06-apps of integration

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8F38E-mail: [email protected]

Chapter 6

Applications of Integration

Calculus 5eSingle Variable

James Stewart

2

Chapter 6 Application of Integration

1. Area between Curves

2. Volumes

3. Volumes by Cylindrical Shells

4. Work

5. Average Value of a Function
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3Chapter 6 Application of Integration

6.1 Areas Between Curves

n Consider the region between two curves y=f

(x) and y =g (x) and between the vertical

linesx= a andx= b , wheref andg arecontinuous functions andf(x) g (x) for all

xin [a, b]:

4


Areas Between Curvesn We divide the region into n strips of equal width and

then we approximate the ith strip by a rectangle

with base xand heightf(xi) g (xi).

The Reimann sum

is an approximation to what we intuitively think of

as the area of the region

=

n

i

iixxgxf

1

)]()([


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Areas Between Curvesn The above approximation appears to become better

and better as n .

Therefore, we define the area as the limiting value

of the Reimann sum

It is the definite integral offg :

The areaA of the region bounded by the curves y=f

(x), y =g (x) and the linesx= a andx= b , wheref

andg are continuous functions andf(x) g (x) for

allxin [a, b] , is

=

=

n

i

iin

xxgxfA1

)]()([lim

=b

adxxgxfA )]()([

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Illustration(1/4)

n Finding the Area between Two Curves

Find the area bounded by the graphs ofy= 3xand y=x29


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Illustration(2/4)

n Solution:

To find the limits, we set the two functions equal

and solve forx. We have

Thus, the curves intersect atx= -4 andx= 3.

The upper boundary is formed by y= 3-x.

So, for each fixed value ofx, the height of a

rectangle is h (x) = (3x)(x29 ).

A

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Illustration(3/4)n Finding the Area between Two Curves That Cross

Find the area bounded by the graphs ofy=x2 and y

= 2x2 for 0x2.


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Illustration(4/4)n Solution:

We will need to compute two integrals,

one on the interval where 2x2x2 and

a second integral on the interval wherex2

2x2

To find the point of intersection exactly, we solvex2

= 2x2 , so that 2x2 = 2 orx2 = 1 orx=1.

A

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Illustration(5/6)

n Example: An Area Computed by Integrating with Respect

to y

Find the area bounded by the graphs ofy=x2, y= 2xand y= 0.


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Illustration(6/6)

n Solution

we must first write these left-and right-hand boundaries

as functions ofy.

To do this, simply solve the equation y=x2 forx. We get

Likewise, y= 2-xis equivalent tox= 2-y.

Finally, these curves intersect where

So, the curves intersect at y= 1 and y= 4. From the

figure, it is clear that y= 1 is the solution we need.

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Example 2n Find the area of the region enclosed by the

parabolas y=x2 and y = 2x-x2


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Example 4n The figure shows velocity

curves for two cars,

A and B, that start side by side

and move along the same road.What does the areas between

the curves represent?

Use the Midpoint Rule with n = 4 intervals

to estimate it.

A

B

sec

ft/s

30292928252320130vA vB

65636056514434210vB95928984766754340vA

1614121086420t

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Example 4n Solution: Use the Midpoint Rule with n = 4 intervals

30292928252320130vA vB

65636056514434210vB

95928984766754340vA

1614121086420t

ft

tdtvv AA

372)93(4

]29282313[)(16

0

==

+++


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Example 5

n Find the area of the region bounded by the

curves y = sin x, y = cos x, x = 0, and x = /2.

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Example 6n Find the area enclosed by the line y=x-1

and the parabola y2 = 2x+ 6


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6.2 Volumes

n What kinds of solids do you presently know

how to compute the volume?

box (V= length width height)

a sphere ( V=(4/3) r3 )

a right circular cylinder ( V=r2h )

a cylinder is anysolid whose cross sections

(perpendicular to some axis running through the

solid) are all the same.

Which are cylinders?

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Volumesn The volume of a right circular cylinder is

n Likewise, in the case of a box, we have

n In general, the volume of any cylinder is

found by

V= (cross-sectional area) (height).


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Definition of Volume(P.384)

Figures 2, 3 A cross-se ction a nd slabsof a solid S

20


Tips1.

2. (cross section)

3.


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Show that the volume of a sphere of radius r is

V=(4/3)r3

Example 1

Approxima ting th e volum e of a sphe re with radius 1

(a) Using 5 disks,V 4.2726

(b) Using 10 disks,V 4.2097

(c) Using 20 disks,V 4.1940

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Example 2


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Example 3

24


Example 7


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Example 8

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6.3 Volumes by Cylindrical Shellsn Cylinder shellCylinder shells


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Volumes by Cylindrical Shells

bawherexfxVa

b


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Example 2

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Example 3


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Example 4

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6.4 Work


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Examples

n Example 1

n Example 3n Example 4

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6.5 Average Value of a Functionn It is easy to calculate the average value of

finitely many numbers y1, y2,, yn:

n But how do we compute the average

temperature during a day ifinfinitely many

temperature readings are possible?

n

yyyy nave

+++=

...21

t

T

Tave


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Average Value of a Function

n Lets try to compute the average value of a

function y=f(x), axb.

Divide the interval [a, b] into n equal

subintervals, each with length x= (ba)/n.

Choose points x1*, xn

* in successivesubintervals and calculate the average:

=

=

++

=++

n

i

i

nn

xxfab

x

ab

xfxf

n

xfxf

1

*

**

1

**

1

)(1

)(...)()(...)(

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Average Value of a Functionn Let n increase, we would be computing the averagevalue of a large number of closely spaced values.

The limiting value is

by the definition of a definite integral.

n Therefore, we define the average value off on the

interval [a, b] as

= =b

a

n

i

in

dxxfab

xxfab

)(1

)(1

lim1

*

=b

aave dxxf

abf )(

1


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Example 1

n Find the average value of the functionf

(x)=1+x2 on the interval [-1, 2].

n Solution:

23

1

3

1

)1()1(2

1)(

1

2

1

3

2

1

2

=

+=

+

=

=

xx

dxxdxxfab

fb

aave

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The Mean Value Theorem for Integrals

n The question arises:

Is there a number c at which the value offis

exactly equal to the average value of the

function, that is,f(c) =fave?

n The Mean Value Theorem for IntegralsIff is continuous on [a, b], then there exists a

number c in [a, b] such that

))(()( abcfdxxfb

a=


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The Mean Value Theoremn The Mean Value Theorem for Integrals is a

consequence ofthe Mean Value Theorem for

Derivatives and the Fundamental Theorem of

Calculus.

n The geometric interpretation of the Mean Value

Theorem for Integrals:

For positive functionf, there is a number c such

that the rectangle with base [a, b] and heightf(c)

has the same area as the region under the graph off

from a to b.

f(c)= fave

xba c

y

y=f(x)

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Example 2n Sincef(x)=1+x2 is continuous on the

interval [-1, 2], the Mean Value Theorem for

Integrals says there is a number c in [-1, 2]

such that

Find c in this case specifically.

n Solution:

))1(2)(()1(2

1

2 =+ cfdxx

1so21Therefore,

2)(1

2 ==+

=

== cc

dxxfab

ffb

aavec


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Example 3

n Show that the average velocity of a car over

a time interval [t1, t2] is the same as theaverage of its velocities during the trip.

Questionn Exercises

n

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