Contributed by:

In this section, we:

1. Introduce a notation for antiderivatives.

2. Review the formulas for antiderivatives.

3. Use the formulas to evaluate definite integrals.

4. Reformulate the second part of the FTC (FTC2) in a way that makes it easier to apply to science and engineering problems.

1. Introduce a notation for antiderivatives.

2. Review the formulas for antiderivatives.

3. Use the formulas to evaluate definite integrals.

4. Reformulate the second part of the FTC (FTC2) in a way that makes it easier to apply to science and engineering problems.

1.
5

2.
In Section 5.3, we saw that the second part of

the Fundamental Theorem of Calculus (FTC)

provides a very powerful method for

evaluating the definite integral of a function.

This is assuming that we can find an antiderivative

of the function.

the Fundamental Theorem of Calculus (FTC)

provides a very powerful method for

evaluating the definite integral of a function.

This is assuming that we can find an antiderivative

of the function.

3.
5.4

Indefinite Integrals and

the Net Change Theorem

In this section, we will learn about:

Indefinite integrals and their applications.

Indefinite Integrals and

the Net Change Theorem

In this section, we will learn about:

Indefinite integrals and their applications.

4.
INDEFINITE INTEGRALS AND NET CHANGE THEOREM

In this section, we:

Introduce a notation for antiderivatives.

Review the formulas for antiderivatives.

Use the formulas to evaluate definite integrals.

Reformulate the second part of the FTC (FTC2)

in a way that makes it easier to apply to science

and engineering problems.

In this section, we:

Introduce a notation for antiderivatives.

Review the formulas for antiderivatives.

Use the formulas to evaluate definite integrals.

Reformulate the second part of the FTC (FTC2)

in a way that makes it easier to apply to science

and engineering problems.

5.
INDEFINITE INTEGRALS

Both parts of the FTC establish

connections between antiderivatives

and definite integrals.

x

Part 1 says that if, f is continuous, then

is an antiderivative of f.

f (t ) dt

a

b

Part 2 says that a f ( x) dx can be found by evaluating

F(b) – F(a), where F is an antiderivative of f.

Both parts of the FTC establish

connections between antiderivatives

and definite integrals.

x

Part 1 says that if, f is continuous, then

is an antiderivative of f.

f (t ) dt

a

b

Part 2 says that a f ( x) dx can be found by evaluating

F(b) – F(a), where F is an antiderivative of f.

6.
INDEFINITE INTEGRALS

We need a convenient notation for

antiderivatives that makes them easy

to work with.

We need a convenient notation for

antiderivatives that makes them easy

to work with.

7.
INDEFINITE INTEGRAL

Due to the relation given by the FTC between

antiderivatives and integrals, the notation

∫ f(x) dx is traditionally used for an

antiderivative of f and is called an indefinite

Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)

Due to the relation given by the FTC between

antiderivatives and integrals, the notation

∫ f(x) dx is traditionally used for an

antiderivative of f and is called an indefinite

Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)

8.
INDEFINITE INTEGRALS

For example, we can write

3 3

2 x d x 2

x dx 3 C because dx 3 C x

Thus, we can regard an indefinite integral

as representing an entire family of functions

(one antiderivative for each value of the constant C).

For example, we can write

3 3

2 x d x 2

x dx 3 C because dx 3 C x

Thus, we can regard an indefinite integral

as representing an entire family of functions

(one antiderivative for each value of the constant C).

9.
INDEFINITE VS. DEFINITE INTEGRALS

You should distinguish carefully between

definite and indefinite integrals.

b

A definite integral f ( x) dx is a number.

a

An indefinite integral ∫ f(x) dx is a function

(or family of functions).

You should distinguish carefully between

definite and indefinite integrals.

b

A definite integral f ( x) dx is a number.

a

An indefinite integral ∫ f(x) dx is a function

(or family of functions).

10.
INDEFINITE VS. DEFINITE INTEGRALS

The connection between them is given

by the FTC2.

