Lesson 21: Antiderivatives (notes)

V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010

Section 4.7 Notes

Antiderivatives

V63.0121.006/016, Calculus I

New York University

April 8, 2010

Announcements

Quiz April 16 on §§4.1–4.4
Final Exam: Monday, May 10, 10:00am

Image credit: Ian Hampton

Announcements
Notes

Quiz April 16 on §§4.1–4.4
Final Exam: Monday, May 10, 10:00am

V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 2 / 32

Outline
Notes

What is an antiderivative?

Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions

Finding Antiderivatives Graphically

Rectilinear motion


1


Objectives
Notes

Given an expression for
function f , find a
differentiable function F
such that F = f (F is called
an antiderivative for f ).
Given the graph of a
function f , find a
differentiable function F
such that F = f
Use antiderivatives to solve
problems in rectilinear
motion


Hard problem, easy check
Notes
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
"


Why the MVT is the MITC
Most Important Theorem In Calculus! Notes

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.


2


When two functions have the same derivative
Notes

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
Notes


Power functions
Combinations


Rectilinear motion


Antiderivatives of power functions
Notes

y f (x) = 2x
f (x) = x 2
Recall that the derivative of a
power function is a power
function. F (x) = ?

Fact (The Power Rule)
If f (x) = x r , then f (x) = rx r −1 .

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


3


Example
Notes
Find an antiderivative for the function f (x) = x 3 .

Solution

Try a power function F (x) = ax r
Then F (x) = arx r −1 , so we want arx r −1 = x 3 .
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
x
1
= 4 · x 4−1 = x 3
4
"
1
Any others? Yes, F (x) = x 4 + C is the most general form.
4


Notes
Fact (The Power Rule for antiderivatives)
If f (x) = x r , 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?
Notes
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.

4


Graph of ln |x|
Notes
y

F (x) = ln(x) ln |x|

f (x) = 1/x
x


Combinations of antiderivatives
Notes
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
Notes
Example
Find an antiderivative for f (x) = 16x + 5.

Solution
1
The expression x 2 is an antiderivative for x, and x is an antiderivative for
2
1. So
1 2
F (x) = 16 · x + 5 · x + C = 8x 2 + 5x + C
2
is the antiderivative of f .

Question
Why do we not need two C ’s?

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

5


Exponential Functions
Notes
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) = e x , then F (x) = e x + C is the antiderivative of f .


Logarithmic functions?
Notes

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).


Notes

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.


6


More Trig
Notes
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.

Outline
Notes


Power functions
Combinations


Rectilinear motion


Problem
Below is the graph of a function f . Draw the graph of an antiderivative for Notes
F.

y

y = f (x)

x
1 2 3 4 5 6


7


Using f to make a sign chart for F
Notes
Assuming F = f , we can make a sign chart for f and f to ﬁnd the
intervals of monotonicity and concavity for 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
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?
Notes
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
It’s harder to tell if/when F F
crosses the axis; more about 1 2 3 4 5 6 shape
IP
max
IP
min

that later.


Outline
Notes


Power functions
Combinations


Rectilinear motion


8


Say what?
Notes

“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
Notes


Problem
Suppose a particle of mass m is acted upon by a constant force F . Find Notes
the position function s(t), the velocity function v (t), and the acceleration
function a(t).

Solution

By Newton’s Second Law (F = ma) a constant force induces a
F
constant acceleration. So a(t) = a = .
m
Since v (t) = a(t), v (t) must be an antiderivative of the constant
function a. So
v (t) = at + C = at + v0
where v0 is the initial velocity.
Since s (t) = v (t), s(t) must be an antiderivative of v (t), meaning

1 1
s(t) = at 2 + v0 t + C = at 2 + v0 t + s0
2 2


9


An earlier Hatsumon
Notes

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 − 5t 2
√ √
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.


Example
Notes
The skid marks made by an automobile indicate that its brakes were fully
applied for a distance of 160 ft before it came to a stop. Suppose that the
car in question has a constant deceleration of 20 ft/s2 under the conditions
of the skid. How fast was the car traveling when its brakes were ﬁrst
applied?

Solution (Setup)

While breaking, the car has acceleration a(t) = −20
Measure time 0 and position 0 when the car starts braking. So
s(0) = 0.
The car stops at time some t1 , when v (t1 ) = 0.
We know that when s(t1 ) = 160.
We want to know v (0), or v0 .


Implementing the Solution
Notes
In general,
1
s(t) = s0 + v0 t + at 2
2
Since s0 = 0 and a = −20, we have

s(t) = v0 t − 10t 2
v (t) = v0 − 20t

for all t. Plugging in t = t1 ,
2
160 = v0 t1 − 10t1
0 = v0 − 20t1

We need to solve these two equations.


10


Solving
Notes
We have
2
v0 t1 − 10t1 = 160 v0 − 20t1 = 0

The second gives t1 = v0 /20, so substitute into the ﬁrst:

v0 v0 2
v0 · − 10 = 160
20 20
or
2
v0 10v02
− = 160
20 400
2 2
2v0 − v0 = 160 · 40 = 6400

So v0 = 80 ft/s ≈ 55 mi/hr


What have we learned today?
Notes

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


Notes

11

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