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In this section, we will learn about: Antiderivatives and how they are useful in solving certain scientific problems.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

4.9

Antiderivatives

In this section, we will learn about:

Antiderivatives and how they are useful

in solving certain scientific problems.

4.9

Antiderivatives

In this section, we will learn about:

Antiderivatives and how they are useful

in solving certain scientific problems.

3.
A physicist who knows the velocity of

a particle might wish to know its position

at a given time.

a particle might wish to know its position

at a given time.

4.
An engineer who can measure the variable

rate at which water is leaking from a tank

wants to know the amount leaked over

a certain time period.

rate at which water is leaking from a tank

wants to know the amount leaked over

a certain time period.

5.
A biologist who knows the rate at which

a bacteria population is increasing might want

to deduce what the size of the population will

be at some future time.

a bacteria population is increasing might want

to deduce what the size of the population will

be at some future time.

6.
In each case, the problem is to find

a function F whose derivative is a known

function f.

If such a function F exists, it is called

an antiderivative of f.

a function F whose derivative is a known

function f.

If such a function F exists, it is called

an antiderivative of f.

7.
A function F is called an antiderivative

of f on an interval I if F’(x) = f(x) for all

x in I.

of f on an interval I if F’(x) = f(x) for all

x in I.

8.
For instance, let f(x) = x2.

It isn’t difficult to discover an antiderivative of f

if we keep the Power Rule in mind.

In fact, if F(x) = ⅓ x3, then F’(x) = x2 = f(x).

It isn’t difficult to discover an antiderivative of f

if we keep the Power Rule in mind.

In fact, if F(x) = ⅓ x3, then F’(x) = x2 = f(x).

9.
However, the function G(x) = ⅓ x3 + 100

also satisfies G’(x) = x2.

Therefore, both F and G are antiderivatives of f.

also satisfies G’(x) = x2.

Therefore, both F and G are antiderivatives of f.

10.
Indeed, any function of the form

H(x)=⅓ x3 + C, where C is a constant,

is an antiderivative of f.

The question arises: Are there any others?

H(x)=⅓ x3 + C, where C is a constant,

is an antiderivative of f.

The question arises: Are there any others?

11.
To answer the question, recall that,

in Section 4.2, we used the Mean Value

It was to prove that, if two functions have identical

derivatives on an interval, then they must differ by

a constant (Corollary 7).

in Section 4.2, we used the Mean Value

It was to prove that, if two functions have identical

derivatives on an interval, then they must differ by

a constant (Corollary 7).

12.
Thus, if F and G are any two antiderivatives

of f, then F’(x) = f(x) = G(x)

So, G(x) – F(x) = C, where C is a constant.

We can write this as G(x) = F(x) + C.

Hence, we have the following theorem.

of f, then F’(x) = f(x) = G(x)

So, G(x) – F(x) = C, where C is a constant.

We can write this as G(x) = F(x) + C.

Hence, we have the following theorem.

13.
ANTIDERIVATIVES Theorem 1

If F is an antiderivative of f on an interval I,

the most general antiderivative of f on I is

F(x) + C

where C is an arbitrary constant.

If F is an antiderivative of f on an interval I,

the most general antiderivative of f on I is

F(x) + C

where C is an arbitrary constant.

14.
Going back to the function f(x) = x2,

we see that the general antiderivative

of f is ⅓ x3 + C.

we see that the general antiderivative

of f is ⅓ x3 + C.

15.
FAMILY OF FUNCTIONS

By assigning specific values to C, we

obtain a family of functions.

Their graphs are

vertical translates

of one another.

This makes sense, as

each curve must have

the same slope at any

given value of x.

By assigning specific values to C, we

obtain a family of functions.

Their graphs are

vertical translates

of one another.

This makes sense, as

each curve must have

the same slope at any

given value of x.

16.
ANTIDERIVATIVES Example 1

Find the most general antiderivative of

each function.

a. f(x) = sin x

b. f(x) = 1/x

c. f(x) = xn, n ≠ -1

Find the most general antiderivative of

each function.

a. f(x) = sin x

b. f(x) = 1/x

c. f(x) = xn, n ≠ -1

17.
ANTIDERIVATIVES Example 1 a

If f(x) = - cos x, then F’(x) = sin x.

So, an antiderivative of sin x is - cos x.

By Theorem 1, the most general antiderivative

is: G(x) = - cos x + C

If f(x) = - cos x, then F’(x) = sin x.

So, an antiderivative of sin x is - cos x.