If f is continuous on [a, b], then

b b

f ( x ) dx f ( x ) dx

a a

The connection between them is given

by the FTC2.

If f is continuous on [a, b], then

b b

f ( x ) dx f ( x ) dx

a a

11.
INDEFINITE INTEGRALS

The effectiveness of the FTC depends

on having a supply of antiderivatives

of functions.

Therefore, we restate the Table of Antidifferentiation

Formulas from Section 4.9, together with a few others,

in the notation of indefinite integrals.

The effectiveness of the FTC depends

on having a supply of antiderivatives

of functions.

Therefore, we restate the Table of Antidifferentiation

Formulas from Section 4.9, together with a few others,

in the notation of indefinite integrals.

12.
INDEFINITE INTEGRALS

Any formula can be verified by differentiating

the function on the right side and obtaining

the integrand.

2

For instance, sec x dx tan x C

because

d

(tan x C ) sec 2 x

dx

Any formula can be verified by differentiating

the function on the right side and obtaining

the integrand.

2

For instance, sec x dx tan x C

because

d

(tan x C ) sec 2 x

dx

13.
TABLE OF INDEFINITE INTEGRALS Table 1

cf ( x) dx c f ( x) dx [ f ( x) g ( x)] dx f ( x) dx g ( x) dx

k dx kx C

n 1

n x 1

dx n 1 C (n 1)

x x dx ln | x | C

x x x ax

e dx e C a dx ln a C

cf ( x) dx c f ( x) dx [ f ( x) g ( x)] dx f ( x) dx g ( x) dx

k dx kx C

n 1

n x 1

dx n 1 C (n 1)

x x dx ln | x | C

x x x ax

e dx e C a dx ln a C

14.
TABLE OF INDEFINITE INTEGRALS Table 1

sin x dx cos x C cos x dx sin x C

2 2

sec x dx tan x C csc x dx cot x C

sec x tan x dx sec x C csc x cot x dx csc x C

1 1 1

x 2 1 dx tan x C 1 dx sin 1 x C

x2

sinh x dx cosh x C cosh x dx sinh x C

sin x dx cos x C cos x dx sin x C

2 2

sec x dx tan x C csc x dx cot x C

sec x tan x dx sec x C csc x cot x dx csc x C

1 1 1

x 2 1 dx tan x C 1 dx sin 1 x C

x2

sinh x dx cosh x C cosh x dx sinh x C

15.
INDEFINITE INTEGRALS

Recall from Theorem 1 in Section 4.9 that

the most general antiderivative on a given

interval is obtained by adding a constant to

a particular antiderivative.

We adopt the convention that, when a formula for

a general indefinite integral is given, it is valid only

on an interval.

Recall from Theorem 1 in Section 4.9 that

the most general antiderivative on a given

interval is obtained by adding a constant to

a particular antiderivative.

We adopt the convention that, when a formula for

a general indefinite integral is given, it is valid only

on an interval.

16.
INDEFINITE INTEGRALS

Thus, we write 1 1

x 2 dx C

x

with the understanding that it is valid on

the interval (0, ∞) or on the interval (-∞, 0).

Thus, we write 1 1

x 2 dx C

x

with the understanding that it is valid on

the interval (0, ∞) or on the interval (-∞, 0).

17.
INDEFINITE INTEGRALS

This is true despite the fact that the general

antiderivative of the function f(x) = 1/x2,

x ≠ 0, is:

1

C1 if x 0

x

F ( x)

1

C2 if x 0

x

This is true despite the fact that the general

antiderivative of the function f(x) = 1/x2,

x ≠ 0, is:

1

C1 if x 0

x

F ( x)

1

C2 if x 0

x

18.
INDEFINITE INTEGRALS Example 1

Find the general indefinite integral

∫ (10x4 – 2 sec2x) dx

Using our convention and Table 1, we have:

∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx

= 10(x5/5) – 2 tan x + C

= 2x5 – 2 tan x + C

You should check this answer by differentiating it.

