At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 23: Antiderivatives (Section 041 handout)
1. Section 4.7
Antiderivatives
V63.0121.041, Calculus I
New York University
November 29, 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
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 2 / 33
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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 3 / 33
Notes
Notes
Notes
1
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
2. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 4 / 33
What is an antiderivative?
Definition
Let f be a function. An antiderivative for f is a function F such that
F = f .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 5 / 33
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 "
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 33
Notes
Notes
Notes
2
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
3. 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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 7 / 33
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)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 8 / 33
Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 9 / 33
Notes
Notes
Notes
3
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
4. 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) = ?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 33
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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 33
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 .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 33
Notes
Notes
Notes
4
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
5. 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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 33
Graph of ln |x|
x
y
f (x) = 1/x
F(x) = ln |x|
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 33
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
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 33
Notes
Notes
Notes
5
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
6. Antiderivatives of Polynomials
Example
Find an antiderivative for f (x) = 16x + 5.
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 .
Question
Why do we not need two C’s?
Answer
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 33
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 .
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 33
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).
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 33
Notes
Notes
Notes
6
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
7. Trigonometric functions
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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 33
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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 33
Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 33
Notes
Notes
Notes
7
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
8. Finding Antiderivatives Graphically
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)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 22 / 33
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?
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 23 / 33
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
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 24 / 33
Notes
Notes
Notes
8
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
9. Outline
What is an antiderivative?
Tabulating Antiderivatives
Power functions
Combinations
Exponential functions
Trigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 33
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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 33
Application: Dead Reckoning
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 27 / 33
Notes
Notes
Notes
9
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
10. 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
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 28 / 33
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.
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 33
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)
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 33
Notes
Notes
Notes
10
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
11. Implementing the Solution
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 31 / 33
Solving
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 33
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) = ???
V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 33 / 33
Notes
Notes
Notes
11
Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010