# Antiderivatives

Electronics - Devices

Oct 10, 2013 (4 years and 7 months ago)

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1

Notes on
Antiderivatives

So far our studies have concentrated on given a function how do we find its
derivative? Now we ask the reverse question: Given a derivative how do we find the
function?

The following definition provides some clarity for this last
question.

Definition
: Let
f

denote a function on an interval
I
. A function
F

is said to be an
antiderivative of

f

on the interval
I

provided
for all values of
.

Since
for
, the function
is an antiderivative of the
function
on the interval
. There are other antiderivatives of

such
as
,
and
. In fact, since
, where
C

is a constant, all
the functions
, where
C

is any constant, are antiderivatives of
.

So the are
infinitely many antiderivatives for the function
. Are there any others?

To answer this question the following theorems provide the basis
for the theory.
In addition these theorems have important consequences.

Rolle
’s Theorem
:
Suppose
is a function which is continuous on the interval
,
differentiable on the interval
and for which
. Then there is a number
,
,

for which
.

The Mean Value Theorem (for Derivatives)
:
Suppose
is a function which is
continuous on the interval
and differentiable

on the interval
.

Then there is a
number
,
, for which
.

Observations
:

The following are consequences of these theorems.

Rolle’s Theorem is a special
case of the Mean Value Theorem.

Rolle’s Theorem guarantees the existence of at least one critical point on the
interval
.

The geometrical interpretation of the Mean Value Theorem is that there is a
tangent line to the curve
which is parallel to the (secant) line joining the
points
and
.

When the function
is interpreted as the position of a particle moving along a
line, then the M
ean Value Theorem says that there is a point in time that the
instantaneous velocity is the same as the average velocity over the time
interval
.

2

The following theorems provide the answer to the questions posed earlier. The proofs

are based on the Mean Value Theorem and are included
at the end of these notes

for
completeness.

Theorem 1
: Suppose
f

is a differentiable function on an interval
I

for which
for
all
. Then
f

is a consta
nt function, that is to say,
for some constant
C
.

Theorem 2
: Suppose
f

is a function defined on an interval
I

and suppose also that
F

and
G

are both antiderivatives of
f

on the interval
I
. Then there is a constant
C

for
which
for all values of
x

in the interval
I
.

One can now answer

the ques
tion originally posed, namely, d
oes the family

of
functions
,
C

a constant, represent all the antiderivatives of the function
on the
interval
? According to Theorem 2, the answer is “YES”!

Indeed the function
is
a known antiderivative of
. If
is any other antideriv
ative of
, Theorem 2
states
for some constant
.

Definition
: Suppose a function
on an interval
I
, the family of all antiderivatives of
on
the interval
I

is called the
indefinite integral of
and this family is denoted by the
symbol
.

Observation
: Notice that Theorem 2 states that if
is a function for
wh
ich
, then
, where C is a constant called the
constant of integration
. Conversely, if
and
are functions for
which
, where C is

a constant, then
.

Example 1
: Verify that the function
satisfies the hypotheses of
Rolle’s Theorem on the interval
. Then find all the numbers on that interval that
satisfies t
he conclusion of Rolle’s Theorem.

Solution
: The given function is a polynomial and therefore continuous everywhere and in
particular on the interval
. Since polynomials are differentiable everywhere, the
given function is differen
tiable on the interval
.

Furthermore
and

. So the given function satisfies the
hypotheses of Rolle’s Theorem.

Find the numbers
so
that
:

(
U
se the Quadratic Formula
)

3

.

There are two possibilities, namely
and
. The second
possibility
is negative and therefore not in the interval
.
However the first
possibility
and therefore in the interval
.

Example 2
:
Verify that the function
satisfies the hypotheses of the Mean
Value Theorem on the interval
. Then find all the numbers
that satisfy the
conclusion of the Mean Value Theorem.

Solution
:

The given function is a combination of
functions that are continuous
everywhere and therefore continuous on the interval
. The given function is also
differentiable on the interval
since
everywhere. So the given funct
ion
satisfies the hypotheses of the Mean Value Theorem.