By Theorem 1, the most general antiderivative

is: G(x) = - cos x + C

18.
ANTIDERIVATIVES Example 1 b

Recall from Section 3.6 that

d 1

(ln x)

dx x

So, on the interval (0, ∞), the general antiderivative

of 1/x is ln x + C.

Recall from Section 3.6 that

d 1

(ln x)

dx x

So, on the interval (0, ∞), the general antiderivative

of 1/x is ln x + C.

19.
ANTIDERIVATIVES Example 1 b

We also learned that

d 1

(ln | x |)

dx x

for all x ≠ 0.

Theorem 1 then tells us that the general antiderivative

of f(x) = 1/x is ln |x| + C on any interval that doesn’t

contain 0.

We also learned that

d 1

(ln | x |)

dx x

for all x ≠ 0.

Theorem 1 then tells us that the general antiderivative

of f(x) = 1/x is ln |x| + C on any interval that doesn’t

contain 0.

20.
ANTIDERIVATIVES Example 1 b

In particular, this is true on each of

the intervals (-∞, 0) and (0, ∞).

So, the general antiderivative of f is:

ln x C1 if x 0

F ( x)

ln( x) C2 if x 0

In particular, this is true on each of

the intervals (-∞, 0) and (0, ∞).

So, the general antiderivative of f is:

ln x C1 if x 0

F ( x)

ln( x) C2 if x 0

21.
ANTIDERIVATIVES Example 1 c

We use the Power Rule to discover

an antiderivative of xn.

In fact, if n ≠ -1, then

n 1 n

d x (n 1) x n

x

dx n 1 n 1

We use the Power Rule to discover

an antiderivative of xn.

In fact, if n ≠ -1, then

n 1 n

d x (n 1) x n

x

dx n 1 n 1

22.
ANTIDERIVATIVES Example 1 c

Therefore, the general antiderivative

n 1

of f(x) = xn is: x

F ( x) C

n 1

This is valid for n ≥ 0 since then f(x) = xn is defined

on an interval.

If n is negative (but n ≠ -1), it is valid on any interval

that doesn’t contain 0.

Therefore, the general antiderivative

n 1

of f(x) = xn is: x

F ( x) C

n 1

This is valid for n ≥ 0 since then f(x) = xn is defined

on an interval.

If n is negative (but n ≠ -1), it is valid on any interval

that doesn’t contain 0.

23.
ANTIDERIVATIVE FORMULA

As in the example, every differentiation

formula, when read from right to left,

gives rise to an antidifferentiation formula.

As in the example, every differentiation

formula, when read from right to left,

gives rise to an antidifferentiation formula.

24.
ANTIDERIVATIVE FORMULA Table 2

Here, we list some particular

Here, we list some particular

25.
ANTIDERIVATIVE FORMULA

Each formula is true because the derivative

of the function in the right column appears

in the left column.

Each formula is true because the derivative

of the function in the right column appears

in the left column.

26.
ANTIDERIVATIVE FORMULA

In particular, the first formula says that the

antiderivative of a constant times a function

is the constant times the antiderivative of

the function.

In particular, the first formula says that the

antiderivative of a constant times a function

is the constant times the antiderivative of

the function.

27.
ANTIDERIVATIVE FORMULA

The second formula says that the

antiderivative of a sum is the sum of the

We use the notation F’ = f, G’ = g.

The second formula says that the

antiderivative of a sum is the sum of the

We use the notation F’ = f, G’ = g.

28.
ANTIDERIVATIVES Example 2

Find all functions g such that

5

2x x

g '( x) 4sin x

x

Find all functions g such that

5

2x x

g '( x) 4sin x

x

29.
ANTIDERIVATIVES Example 2

First, we rewrite the given function:

5

2x x 4 1

g '( x) 4sin x 4sin x 2 x

x x x

Thus, we want to find an antiderivative of:

4 1 2

g '( x) 4sin x 2 x x

First, we rewrite the given function:

5

2x x 4 1

g '( x) 4sin x 4sin x 2 x

x x x

Thus, we want to find an antiderivative of:

4 1 2

g '( x) 4sin x 2 x x

30.
ANTIDERIVATIVES Example 2

Using the formulas in Table 2 together

with Theorem 1, we obtain:

5 12

x x

g ( x) 4( cos x) 2 1 C

5 2

5

4 cos x x 2 x C

2

5

Using the formulas in Table 2 together

with Theorem 1, we obtain:

5 12

x x

g ( x) 4( cos x) 2 1 C

5 2

5

4 cos x x 2 x C

2

5

31.
In applications of calculus, it is very common

to have a situation as in the example—

where it is required to find a function, given

knowledge about its derivatives.

to have a situation as in the example—

where it is required to find a function, given

knowledge about its derivatives.

