7 Integration by Parts Integration Using Tables of Integrals Numerical Integration Improper Integrals Applications of Calculus to Probability

77

Integration by PartsIntegration by Parts Integration Using Tables of IntegralsIntegration Using Tables of Integrals Numerical IntegrationNumerical Integration Improper IntegralsImproper Integrals Applications of Calculus to ProbabilityApplications of Calculus to Probability

Additional Topics in IntegrationAdditional Topics in Integration

7.17.1Integration by PartsIntegration by Parts

ln x x dx udv uv vdu ln x x dx udv uv vdu 2 21 1 1ln

2 2x x x dx

x 2 21 1 1

ln2 2

x x x dxx

21 1ln

2 2x x xdx 21 1

ln2 2

x x xdx

u dv uv v du u dv uv v du

The Method of Integration by PartsThe Method of Integration by Parts

Integration by parts formulaIntegration by parts formula

ExampleExample

EvaluateEvaluate

SolutionSolution Let Let uu == x x andand dvdv == e ex x dxdx So that So that dudu = = dxdx andand v v == e exx

Therefore,Therefore,

xxe dx xxe dx

xxe dx udv xxe dx udv uv vdu uv vdu

x xxe e dx x xxe e dx

( 1)

x x

x

xe e C

x e C

( 1)

x x

x

xe e C

x e C

Example 1, page 484

Guidelines for Integration by PartsGuidelines for Integration by Parts

Choose Choose uu an an dvdv so that so that

1.1. dudu is is simplersimpler than than uu..

2.2. dvdv is is easyeasy toto integrate integrate..

ExampleExample EvaluateEvaluate

SolutionSolution Let Let uu == lnln xx andand dvdv == x x dxdx

So thatSo that and and


ln x x dx ln x x dx

ln x x dx udv uv vdu ln x x dx udv uv vdu 2 21 1 1ln

2 2x x x dx

x 2 21 1 1

ln2 2

x x x dxx

21 1ln

2 2x x xdx 21 1

ln2 2

x x xdx

1du dx

x

1du dx

x 21

2v x 21

2v x

2 2 21 1 1ln (2 ln 1)

2 4 4x x x C x x C 2 2 21 1 1

ln (2 ln 1)2 4 4

x x x C x x C

Example 2, page 485


SolutionSolution Let Let uu == xe xexx and and

So thatSo that and and


2( 1)

xxedx

x 2( 1)

xxedx

x

2( 1)

xxedx udv uv vdu

x

2( 1)

xxedx udv uv vdu

x

1 1

( 1)1 1

x xxe e x dxx x

1 1

( 1)1 1

x xxe e x dxx x

1 1

x xx xxe xe

e dx e Cx x

1 1

x xx xxe xe

e dx e Cx x

( )

( 1)

x x

x

du xe e dx

e x dx

( )

( 1)

x x

x

du xe e dx

e x dx

1

1v

x

1

1v

x

2

1

( 1)dv dx

x

2

1

( 1)dv dx

x

Example 3, page 486


SolutionSolution Let Let uu == x x22 andand dvdv == e ex x dxdx So that So that dudu = 2= 2xdxxdx andand v v == e exx


2 xx e dx 2 xx e dx

2 xx e dx udv uv vdu 2 xx e dx udv uv vdu 2 (2 )x xx e e x dx 2 (2 )x xx e e x dx 2 2x xx e xe dx 2 2x xx e xe dx

2 ( 1)x xxe x e C 2 ( 1)x xxe x e C 2( 2 2)xe x x C 2( 2 2)xe x x C

(From first example)(From first example)

Example 4, page 486

Applied Example:Applied Example: Oil Production Oil Production

The estimated The estimated raterate at which at which oiloil will be will be producedproduced from an from an oil well oil well tt yearsyears after production has begun is given by after production has begun is given by

thousand barrels per year.thousand barrels per year. Find an expression that describes the Find an expression that describes the total productiontotal production of of

oil at the end of oil at the end of yearyear tt..

0.1( ) 100 tR t te 0.1( ) 100 tR t te

Applied Example 5, page 487


SolutionSolution Let Let TT((tt)) denote the denote the total productiontotal production of oil from the well at of oil from the well at

the the end of yearend of year tt ( (tt 0) 0).. Then, the Then, the rate of oil productionrate of oil production will be given by will be given by TT ′′((tt))

thousand barrels per year.thousand barrels per year.