Find
so that
:

. So this number lies in

the interval
.

Example
3
:
Consider the function
. Show that there is no number
in the
interval
for which
. Why does this not

contradict the Mean
Value Theorem?

Solution
:

. Since this equation
which has no real solutions, there is no such number
. This does not contradict the Mean
Value Theorem because the given function has a discontinuity at
and is not
continuous on the interval
. (
It is not differentiable on
either
.
)

4

Example
4
:
Show that the equation
has exactly one real root.

Solution
:

The function
is a combination of continuous functions and
therefore continuous everywhere. Furthermore
and
. So
by the Intermediate Value Theorem there is at least one real root on the interval
.

Suppose there are two real roots,
.

(
We will reach a contradiction.
) For all values
o
f
x
,
. So the function
is continuous on the
interval
and differentiable on the interval
. Further
and
.
So by Rolle’s Theorem there must exist a number
,
, for which
.
So
. This is a contradiction.

So there cannot be two real roots of
the equation.

Ex
ample
5
:
Evaluate each of the following indefinite integrals.

a)

c)

b)

d)

Solution
:

a) Since
,

the function
is an antiderivative of
.

Therefore
.

b)

Since
, the function
is an antiderivative of
.

Therefore
.

c)

Since
, the function
is an antiderivative of
.

Therefore

d)

Since
, the function
is an anti
derivative of
.

Therefore
.

The proofs of the t
heorems are now presented. It is not important for you to
memorize these proofs, but
as
you read them
you can

see why
the theorems hold true
.

Rolle’s Theor
em
: Suppose
is a function which is continuous on the interval
,
differentiable on the interval
and for which
. Then there is a number
,
, for which
.

Proof
:
There are three cases to consider.

5

Case 1: Suppose
is a constant function. Then
for all values of
x

in the
interval
. Select a number
. Then
.

Case 2: Suppose
for some value of
. Since
is continuous on the
interval
, the function
has a maximum value, call it
, on the interval
.
Then the number
c

is a critical point for the function
and, since
is differentiable on the
interval
,
.

Case 3: Suppose
for some value of
. Since
is continuous on the
interval
, the function
has a
minimum

value, call it
, on the interval
.
Then the number
c

is a critical point for the function
and, since
is differentiable on the
interval
,
.

The Mean Value Theorem (for Derivatives)
: Suppose
is a function which is
continuous on the interval
and differentiable on the interval
.

Then there is a
number
,
, for which
.

Proof
: Define the function
which is a
combination of the given function and a polynomial. Since
is continuous

on the
interval
, so is
. Since
is

differentiable on the interval
, so is
.
Furthermore
and

.

So by Rolle’s Theorem, there is a
number
,
, for which
. Since
, it follows
that
and
.

Theorem 1
: Suppose
f

is a differentiable function on an interval
I

for which
for
all
. Then
f

is a constant function, that is to say,
for some constant
C
.

Proof
: Let
be a fixed point in the interval
I

and let x denote any other point in the
interval
I
. For demonstration purposes, suppo
se

and note that the interval
is
contained in the interval
I
. Since f is differentiable on the interval
I
, it is continuous on
the interval
I
. Therefore
f

is continuous on the interval
and differentiable on the
interval
. So by the Mean Value Theorem for Derivatives there is a point
in

6

the interval
for which
. Since
,

and it
follows that
. So
. Since the point
x

is
arbitrary, f is a constant function.

Theorem 2
: Suppose
f

is a function defined on an interval
I

and suppose a
lso that
F

and
G

are both antiderivatives of
f

on the interval
I
. Then there is a constant
C

for
which
for all values of
x

in the interval
I
.

Proof
: Define
for
. Since
F

and
G

ar
e antiderivatives of
f

on the
interval
I
, the definition states that
and
for all
.
Then
for all
. By Theorem 1,

for
all
for some constant
C
. Therefore

or
.