Indefinite Integrals and the Net Change Theorem

8/14/2019 Indefinite Integrals and the Net Change Theorem

1/34

5.4

Indefinite Integrals and

the Net Change Theorem

Wednesday, February 10, 2010

In this section, we will learn about:Indefinite integrals and their applications.

INTEGRALS


2/34

INDEFINITE INTEGRALS

Both parts of the FTC establish

connections between antiderivatives

and definite integrals.

Part 1 says that if, fis continuous, thenis an antiderivative of f.

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

( )x

a f t dt

( )

b

a f x dx


3/34

Due to the relation given by the FTC between

antiderivatives and integrals, the notation

f(x) dxis traditionally used for an

antiderivative of fand is called an indefiniteintegral.

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

INDEFINITE INTEGRAL


4/34


For example, we can write

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

3 32 2

because3 3

x d x x dx C C x

dx


5/34

INDEFINITE VS. DEFINITE INTEGRALS

You should distinguish carefully between

definite and indefinite integrals.

A definite integral is a number.

An indefinite integral f(x) dxis a function(or family of functions).

( )

b

a f x dx


6/34


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

d x C x

dx


7/34

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

x

k dx kx C x dx C nn

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

Table 1


8/34


Thus, we write

with the understanding that it is valid on

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

21 1dx Cxx


9/34


This is true despite the fact that the general

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

x 0, is:

1

2

1 if 0

( )1

if 0

C xx

F x

C xx


10/34


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.

Example 1


11/34


Evaluate

This indefinite integral isnt immediately apparentin Table 1.

So, we use trigonometric identities to rewritethe function before integrating:

Example 2

2

cos 1 cos

sin sinsin

csc cot csc

d d

d C

2

cos

sin d


12/34


Evaluate

Using FTC2 and Table 1, we have:

Compare this with Example 2 b in Section 5.2

Example 33

3

0

( 6 ) x x dx

34 23

3

00

4 2 4 21 1

4 4

81

4

( 6 ) 64 2

3 3 3 0 3 0

27 0 0 6.75

x x x x dx


13/34


Find

Example 4

12

0( 12sin ) x x dx


14/34

The FTC gives:

This is the exact value of the integral.

INDEFINITE INTEGRALS Example 4

2

12 12

00

212

12sin |12( cos )2

(12) 12(cos12 cos0)

72 12cos12 12

60 12cos12

x

x x dx x


15/34


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

Example 4


16/34

The figure shows the graph of the integrand

in the example.

We know from Section 5.2

that the value of theintegral can beinterpreted as the sumof the areas labeled

with a plus sign minusthe area labeled witha minus sign.



17/34


Evaluate

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

Example 5

2 29

21

2 1t t tdtt

2 2

9 9 1 2 221 1

2 1 (2 )t t t dt t t dt t


18/34

Then,9

1 2 2

1

93 2 1

32

19

3 2

1

3 2 3 22 1 2 1

3 9 3 1

1 2 4

9 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 Example 5


19/34

APPLICATIONS

The FTC2 says that, if fis continuous on

[a, b], then

where Fis 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


20/34

We know F(x) represents the rate of change

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

the change in ywhen xchanges from ato b.

Note that ycould, for instance, increase,then decrease, then increase again.

Although ymight change in both directions,F(b) F(a) represents the net change in y.

APPLICATIONS


21/34

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

a

F x dx F b F a


22/34

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.


23/34

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 waterin the reservoir between time t1 and time t2.

2

12 1

'( ) ( ) ( )t

tV t dt V t V t

NET CHANGE THEOREM


24/34

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 Cfrom time t1 to time t2.

2

12 1

[ ][ ]( ) [ ]( )

t

t

d Cdt C t C t

dt

NET CHANGE THEOREM


25/34

If the mass of a rod measured from the left

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

is (x) = m(x).

So,

is the mass of the segment of the rodthat lies between x= aand x= b.

( ) ( ) ( )b

ax dx m b m a

NET CHANGE THEOREM


26/34

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

then

is the net change in population during the timeperiod from t1 to t2.

The population increases when births happenand decreases when deaths occur.

The net change takes into account both birthsand deaths.

2

12 1

( ) ( )t

t

dndt n t n t

dt

NET CHANGE THEOREM


27/34

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

a commodity, then the marginal cost is

the derivative C(x).

So,

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

2

12 1

'( ) ( ) ( )x

xC x dx C x C x

NET CHANGE THEOREM


28/34

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

Equation 2NET CHANGE THEOREM


29/34

In both cases, the distance is computed by

integrating |v(t)|, the speed.

Therefore,

Equation 3

2

1

| ( ) | total distance traveledt

tv t dt

NET CHANGE THEOREM


30/34

The figure shows how both displacement and

distance traveled can be interpreted in terms

of areas under a velocity curve.

NET CHANGE THEOREM


31/34

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


32/34

A particle moves along a line so that its

velocity at time tis:

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

a) Find the displacement of the particle duringthe time period 1 t 4.

b) Find the distance traveled during this time period.

Example 6NET CHANGE THEOREM


33/34

By Equation 2, the displacement is:

This means that the particle moved 4.5 mtoward the left.

Example 6 a

4 42

1 1

43 2

1

(4) (1) ( ) ( 6)

963 2 2

s s v t dt t t dt

t t t

NET CHANGE THEOREM


34/34

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 6

3 2 3 261

10.17 m6

v t dt v t dt v t dt

t t dt t t dt

t t t t t t

Example 6 bNET CHANGE THEOREM

Documents

Indefinite Integrals and the Net Change Theorem