indefinite integrals both parts of the ftc establish connections between antiderivatives and...

INDEFINITE INTEGRALS

Both parts of the FTC establish

connections between antiderivatives

and definite integrals.

Part 1 says that if, f is continuous, then is an antiderivative of f.

Part 2 says that can be found by evaluating F(b) – F(a), where F is an antiderivative of f.

( )x

af t dt

( )b

af x dx

INDEFINITE INTEGRALS

We need a convenient notation for

antiderivatives that makes them easy

to work with.

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

integral.

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

INDEFINITE INTEGRAL

INDEFINITE INTEGRALS

For example, we can write

Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).

3 32 2because

3 3

x d xx dx C C x

dx

INDEFINITE VS. DEFINITE INTEGRALS

You should distinguish carefully between

definite and indefinite integrals.

A definite integral is a number.

An indefinite integral ∫ f(x) dx is a function (or family of functions).

( )b

af x dx

The connection between them is given

by the FTC2.

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

( ) ( )bb

a af x dx f x dx

INDEFINITE VS. DEFINITE INTEGRALS

INDEFINITE INTEGRALS

The effectiveness of the FTC depends

on having a supply of antiderivatives

of functions.

INDEFINITE INTEGRALS

Any formula can be verified by differentiating

the function on the right side and obtaining

the integrand.

For instance, 2

2

sec tan

because

(tan ) sec

x dx x C

dx C x

dx

TABLE OF INDEFINITE INTEGRALS

1

2 2

( ) ( ) [ ( ) ( )]

( ) ( )

( 1)1

sin cos cos sin

sec tan csc cot

sec tan sec csc cot csc

nn

cf x dx c f x dx f x g x dx

f x dx g x dx

xk dx kx C x dx C n

n

x dx x C x dx x C

x dx x C x dx x C

x x dx x C x x dx x C

INDEFINITE INTEGRALS

We write

with the understanding that it is valid on

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

2

1 1dx C

xx

INDEFINITE INTEGRALS

This is true despite the fact that the general

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

x ≠ 0, is:

1

2

1if 0

( )1

if 0

C xxF x

C xx

INDEFINITE INTEGRALS

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.

INDEFINITE INTEGRALS

Evaluate

This indefinite integral isn’t immediately apparent in Table 1.

So, we use trigonometric identities to rewrite the function before integrating:

2

cos 1 cos

sin sinsin

csc cot csc

d d

d C

2

cos

sind

INDEFINITE INTEGRALS

Evaluate

Using FTC2 and Table 1, we have:

3 3

0( 6 )x x dx

34 23 3

00

4 2 4 21 14 4

814

( 6 ) 64 2

3 3 3 0 3 0

27 0 0 6.75

x xx x dx

INDEFINITE INTEGRALS

Find

12

0( 12sin ) x x dx

The FTC gives:

This is the exact value of the integral.

INDEFINITE INTEGRALS

212 12

00

212

12sin 12( cos )2

(12) 12(cos12 cos0)

72 12cos12 12

60 12cos12

xx x dx x

INDEFINITE INTEGRALS

If a decimal approximation is desired, we can

use a calculator to approximate cos 12.

Doing so, we get:

12

012sin 70.1262 x x dx

The figure shows the graph of the integrand

in the example.

We know 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.

INDEFINITE INTEGRALS

INDEFINITE INTEGRALS

Evaluate

First, we need to write the integrand in a simpler form by carrying out the division:

2 29

21

2 1t t tdt

t

2 29 9 1 2 221 1

2 1(2 )

t t tdt t t dt

t

Then,9 1 2 2

1

93 2 1

32 1

93 2

1

3 2 3 22 1 2 13 9 3 1

1 2 49 3 9

(2 )

21

2 12

3

(2 9 9 ) (2 1 1 )

18 18 2 1 32

t t dt

t tt

t tt

INDEFINITE INTEGRALS

APPLICATIONS

The FTC2 says that, if f is continuous on

[a, b], then

where F is any antiderivative of f.

This means that F’ = f. So, the equation can be rewritten as:

( ) ( ) ( )b

af x dx F b F a

'( ) ( ) ( )b

aF x dx F b F a

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.

APPLICATIONS

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

aF x dx F b F a

NET CHANGE THEOREM

This principle can be applied to all the rates

of change in the natural and social sciences.

The following are a few instances of the idea.

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.

So,

is the change in the amount of water in the reservoir between time t1 and time t2.

2

12 1'( ) ( ) ( )

t

tV t dt V t V t

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.

So,

is the change in the concentration of C from time t1 to time t2.

2

12 1

[ ][ ]( ) [ ]( )

t

t

d Cdt C t C t

dt

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).

So,

is the mass of the segment of the rod that lies between x = a and x = b.

( ) ( ) ( )b

ax dx m b m a

NET CHANGE THEOREM

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

then

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.

2

12 1( ) ( )

t

t

dndt n t n t

dt

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).

So,

is the increase in cost when production is increased from x1 units to x2 units.

2

12 1'( ) ( ) ( )

x

xC x dx C x C x

NET CHANGE THEOREM

If an object moves along a straight line

with position function s(t), then its velocity

is v(t) = s’(t).

So,

is the net change of position, or displacement, of the particle during the time period from t1 to t2.

2

12 1( ) ( ) ( )

t

tv t dt s t s t

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)

NET CHANGE THEOREM

In both cases, the distance is computed by

integrating |v(t)|, the speed.

Therefore,

2

1

| ( ) | total distance traveledt

tv t dt

NET CHANGE THEOREM

The figure shows how both displacement and

distance traveled can be interpreted in terms

of areas under a velocity curve.

NET CHANGE THEOREM

The acceleration of the object is

a(t) = v’(t).

So,

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

2

12 1( ) ( ) ( )

t

ta t dt v t v t

NET CHANGE THEOREM

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.

NET CHANGE THEOREM

By Equation 2, the displacement is:

This means that the particle moved 4.5 m toward the left.

Example 6 a

4 4 2

1 1

43 2

1

(4) (1) ( ) ( 6)

96

3 2 2

s s v t dt t t dt

t tt

NET CHANGE THEOREM

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]

NET CHANGE THEOREM

So, from Equation 3, the distance traveled is:

4 3 4

1 1 3

3 42 2

1 3

3 43 2 3 2

1 3

( ) [ ( )] ( )

( 6) ( 6)

6 63 2 3 2

6110.17 m

6

v t dt v t dt v t dt

t t dt t t dt

t t t tt t

NET CHANGE THEOREM

We can use a Ti-calculator to find this definite integral.Use math 9 on Ti-83 & 84.

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.

Power is the rate of change of energy : P(t) = E’(t)

Power is the rate of change of energy:

P(t) = E’(t)

So, by the Net Change Theorem,

is the total amount of energy used that day.

24 24

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

NET CHANGE THEOREM

We approximate the value of

the integral using the Midpoint Rule

with 12 subintervals and ∆t = 2,

as follows.

NET CHANGE THEOREM

The energy used was approximately 15,840 megawatt-hours.

24

0( )

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

(440 400 420 620 790 840 850

840 810 690 670 550)(2)

15,840

P t dt

P P P P P t

NET CHANGE THEOREM

How did we know what

units to use for energy in

the example?

NET CHANGE THEOREM

The integral is defined as the limit

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

0( )P t dt

NET CHANGE THEOREM

In general, the unit of measurement for

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

the unit for x.

( )b

af x dx

NET CHANGE THEOREM