Find the general indefinite integral

∫ (10x4 – 2 sec2x) dx

Using our convention and Table 1, we have:

∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx

= 10(x5/5) – 2 tan x + C

= 2x5 – 2 tan x + C

You should check this answer by differentiating it.

19.
INDEFINITE INTEGRALS Example 2

cos

Evaluate 2 d

sin

This indefinite integral isn’t immediately apparent

in Table 1.

So, we use trigonometric identities to rewrite

the function before integrating:

cos 1 cos

sin 2 d sin sin d

csc cot d csc C

cos

Evaluate 2 d

sin

This indefinite integral isn’t immediately apparent

in Table 1.

So, we use trigonometric identities to rewrite

the function before integrating:

cos 1 cos

sin 2 d sin sin d

csc cot d csc C

20.
INDEFINITE INTEGRALS Example 3

3

3

Evaluate ( x

0

6 x) dx

Using FTC2 and Table 1, we have:

4 2 3

3

3 x x

0 ( x 6 x) dx 4 6 2

0

3 3 3

1

4

4 2

1

4

4

0 3 0 2

814 27 0 0 6.75

Compare this with Example 2 b in Section 5.2

3

3

Evaluate ( x

0

6 x) dx

Using FTC2 and Table 1, we have:

4 2 3

3

3 x x

0 ( x 6 x) dx 4 6 2

0

3 3 3

1

4

4 2

1

4

4

0 3 0 2

814 27 0 0 6.75

Compare this with Example 2 b in Section 5.2

21.
INDEFINITE INTEGRALS Example 4

2 3 3

0

2 x 6 x 2 dx

x 1

and interpret the result in terms of areas.

2 3 3

0

2 x 6 x 2 dx

x 1

and interpret the result in terms of areas.

22.
INDEFINITE INTEGRALS Example 4

The FTC gives:

4 2

2 3 3 x x 1 2

0 2 x 6 x x 2 1 dx 2 4 6 2 3 tan x 0

4 2 1 2

x 3x 3 tan x

1

2 0

4 2 1

(2 ) 3(2 ) 3 tan 2 0

1

2

4 3 tan 1 2

This is the exact value of the integral.

The FTC gives:

4 2

2 3 3 x x 1 2

0 2 x 6 x x 2 1 dx 2 4 6 2 3 tan x 0

4 2 1 2

x 3x 3 tan x

1

2 0

4 2 1

(2 ) 3(2 ) 3 tan 2 0

1

2

4 3 tan 1 2

This is the exact value of the integral.

23.
INDEFINITE INTEGRALS Example 4

If a decimal approximation is desired, we can

use a calculator to approximate tan-1 2.

Doing so, we get:

2 3 3

0

2 x 6 x 2 dx 0.67855

x 1

If a decimal approximation is desired, we can

use a calculator to approximate tan-1 2.

Doing so, we get:

2 3 3

0

2 x 6 x 2 dx 0.67855

x 1

24.
INDEFINITE INTEGRALS

The figure shows the graph of the integrand

in the example.

We know from Section 5.2

that the value of the

integral can be

interpreted as the sum

of the areas labeled

with a plus sign minus

the area labeled with

a minus sign.

The figure shows the graph of the integrand

in the example.

We know from Section 5.2

that the value of the

integral can be

interpreted as the sum

of the areas labeled

with a plus sign minus

the area labeled with

a minus sign.

25.
INDEFINITE INTEGRALS Example 5

2 2

Evaluate 9 2t t t 1

1 2

dt

t

First, we need to write the integrand in a simpler

form by carrying out the division:

9 2t 2 t 2 t 1 9

12 2

1 2

dt (2 t t ) dt

t 1

2 2

Evaluate 9 2t t t 1

1 2

dt

t

First, we need to write the integrand in a simpler

form by carrying out the division:

