Lesson 23: Antiderivatives (slides)

Sec on 4.7
An deriva ves
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University

April 19, 2011

.

Announcements
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
I am teaching Calc II MW
2:00pm and Calc III TR
2:00pm both Fall ’11 and
Spring ’12

Objectives
Given a ”simple“ elementary
func on, ﬁnd a func on whose
deriva ve is that func on.
Remember that a func on
whose deriva ve is zero along
an interval must be zero along
that interval.
Solve problems involving
rec linear mo on.

Outline
What is an an deriva ve?
Tabula ng An deriva ves
Power func ons
Combina ons
Exponen al func ons
Trigonometric func ons
An deriva ves of piecewise func ons
Finding An deriva ves Graphically
Rec linear mo on

What is an antiderivative?

Deﬁni on
Let f be a func on. An an deriva ve for f is a func on F such that
F′ = f.

Who cares?
Ques on
Why would we want the an deriva ve of a func on?

Answers
For the challenge of it
For applica ons when the deriva ve of a func on is known but
the original func on is not
Biggest applica on will be a er the Fundamental Theorem of
Calculus (Chapter 5)

Hard problem, easy check

Example
Find an an deriva ve for f(x) = ln x.


Example
Find an an deriva ve for f(x) = ln x.

Solu on
???

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

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

Solu on

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. By MVT there exists a
point z in (x, y) such that

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.

Functions with the same derivative
Theorem
Suppose f and g are two diﬀeren able func ons 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)

Antiderivatives of power functions
Recall that the deriva ve of a y
power func on is a power f(x) = x2
func on.
Fact (The Power Rule)
If f(x) = xr , then f′ (x) = rxr−1 .

.
x

′
Recall that the deriva ve of a yf (x) = 2x
func on.
Fact (The Power Rule)

.
x

′
func on.
Fact (The Power Rule) F(x) = ?


.
x

′
func on.
Fact (The Power Rule) F(x) = ?

So in looking for
an deriva ves of power .
x
func ons, try power
func ons!

Example
Find an an deriva ve for the func on f(x) = x3 .

Solu on

Example

Solu on
Try a power func on F(x) = axr

Example

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

Example

Solu on
Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
1
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = .
4

Example

Solu on

1
So F(x) = x4 is an an deriva ve.
4

Example

Solu on

1
4
( )
Check:
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4

Example

Solu on

1
4
( )
Check:
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4

Any others?

Example

Solu on

1
4
( )
Check:
d 1 4
dx 4
1
x = 4 · x4−1 = x3
4

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

General power functions
Fact (The Power Rule for an deriva ves)
If f(x) = xr , then
1 r+1
F(x) = x
r+1
is an an deriva ve for f…

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

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

Fact
1
If f(x) = x−1 = , then F(x) = ln |x| + C is an an deriva ve for f.
x

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 posi ve numbers.

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

d
If x 0, ln |x|
dx

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

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

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

d d 1
If x 0, 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

d
If x 0, 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

d d
If x 0, 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

d d 1
If x 0, 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

d d 1 1
If x 0, 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 an deriva ve with the larger domain.

Graph of ln |x|
y

. x = 1/x
f(x)

Graph of ln |x|
y

F(x) = ln(x)

. x = 1/x
f(x)

Graph of ln |x|
y

F(x) = ln |x|

. x = 1/x
f(x)

Combinations of antiderivatives

Fact (Sum and Constant Mul ple Rule for An deriva ves)

If F is an an deriva ve of f and G is an an deriva ve of g, then
F + G is an an deriva ve of f + g.
If F is an an deriva ve of f and c is a constant, then cF is an
an deriva ve of cf.

Combinations of antiderivatives
Proof.
These follow from the sum and constant mul ple rule for
deriva ves:
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 an deriva ve for f(x) = 16x + 5.

Example
Find an an deriva ve for f(x) = 16x + 5.

Solu on
1
The expression x2 is an an deriva ve for x, and x is an
2
an deriva ve for 1. So
( )
1 2
F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C
2
is the an deriva ve of f.


Ques on
Do we need two C’s or just one?


Ques on
Do we need two C’s or just one?

Answer
Just one. A combina on of two arbitrary constants is s ll 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 an deriva ve of f.
ln a

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 an deriva ve of f.
ln a
Proof.
Check it yourself.


In par cular,


In par cular,
Fact
If f(x) = ex , then F(x) = ex + C is the an deriva ve of f.

Logarithmic functions?
Remember we found F(x) = x ln x − x is an an deriva ve of
f(x) = ln x.

f(x) = ln x.
This is not obvious. See Calc II for the full story.

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 an deriva ve of f(x).

Trigonometric functions
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 func on F(x) = − cos x + C is the an deriva ve of
f(x) = sin x.