Thus,Thus,

So,So,

0.1( ) ( ) 100 tT t R t te 0.1( ) ( ) 100 tT t R t te

0.1

0.1

( ) 100

100

t

t

T t te dt

te dt

0.1

0.1

( ) 100

100

t

t

T t te dt

te dt



SolutionSolution Use Use integration by partsintegration by parts to evaluate the integral. to evaluate the integral.

LetLet andand

So thatSo that andand


u tu t

0.1 0.1 0.1( ) 100 100 10 10t t tT t te dt te e dt 0.1 0.1 0.1( ) 100 100 10 10t t tT t te dt te e dt

0.1tdv e dt 0.1tdv e dt

du dtdu dt 0.1

0.1

1

0.1

10

t

t

v e

e

0.1

0.1

1

0.1

10

t

t

v e

e

0.1 0.1

0.1

100 10 100

1000 ( 10)

t t

t

te e C

e t C

0.1 0.1

0.1

100 10 100

1000 ( 10)

t t

t

te e C

e t C



SolutionSolution To determine the value of To determine the value of CC, note that the total quantity of , note that the total quantity of

oil produced at the end of oil produced at the end of yearyear 00 is is nilnil, so , so TT(0) = 0(0) = 0.. This gives,This gives,

Thus, the required Thus, the required production functionproduction function is given by is given by

0.1(0)(0) 1000 (0 10) 0

1000(10) 0

10,000

T e C

C

C

0.1(0)(0) 1000 (0 10) 0

1000(10) 0

10,000

T e C

C

C

0.1( ) 1000 ( 10)tT t e t C 0.1( ) 1000 ( 10)tT t e t C

0.1( ) 1000 ( 10) 10,000tT t e t 0.1( ) 1000 ( 10) 10,000tT t e t


7.27.2Integration Using Tables of IntegralsIntegration Using Tables of Integrals

2

2( 2 )

3

udubu a a bu C

ba bu

2

2( 2 )

3

udubu a a bu C

ba bu

4

2 2 2 2 2 2 2( 2 ) ln8 8

u au a u du a u u a u C

42 2 2 2 2 2 2( 2 ) ln

8 8

u au a u du a u u a u C

11n au n au n aunu e du u e u e du

a a 11n au n au n aun

u e du u e u e dua a

2

1[ ln ]

udua bu a a bu C

a bu b

2

1[ ln ]

udua bu a a bu C

a bu b

A Table of IntegralsA Table of Integrals

We have covered We have covered several techniquesseveral techniques for finding the for finding the antiderivativesantiderivatives of functions. of functions.

There are There are many more such techniquesmany more such techniques and extensive and extensive integration formulasintegration formulas have been developed for them. have been developed for them.

You can find a You can find a table of integralstable of integrals on on pagespages 491491 and and 492492 of of the text that include some such formulas for your benefit.the text that include some such formulas for your benefit.

We will now consider some We will now consider some examplesexamples that illustrate how that illustrate how this table can be used to evaluate an integral.this table can be used to evaluate an integral.

ExamplesExamples

Use the Use the table of integralstable of integrals to find to find

SolutionSolution We first We first rewriterewrite

SinceSince is of the formis of the form , with , with aa = 3 = 3, , bb = 1 = 1, , and and u u = = xx, we use , we use Formula (5)Formula (5),,

obtainingobtaining

2

3

x dx

x 2

3

x dx

x

3 x3 x a bua bu

22

3 3

x dx x dx

x x

2

23 3

x dx x dx

x x

2

2( 2 )

3

u dubu a a bu C

ba bu

2

2( 2 )

3

u dubu a a bu C

ba bu

2

22 2 ( 2 3) 3

3(1)3

x dxx x C

x

2

22 2 ( 2 3) 3

3(1)3

x dxx x C

x

4( 6) 3

3x x C

4( 6) 3

3x x C

Example 1, page 493

ExamplesExamples


SolutionSolution We first We first rewriterewrite 33 as as , so that, so that has the has the

formform

with with and and u u == x x.. Using Using Formula (8)Formula (8),,

obtainingobtaining

2 23x x dx 2 23x x dx

23 x 23 x2( 3)2( 3)

42 2 2 2 2 2 2 2 2( 2 ) ln

8 8

u au a u du a u a u u a u C

42 2 2 2 2 2 2 2 2( 2 ) ln

8 8

u au a u du a u a u u a u C

3a 3a 2 2a u2 2a u

2 2 2 2 293 (3 2 ) 3 ln 3

8 8

xx x dx x x x x C 2 2 2 2 29

3 (3 2 ) 3 ln 38 8

xx x dx x x x x C

Example 2, page 493

ExamplesExamples


SolutionSolution We can use We can use Formula (24)Formula (24),,

Letting Letting nn = 2 = 2, , aa = – ½ = – ½, and , and u u == x x, we have, we have