32.
DIFFERENTIAL EQUATIONS

An equation that involves the derivatives of

a function is called a differential equation.

These will be studied in some detail in Chapter 9.

For the present, we can solve some elementary

differential equations.

An equation that involves the derivatives of

a function is called a differential equation.

These will be studied in some detail in Chapter 9.

For the present, we can solve some elementary

differential equations.

33.
DIFFERENTIAL EQUATIONS

The general solution of a differential equation

involves an arbitrary constant (or constants),

as in Example 2.

However, there may be some extra conditions

given that will determine the constants and, therefore,

uniquely specify the solution.

The general solution of a differential equation

involves an arbitrary constant (or constants),

as in Example 2.

However, there may be some extra conditions

given that will determine the constants and, therefore,

uniquely specify the solution.

34.
DIFFERENTIAL EQUATIONS Example 3

Find f if

f’(x) = ex + 20(1 + x2)-1 and f(0) = – 2

The general antiderivative of

x 20

f '( x) e

1 x2

x 1

is f '( x) e 20 tan x C

Find f if

f’(x) = ex + 20(1 + x2)-1 and f(0) = – 2

The general antiderivative of

x 20

f '( x) e

1 x2

x 1

is f '( x) e 20 tan x C

35.
DIFFERENTIAL EQUATIONS Example 3

To determine C, we use the fact that

f(0) = – 2 :

f(0) = e0 + 20 tan-10 + C = – 2

Thus, we have: C = – 2 – 1 = – 3

So, the particular solution is:

f(x) = ex + 20 tan-1x –

3

To determine C, we use the fact that

f(0) = – 2 :

f(0) = e0 + 20 tan-10 + C = – 2

Thus, we have: C = – 2 – 1 = – 3

So, the particular solution is:

f(x) = ex + 20 tan-1x –

3

36.
DIFFERENTIAL EQUATIONS Example 4

Find f if f’’(x) = 12x2 + 6x – 4,

f(0) = 4, and f(1) = 1.

Find f if f’’(x) = 12x2 + 6x – 4,

f(0) = 4, and f(1) = 1.

37.
DIFFERENTIAL EQUATIONS Example 4

The general antiderivative of

f’’(x) = 12x2 + 6x – 4 is:

3 2

x x

f '( x) 12 6 4 x C

3 2

3 2

4 x 3x 4 x C

The general antiderivative of

f’’(x) = 12x2 + 6x – 4 is:

3 2

x x

f '( x) 12 6 4 x C

3 2

3 2

4 x 3x 4 x C

38.
DIFFERENTIAL EQUATIONS Example 4

Using the antidifferentiation rules once

more, we find that:

4 3 2

x x x

f ( x) 4 3 4 Cx D

4 3 2

4 3 2

x x 2 x Cx D

Using the antidifferentiation rules once

more, we find that:

4 3 2

x x x

f ( x) 4 3 4 Cx D

4 3 2

4 3 2

x x 2 x Cx D

39.
DIFFERENTIAL EQUATIONS Example 4

To determine C and D, we use the given

conditions that f(0) = 4 and f(1) = 1.

As f(0) = 0 + D = 4, we have: D = 4

As f (1) = 1 + 1 – 2 + C + 4 = 1, we have: C = –3

To determine C and D, we use the given

conditions that f(0) = 4 and f(1) = 1.

As f(0) = 0 + D = 4, we have: D = 4

As f (1) = 1 + 1 – 2 + C + 4 = 1, we have: C = –3

40.
DIFFERENTIAL EQUATIONS Example 4

Therefore, the required function

f(x) = x4 + x3 – 2x2 – 3x + 4

Therefore, the required function

f(x) = x4 + x3 – 2x2 – 3x + 4

41.
If we are given the graph of a function f,

it seems reasonable that we should be able

to sketch the graph of an antiderivative F.

it seems reasonable that we should be able

to sketch the graph of an antiderivative F.

42.
Suppose we are given that F(0) = 1.

We have a place to start—the point (0, 1).

The direction in which we move our pencil

is given at each stage by the derivative F’(x) = f(x).

We have a place to start—the point (0, 1).

