Lesson 23: Antiderivatives (Section 021 handout)

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
Announcements
Quiz 5 in recitation this
week on §§4.1–4.4
V63.0121.021, Calculus I (NYU) Section 4.7 Antiderivatives November 30, 2010 2 / 35
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.
Notes
Notes
Notes
1
Section 4.7 : AntiderivativesV63.0121.021, Calculus I November 30, 2010

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 .
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
(x ln x − x) = 1 · ln x + x ·
1
x
− 1 = ln x "
Notes
Notes
Notes
2

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
Notes
Notes
Notes
3

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
.
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
.
Solution
Try a power function F(x) = axr
Then F (x) = arxr−1
, so we want arxr−1
= x3
.
r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =
1
4
.
So F(x) =
1
4
x4
is an antiderivative.
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 . . . as long as r = −1.
Fact
If f (x) = x−1
=
1
x
, then
F(x) = ln |x| + C
is an antiderivative for f .
Notes
Notes
Notes
4

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.
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
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 .
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
Notes
Notes
Notes
5

Antiderivatives of Polynomials
Example
Find an antiderivative for f (x) = 16x + 5.
Solution
Question
Do we need two C’s or just one?
Answer
Exponential Functions
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 .
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.
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).
Notes
Notes
Notes
6

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.
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.
Solution
Notes
Notes
Notes
7

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)
Notes
Notes
Notes
8

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
F1 2 3 4 5 6
+ + − − +
max min
f = F
F1 2 3 4 5 6
++ −− −− ++ ++
IP IP
F
shape1 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
We start with F(1) = 0.
Using the sign chart, we
draw arcs with the speciﬁed
monotonicity and 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
shape1 2 3 4 5 6
IP
max
IP
min
Outline
Power functions
Combinations
Rectilinear motion
Notes
Notes
Notes
9

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).
Solution
By Newton’s Second Law (F = ma) a constant force induces a
constant acceleration. So a(t) = a =
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
Notes
Notes
Notes
10

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?
Solution (Setup)
Implementing the Solution
Notes
Notes
Notes
11

Solving
Summary
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) = ???
Notes
Notes
Notes
12

Lesson 23: Antiderivatives (Section 021 handout)

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