9 2t 2 t 2 t 1 9

12 2

1 2

dt (2 t t ) dt

t 1

26.
INDEFINITE INTEGRALS Example 5

9

12 2

Then,

(2 t

1

t )dt

9

32 1

t t

2t

3

2 1 1

9

2 3 2 1

2t t

3 t 1

32 32

(2 9 9 2

3 ) (2 1 1 11 )

1

9

2

3

18 18 19 2 2

3 1 32 94

9

12 2

Then,

(2 t

1

t )dt

9

32 1

t t

2t

3

2 1 1

9

2 3 2 1

2t t

3 t 1

32 32

(2 9 9 2

3 ) (2 1 1 11 )

1

9

2

3

18 18 19 2 2

3 1 32 94

27.
The FTC2 says that, if f is continuous on

[a, b], then b

f ( x) dx F (b) F (a)

a

where F is any antiderivative of f.

This means that F’ = f.

So, the equation can be rewritten as:

b

F '( x) dx F (b) F (a)

a

[a, b], then b

f ( x) dx F (b) F (a)

a

where F is any antiderivative of f.

This means that F’ = f.

So, the equation can be rewritten as:

b

F '( x) dx F (b) F (a)

a

28.
We know F’(x) represents the rate of change

of y = F(x) with respect to x and F(b) – F(a) is

the change in y when x changes from a to b.

Note that y could, for instance, increase,

then decrease, then increase again.

Although y might change in both directions,

F(b) – F(a) represents the net change in y.

of y = F(x) with respect to x and F(b) – F(a) is

the change in y when x changes from a to b.

Note that y could, for instance, increase,

then decrease, then increase again.

Although y might change in both directions,

F(b) – F(a) represents the net change in y.

29.
NET CHANGE THEOREM

So, we can reformulate the FTC2 in words,

as follows.

The integral of a rate of change is

the net change: b

F '( x) dx F (b) F (a)

a

So, we can reformulate the FTC2 in words,

as follows.

The integral of a rate of change is

the net change: b

F '( x) dx F (b) F (a)

a

30.
NET CHANGE THEOREM

This principle can be applied to all the rates

of change in the natural and social sciences

that we discussed in Section 3.7

The following are a few instances of the idea.

This principle can be applied to all the rates

of change in the natural and social sciences

that we discussed in Section 3.7

The following are a few instances of the idea.

31.
NET CHANGE THEOREM

If V(t) is the volume of water in a reservoir at

time t, its derivative V’(t) is the rate at which

water flows into the reservoir at time t.

t2

So,

V '(t ) dt V (t

t1 2 ) V (t1 )

is the change in the amount of water

in the reservoir between time t1 and time t2.

If V(t) is the volume of water in a reservoir at

time t, its derivative V’(t) is the rate at which

water flows into the reservoir at time t.

t2

So,

V '(t ) dt V (t

t1 2 ) V (t1 )

is the change in the amount of water

in the reservoir between time t1 and time t2.

32.
NET CHANGE THEOREM

If [C](t) is the concentration of the product of

a chemical reaction at time t, then the rate of

reaction is the derivative d[C]/dt.

t2d [C ]

So,

t1 dt dt [C ](t2 ) [C ](t1 )

is the change in the concentration of C

from time t1 to time t2.

If [C](t) is the concentration of the product of

a chemical reaction at time t, then the rate of

reaction is the derivative d[C]/dt.

t2d [C ]

So,

t1 dt dt [C ](t2 ) [C ](t1 )

is the change in the concentration of C

from time t1 to time t2.

33.
NET CHANGE THEOREM

If the mass of a rod measured from the left

end to a point x is m(x), then the linear density

is ρ(x) = m’(x).

b

So,

( x) dx m(b) m(a)

a

is the mass of the segment of the rod

that lies between x = a and x = b.

If the mass of a rod measured from the left

end to a point x is m(x), then the linear density

is ρ(x) = m’(x).

b

So,

( x) dx m(b) m(a)

a

is the mass of the segment of the rod

that lies between x = a and x = b.

34.
NET CHANGE THEOREM

If the rate of growth of a population is dn/dt,

t2 dn

dt n(t2 ) n(t1 )

t1 dt

is the net change in population during the time

period from t1 to t2.