The direction in which we move our pencil

is given at each stage by the derivative F’(x) = f(x).

43.
In the next example, we use the principles

of this chapter to show how to graph F even

when we don’t have a formula for f.

This would be the case, for instance, when

f(x) is determined by experimental data.

of this chapter to show how to graph F even

when we don’t have a formula for f.

This would be the case, for instance, when

f(x) is determined by experimental data.

44.
GRAPH Example 5

The graph of a function f is given.

Make a rough sketch of an antiderivative F,

given that F(0) = 2.

The graph of a function f is given.

Make a rough sketch of an antiderivative F,

given that F(0) = 2.

45.
GRAPH Example 5

We are guided by the fact that

the slope of y = F(x) is f(x).

We are guided by the fact that

the slope of y = F(x) is f(x).

46.
GRAPH Example 5

We start at (0, 2) and

draw F as an initially

decreasing function

since f(x) is negative

when 0 < x < 1.

We start at (0, 2) and

draw F as an initially

decreasing function

since f(x) is negative

when 0 < x < 1.

47.
GRAPH Example 5

Notice f(1) = f(3) = 0.

So, F has horizontal

tangents when x = 1

and x = 3.

For 1 < x < 3, f(x)

is positive.

Thus, F is increasing.

Notice f(1) = f(3) = 0.

So, F has horizontal

tangents when x = 1

and x = 3.

For 1 < x < 3, f(x)

is positive.

Thus, F is increasing.

48.
GRAPH Example 5

We see F has a local

minimum when x = 1

and a local maximum

when x = 3.

For x > 3, f(x) is

negative.

Thus, F is decreasing

on (3, ∞).

We see F has a local

minimum when x = 1

and a local maximum

when x = 3.

For x > 3, f(x) is

negative.

Thus, F is decreasing

on (3, ∞).

49.
GRAPH Example 5

Since f(x) → 0 as

x → ∞, the graph

of F becomes flatter

as x → ∞.

Since f(x) → 0 as

x → ∞, the graph

of F becomes flatter

as x → ∞.

50.
GRAPH Example 5

Also, F’’(x) = f’(x)

changes from positive

to negative at x = 2

and from negative to

positive at x = 4.

So, F has inflection

points when x = 2

and x = 4.

Also, F’’(x) = f’(x)

changes from positive

to negative at x = 2

and from negative to

positive at x = 4.

So, F has inflection

points when x = 2

and x = 4.

51.
RECTILINEAR MOTION

Antidifferentiation is particularly useful

in analyzing the motion of an object

moving in a straight line.

Antidifferentiation is particularly useful

in analyzing the motion of an object

moving in a straight line.

52.
RECTILINEAR MOTION

Recall that, if the object has position

function s = f(t), then the velocity function

is v(t) = s’(t).

This means that the position function is

an antiderivative of the velocity function.

Recall that, if the object has position

function s = f(t), then the velocity function

is v(t) = s’(t).

This means that the position function is

an antiderivative of the velocity function.

53.
RECTILINEAR MOTION

Likewise, the acceleration function

is a(t) = v’(t).

So, the velocity function is an antiderivative

of the acceleration.

Likewise, the acceleration function

is a(t) = v’(t).

So, the velocity function is an antiderivative

of the acceleration.

54.
RECTILINEAR MOTION

If the acceleration and the initial values

s(0) and v(0) are known, then the position

function can be found by antidifferentiating

If the acceleration and the initial values

s(0) and v(0) are known, then the position

function can be found by antidifferentiating

55.
RECTILINEAR MOTION Example 6

A particle moves in a straight line and has

acceleration given by a(t) = 6t + 4.

Its initial velocity is v(0) = -6 cm/s and its initial

displacement is s(0) = 9 cm.

Find its position function s(t).

A particle moves in a straight line and has

acceleration given by a(t) = 6t + 4.

Its initial velocity is v(0) = -6 cm/s and its initial

displacement is s(0) = 9 cm.

Find its position function s(t).

56.
RECTILINEAR MOTION Example 6

As v’(t) = a(t) = 6t + 4, antidifferentiation

2

t

v(t ) 6 4t C

2

2

3t 4t C

As v’(t) = a(t) = 6t + 4, antidifferentiation

2

t

v(t ) 6 4t C

2

2

3t 4t C

57.
RECTILINEAR MOTION Example 6

Note that v(0) = C.

However, we are given that v(0) = – 6,

so C = – 6.