Fact

d d
dx dx

So to turn these around,
Fact
The func on F(x) = − cos x + C is the an deriva ve of
f(x) = sin x.
The func on F(x) = sin x + C is the an deriva ve of
f(x) = cos x.

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

More Trig
Example

Solu on
???

More Trig
Example

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

More Trig
Example

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

Check

d 1 d
= · sec x
dx sec x dx

More Trig
Example

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

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

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

Check

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

More Trig
Example

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

Check

d
=
1
·
d
dx sec x dx
sec x =
1
sec x
· sec x tan x = tan x
More about this later.

Antiderivatives of piecewise functions

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

Antiderivatives of piecewise functions
Solu on
We can an diﬀeren ate 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
con nuous (at 1).

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

Finding Antiderivatives Graphically

y
Problem
Pictured is the graph of a y = f(x)
func on f. Draw the graph of .
an an deriva ve for f. 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:
y f = F′
.
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

y + f = F′
.
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

y + + f = F′
.
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

y + + − f = F′
.
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

y + + − − f = F′
.
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

′
y . + + − − + f=F
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

′
y . + + − − + f=F
1↗2 3 4 5 6 F

f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F

F
1 2 3 4 5 6 shape

′
y . + + − − + f=F
1↗2↗3 4 5 6 F

f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F

F
1 2 3 4 5 6 shape

′
y . + + − − + f=F
1↗2↗3↘4 5 6 F

f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F

F
1 2 3 4 5 6 shape

′
y . + + − − + f=F
1↗2↗3↘4↘5 6 F

f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F

F
1 2 3 4 5 6 shape

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F

f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F

F
1 2 3 4 5 6 shape

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
max
f′ = F′′
.
1 2 3 4 5 6
x 1 2 3 4 5 6F

F
1 2 3 4 5 6 shape

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
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

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

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

′
y . + + − − + f=F
1↗2↗3↘4↘5↗6 F
max min
′ ′′
++ −−−− ++ ++ f = F
. ⌣ ⌢ ⌢ ⌣ ⌣
1 2 3 4 5 6
x 1 2 3 4 5 6F
IP IP
? ? ? ? ? ?F
1 2 3 4 5 6 shape
The only ques on le is: What are the func on values?

Could you repeat the question?
Problem
Below is the graph of a func on f. Draw the graph of the
an deriva ve for f with F(1) = 0.
Solu on y

f
.
x
1 2 3 4 5 6

F
1 2 3 4 5 6 shape

IP
max
IP
min

Problem
Solu on y

We start with F(1) = 0. f
.
x
1 2 3 4 5 6

F
1 2 3 4 5 6 shape

IP
max
IP
min

Problem
Solu on y

.
Using the sign chart, we draw arcs x
1 2 3 4 5 6
with the speciﬁed monotonicity and
concavity F
1 2 3 4 5 6 shape

IP
max
IP
min

Problem
Solu on y

.
Using the sign chart, we draw arcs x
1 2 3 4 5 6
with the speciﬁed monotonicity and
concavity F
It’s harder to tell if/when F crosses
1 2 3 4 5 6 shape

IP
max
IP
min
the axis; more about that later.

Say what?

“Rec linear mo on” just means mo on along a line.
O en we are given informa on about the velocity or
accelera on of a moving par cle and we want to know the
equa ons of mo on.

Problem
Suppose a par cle of mass m is acted upon by a constant force F.
Find the posi on func on s(t), the velocity func on v(t), and the
accelera on func on a(t).

Problem

Solu on
By Newton’s Second Law (F = ma) a constant force induces a
F
constant accelera on. So a(t) = a = .
m

Problem

Solu on
By Newton’s Second Law (F = ma) a constant force induces a
F
constant accelera on. So a(t) = a = .
m
′
Since v (t) = a(t), v(t) must be an an deriva ve of the
constant func on a. So

v(t) = at + C = at + v0

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?

Solu on
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,

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 before it came to a stop.
Suppose that the car in ques on has a constant decelera on of
20 ft/s2 under the condi ons of the skid. How fast was the car
traveling when its brakes were ﬁrst applied?

stopping distance
Example

Solu on (Setup)

While braking, the car has accelera on a(t) = −20

stopping distance
Example

Solu on (Setup)

While braking, the car has accelera on a(t) = −20
Measure me 0 and posi on 0 when the car starts braking. So

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.

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

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

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

The second gives t1 = v0 /20, so subs tute into the ﬁrst:
v0 ( v )2
0
v0 · − 10 = 160
20 20
Solve:
v2
0 10v2
− 0
= 160
20 400
2v2 − v2 = 160 · 40 = 6400
0 0

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

Final Thoughts

An deriva ves are a
useful concept, especially
in mo on y
We can graph an f
an deriva ve from the .
x
graph of a func on 1 2 3 4 5 6F
We can compute
an deriva ves, but not
f(x) = e−x
2
always

Lesson 23: Antiderivatives (slides)

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