The population increases when births happen

and decreases when deaths occur.

The net change takes into account both births

and deaths.

If the rate of growth of a population is dn/dt,

t2 dn

dt n(t2 ) n(t1 )

t1 dt

is the net change in population during the time

period from t1 to t2.

The population increases when births happen

and decreases when deaths occur.

The net change takes into account both births

and deaths.

35.
NET CHANGE THEOREM

If C(x) is the cost of producing x units of

a commodity, then the marginal cost is

the derivative C’(x).

x2

So,

C '( x) dx C ( x ) C ( x )

x1 2 1

is the increase in cost when production

is increased from x1 units to x2 units.

If C(x) is the cost of producing x units of

a commodity, then the marginal cost is

the derivative C’(x).

x2

So,

C '( x) dx C ( x ) C ( x )

x1 2 1

is the increase in cost when production

is increased from x1 units to x2 units.

36.
NET CHANGE THEOREM Equation 2

If an object moves along a straight line

with position function s(t), then its velocity

is v(t) = s’(t).

t2

So,

v(t ) dt s(t

t1 2 ) s (t1 )

is the net change of position, or displacement,

of the particle during the time period from t1 to t2.

If an object moves along a straight line

with position function s(t), then its velocity

is v(t) = s’(t).

t2

So,

v(t ) dt s(t

t1 2 ) s (t1 )

is the net change of position, or displacement,

of the particle during the time period from t1 to t2.

37.
NET CHANGE THEOREM

In Section 5.1, we guessed that this was

true for the case where the object moves in

the positive direction.

Now, however, we have proved that it is

always true.

In Section 5.1, we guessed that this was

true for the case where the object moves in

the positive direction.

Now, however, we have proved that it is

always true.

38.
NET CHANGE THEOREM

If we want to calculate the distance the object

travels during that time interval, we have to

consider the intervals when:

v(t) ≥ 0 (the particle moves to the right)

v(t) ≤ 0 (the particle moves to the left)

If we want to calculate the distance the object

travels during that time interval, we have to

consider the intervals when:

v(t) ≥ 0 (the particle moves to the right)

v(t) ≤ 0 (the particle moves to the left)

39.
NET CHANGE THEOREM Equation 3

In both cases, the distance is computed by

integrating |v(t)|, the speed.

t2

| v(t ) | dt total distance traveled

t1

In both cases, the distance is computed by

integrating |v(t)|, the speed.

t2

| v(t ) | dt total distance traveled

t1

40.
NET CHANGE THEOREM

The figure shows how both displacement and

distance traveled can be interpreted in terms

of areas under a velocity curve.

The figure shows how both displacement and

distance traveled can be interpreted in terms

of areas under a velocity curve.

41.
NET CHANGE THEOREM

The acceleration of the object is

a(t) = v’(t).

t2

So,

a(t ) dt v(t

t1 2 ) v(t1 )

is the change in velocity from time t1 to time t2.

The acceleration of the object is

a(t) = v’(t).

t2

So,

a(t ) dt v(t

t1 2 ) v(t1 )

is the change in velocity from time t1 to time t2.

42.
NET CHANGE THEOREM Example 6

A particle moves along a line so that its

velocity at time t is:

v(t) = t2 – t – 6 (in meters per second)

a) Find the displacement of the particle during

the time period 1 ≤ t ≤ 4.

b) Find the distance traveled during this time period.

A particle moves along a line so that its

velocity at time t is:

v(t) = t2 – t – 6 (in meters per second)

a) Find the displacement of the particle during

the time period 1 ≤ t ≤ 4.

b) Find the distance traveled during this time period.

43.
NET CHANGE THEOREM Example 6 a

By Equation 2, the displacement is:

4 4

2

s (4) s (1) v(t ) dt (t t 6) dt

1 1

3 2 4

t t 9

6t

3 2 1 2

This means that the particle moved 4.5 m

toward the left.