Therefore, we have:

v(t) = 3t2 + 4t – 6

Note that v(0) = C.

However, we are given that v(0) = – 6,

so C = – 6.

Therefore, we have:

v(t) = 3t2 + 4t – 6

58.
RECTILINEAR MOTION Example 6

As v(t) = s’(t), s is the antiderivative

of v: t 3

t 2

s (t ) 3 4 6t D

3 2

3 2

t 2t 6t D

This gives s(0) = D.

We are given that s(0) = 9, so D = 9.

As v(t) = s’(t), s is the antiderivative

of v: t 3

t 2

s (t ) 3 4 6t D

3 2

3 2

t 2t 6t D

This gives s(0) = D.

We are given that s(0) = 9, so D = 9.

59.
RECTILINEAR MOTION Example 6

The required position function

s(t) = t3 + 2t 2 – 6t + 9

The required position function

s(t) = t3 + 2t 2 – 6t + 9

60.
RECTILINEAR MOTION

An object near the surface of the earth is

subject to a gravitational force that produces

a downward acceleration denoted by g.

For motion close to the ground, we may

assume that g is constant.

Its value is about 9.8 m/s2 (or 32 ft/s2).

An object near the surface of the earth is

subject to a gravitational force that produces

a downward acceleration denoted by g.

For motion close to the ground, we may

assume that g is constant.

Its value is about 9.8 m/s2 (or 32 ft/s2).

61.
RECTILINEAR MOTION Example 7

A ball is thrown upward with a speed of

48 ft/s from the edge of a cliff 432 ft above

the ground.

Find its height above the ground t seconds later.

When does it reach its maximum height?

When does it hit the ground?

A ball is thrown upward with a speed of

48 ft/s from the edge of a cliff 432 ft above

the ground.

Find its height above the ground t seconds later.

When does it reach its maximum height?

When does it hit the ground?

62.
RECTILINEAR MOTION Example 7

The motion is vertical, and we choose

the positive direction to be upward.

At time t, the distance above the ground is s(t)

and the velocity v(t) is decreasing.

So, the acceleration must be negative

and we have: dv

a (t ) 32

dt

The motion is vertical, and we choose

the positive direction to be upward.

At time t, the distance above the ground is s(t)

and the velocity v(t) is decreasing.

So, the acceleration must be negative

and we have: dv

a (t ) 32

dt

63.
RECTILINEAR MOTION Example 7

Taking antiderivatives, we have

v(t) = – 32t + C

To determine C, we use the information

that v(0) = 48.

This gives: 48 = 0 + C

So, v(t) = – 32t + 48

Taking antiderivatives, we have

v(t) = – 32t + C

To determine C, we use the information

that v(0) = 48.

This gives: 48 = 0 + C

So, v(t) = – 32t + 48

64.
RECTILINEAR MOTION Example 7

The maximum height is reached

v(t) = 0, that is, after 1.5 s

The maximum height is reached

v(t) = 0, that is, after 1.5 s

65.
RECTILINEAR MOTION Example 7

As s’(t) = v(t), we antidifferentiate again

and obtain:

s(t) = – 16t2 + 48t + D

Using the fact that s(0) = 432, we have

432 = 0 + D. So,

s(t) = – 16t2 + 48t + 432

As s’(t) = v(t), we antidifferentiate again

and obtain:

s(t) = – 16t2 + 48t + D

Using the fact that s(0) = 432, we have

432 = 0 + D. So,

s(t) = – 16t2 + 48t + 432

66.
RECTILINEAR MOTION Example 7

The expression for s(t) is valid until the ball

hits the ground.

This happens when s(t) = 0, that is, when

– 16t2 + 48t + 432 = 0

Equivalently: t2 – 3t – 27 = 0

The expression for s(t) is valid until the ball

hits the ground.

This happens when s(t) = 0, that is, when

– 16t2 + 48t + 432 = 0

Equivalently: t2 – 3t – 27 = 0

67.
RECTILINEAR MOTION Example 7

Using the quadratic formula to solve this

equation, we get: 3 3 13

t

2

We reject the solution with the minus sign—as

it gives a negative value for t.

Using the quadratic formula to solve this

equation, we get: 3 3 13

t

2

We reject the solution with the minus sign—as

it gives a negative value for t.

68.
RECTILINEAR MOTION Example 7

Therefore, the ball hits the ground

3(1 + 13 )/2 ≈ 6.9 s

Therefore, the ball hits the ground

3(1 + 13 )/2 ≈ 6.9 s