
.
−−
.
++
.
++
.
⌣
.
⌢
.
⌢
.
⌣
.
⌣
.
IP
.
IP
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
?
.
?
.
?
.
?
.
?
.
?
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.
Solution
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min

. . . . . .
Problem
Solution
We start with F(1) = 0. ..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
.

. . . . . .
Problem
Solution
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the
specified monotonicity and
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
.

. . . . . .
Problem
Solution
draw arcs with the
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
..

. . . . . .
Problem
Solution
draw arcs with the
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
...

. . . . . .
Problem
Solution
draw arcs with the
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
....

. . . . . .
Problem
Solution
draw arcs with the
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
.....

. . . . . .
Problem
Solution
draw arcs with the
concavity
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
......

. . . . . .
Problem
Solution
draw arcs with the
concavity
It’s harder to tell if/when F
crosses the axis; more
about that later.
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.
F
.
shape
..
1
..
2
..
3
..
4
..
5
..
6
......
IP
.
max
.
IP
.
min
......

. . . . . .
Outline
Power functions
Combinations
Rectilinear motion

. . . . . .
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.

. . . . . .
Application: Dead Reckoning

. . . . . .
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).

. . . . . .
Problem
Solution
By Newton’s Second Law (F = ma) a constant force induces a
constant acceleration. So a(t) = a =
F
m
.

. . . . . .
Problem
Solution
F
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.

. . . . . .
Problem
Solution
F
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
s(t) =
1
2
at2
+ v0t + C =
1
2
at2
+ v0t + s0

. . . . . .
An earlier Hatsumon
Example
Drop a ball off the roof of the Silver Center. What is its velocity when it
hits the ground?

. . . . . .
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.

. . . . . .
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?

. . . . . .
Example
were first applied?
Solution (Setup)
While braking, the car has acceleration a(t) = −20

. . . . . .
Example
were first applied?
Solution (Setup)
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.

. . . . . .
Example
were first applied?
Solution (Setup)
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.

. . . . . .
Implementing the Solution
In general,
s(t) = s0 + v0t +
1
2
at2
Since s0 = 0 and a = −20, we have
s(t) = v0t − 10t2
v(t) = v0 − 20t
for all t.

. . . . . .
Implementing the Solution
In general,
s(t) = s0 + v0t +
1
2
at2
Since s0 = 0 and a = −20, we have
s(t) = v0t − 10t2
v(t) = v0 − 20t
for all t. Plugging in t = t1,
160 = v0t1 − 10t2
1
0 = v0 − 20t1
We need to solve these two equations.

. . . . . .
Solving
We have
v0t1 − 10t2
1 = 160 v0 − 20t1 = 0

. . . . . .
Solving
We have
v0t1 − 10t2
1 = 160 v0 − 20t1 = 0
The second gives t1 = v0/20, so substitute into the first:
v0 ·
v0
20
− 10
( v0
20
)2
= 160
Solve:
v2
0
20
−
10v2
0
400
= 160
2v2
0 − v2
0 = 160 · 40 = 6400

. . . . . .
Solving
We have
v0t1 − 10t2
1 = 160 v0 − 20t1 = 0
The second gives t1 = v0/20, so substitute into the first:
v0 ·
v0
20
− 10
( v0
20
)2
= 160
Solve:
v2
0
20
−
10v2
0
400
= 160
2v2
0 − v2
0 = 160 · 40 = 6400
So v0 = 80 ft/s ≈ 55 mi/hr

. . . . . .
Summary of Antiderivatives so far
f(x) F(x)
xr
, r ̸= 1
1
r + 1
xr+1
+ C
1
x
= x−1
ln |x| + C
ex
ex
+ C
ax 1
ln a
ax
+ C
ln x x ln x − x + C
loga x
x ln x − x
ln a
+ C
sin x − cos x + C
cos x sin x + C
tan x ln | tan x| + C

. . . . . .
Final Thoughts
Antiderivatives are a useful
concept, especially in
motion
We can graph an
antiderivative from the
graph of a function
We can compute
antiderivatives, but not
always
..
x
.
y
..
1
..
2
..
3
..
4
..
5
..
6
.......
f
.......
F
f(x) = e−x2
f′
(x) = ???

Lesson 23: Antiderivatives (Section 021 slides)

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Lesson 23: Antiderivatives (Section 021 slides)