By Equation 2, the displacement is:

4 4

2

s (4) s (1) v(t ) dt (t t 6) dt

1 1

3 2 4

t t 9

6t

3 2 1 2

This means that the particle moved 4.5 m

toward the left.

44.
NET CHANGE THEOREM Example 6 b

Note that

v(t) = t2 – t – 6 = (t – 3)(t + 2)

Thus,

v(t) ≤ 0 on the interval [1, 3] and v(t) ≥ 0 on [3, 4]

Note that

v(t) = t2 – t – 6 = (t – 3)(t + 2)

Thus,

v(t) ≤ 0 on the interval [1, 3] and v(t) ≥ 0 on [3, 4]

45.
NET CHANGE THEOREM Example 6 b

So, from Equation 3, the distance traveled is:

4 3 4

v(t ) dt [ v(t )] dt v(t ) dt

1 1 3

3 4

2 2

( t t 6) dt (t t 6) dt

1 3

3 2 3 3 2 4

t t t t

6t 6t

3 2 1 3 2 3

61

10.17 m

6

So, from Equation 3, the distance traveled is:

4 3 4

v(t ) dt [ v(t )] dt v(t ) dt

1 1 3

3 4

2 2

( t t 6) dt (t t 6) dt

1 3

3 2 3 3 2 4

t t t t

6t 6t

3 2 1 3 2 3

61

10.17 m

6

46.
NET CHANGE THEOREM Example 7

The figure shows the power consumption in

San Francisco for a day in September.

P is measured in megawatts.

t is measured in hours starting at midnight.

Estimate the

energy used

on that day.

The figure shows the power consumption in

San Francisco for a day in September.

P is measured in megawatts.

t is measured in hours starting at midnight.

Estimate the

energy used

on that day.

47.
NET CHANGE THEOREM Example 7

Power is the rate of change of energy:

P(t) = E’(t)

So, by the Net Change Theorem,

24 24

P(t ) dt E '(t ) dt E (24) E (0)

0 0

is the total amount of energy used that day.

Power is the rate of change of energy:

P(t) = E’(t)

So, by the Net Change Theorem,

24 24

P(t ) dt E '(t ) dt E (24) E (0)

0 0

is the total amount of energy used that day.

48.
NET CHANGE THEOREM Example 7

We approximate the value of

the integral using the Midpoint Rule

with 12 subintervals and ∆t = 2,

as follows.

We approximate the value of

the integral using the Midpoint Rule

with 12 subintervals and ∆t = 2,

as follows.

49.
NET CHANGE THEOREM Example 7

24

P(t ) dt

0

[ P (1) P(3) P(5) ... P(21) P(23)]t

(440 400 420 620 790 840 850

840 810 690 670 550)(2)

The energy used was approximately

15,840 megawatt-hours.

24

P(t ) dt

0

[ P (1) P(3) P(5) ... P(21) P(23)]t

(440 400 420 620 790 840 850

840 810 690 670 550)(2)

The energy used was approximately

15,840 megawatt-hours.

50.
NET CHANGE THEOREM

How did we know what

units to use for energy in

the example?

How did we know what

units to use for energy in

the example?

51.
NET CHANGE THEOREM

24

The integral P(t ) dt is defined as the limit

0

of sums of terms of the form P(ti*) ∆t.

Now, P(ti*) is measured in megawatts and

∆t is measured in hours.

So, their product is measured in megawatt-hours.

The same is true of the limit.

24

The integral P(t ) dt is defined as the limit

0

of sums of terms of the form P(ti*) ∆t.

Now, P(ti*) is measured in megawatts and

∆t is measured in hours.

So, their product is measured in megawatt-hours.

The same is true of the limit.

52.
NET CHANGE THEOREM

In general, the unit of measurement for

b

f ( x) dx

a

is the product of the unit for f(x) and

the unit for x.

In general, the unit of measurement for

b

f ( x) dx

a

is the product of the unit for f(x) and

the unit for x.