2 ( 1/2) xx e dx 2 ( 1/2) xx e dx

11n au n au n aunu e du u e u e du

a a 11n au n au n aun

u e du u e u e dua a

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

1 12 2

1 2

( ) ( )x x xx e dx x e xe dx

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

1 12 2

1 2

( ) ( )x x xx e dx x e xe dx

2 ( 1/2) ( 1/2)2 4x xx e xe dx 2 ( 1/2) ( 1/2)2 4x xx e xe dx

Example 5, page 494

ExamplesExamples


SolutionSolution We haveWe have

Using Using Formula (24)Formula (24) again, with again, with nn = 1 = 1, , aa = – ½ = – ½, and , and u u == x x, , we getwe get

2 (1/2) xx e dx 2 (1/2) xx e dx

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

1 12 2

1 12 4

( ) ( )x x x xx e dx x e xe e dx

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

1 12 2

1 12 4

( ) ( )x x x xx e dx x e xe e dx

( 1/2) 22 ( 4 8)xe x x C ( 1/2) 22 ( 4 8)xe x x C

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

12

12 8

( )x x xx e xe e C

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

12

12 8

( )x x xx e xe e C

2 ( 1/2) ( 1/2) ( 1/2)2 8x x xx e xe e dx 2 ( 1/2) ( 1/2) ( 1/2)2 8x x xx e xe e dx

2 (1/2) 2 ( 1/2) ( 1/2)2 4x x xx e dx x e xe dx 2 (1/2) 2 ( 1/2) ( 1/2)2 4x x xx e dx x e xe dx

Example 5, page 494

Applied Example:Applied Example: Mortgage Rates Mortgage Rates

A study prepared for the National Association of realtors A study prepared for the National Association of realtors estimated that the estimated that the mortgage ratemortgage rate over the next over the next tt monthsmonths will bewill be

percent per year.percent per year. If the prediction holds true, what will be the If the prediction holds true, what will be the average average

mortgage ratemortgage rate over the over the 1212 monthsmonths??

6 75( ) (0 24)

10

tr t t

t

6 75( ) (0 24)

10

tr t t

t



SolutionSolution The The average mortgage rateaverage mortgage rate over the next over the next 1212 monthsmonths will be will be

given bygiven by12

0

1 6 75

12 0 10

tA dt

t

12

0

1 6 75

12 0 10

tA dt

t

12 12

0 0

1 6 75

12 10 10

tdt dt

t t

12 12

0 0

1 6 75

12 10 10

tdt dt

t t

12 12

0 0

1 25 1

2 10 4 10

tdt dt

t t

12 12

0 0

1 25 1

2 10 4 10

tdt dt

t t




Use Use Formula (1)Formula (1)

to evaluate the to evaluate the first first integralintegral

or approximately or approximately 6.99%6.99% per yearper year..

2

1[ ln ]

udua bu a a bu C

a bu b

2

1[ ln ]

udua bu a a bu C

a bu b

12 12

0 0

1 25 1

2 10 4 10

tA dt dt

t t

12 12

0 0

1 25 1

2 10 4 10

tA dt dt

t t

12 12

0 0

1 25[10 10ln(10 )] ln(10 )

2 4A t t t

12 12

0 0

1 25[10 10ln(10 )] ln(10 )

2 4A t t t

1 25[(22 10ln 22) (10 10ln10)] [ln 22 ln10]

2 4

1 25[(22 10ln 22) (10 10ln10)] [ln 22 ln10]

2 4

6.996.99


7.37.3Numerical IntegrationNumerical Integration

x

R1 R2 R3 R4 R5 R6

b

( )f x( )f x

a

y

b ax

n

b ax

n

x

0 11

( ) ( )

2

f x f xR x

0 1

1

( ) ( )

2

f x f xR x

Approximating Definite IntegralsApproximating Definite Integrals

Sometimes, it is necessary to evaluate Sometimes, it is necessary to evaluate definite integralsdefinite integrals based on based on empirical dataempirical data where there is where there is no algebraic ruleno algebraic rule defining the defining the integrandintegrand..

Other situations also arise in which an Other situations also arise in which an integrable functionintegrable function has has an antiderivative that cannot be foundan antiderivative that cannot be found in terms of in terms of elementary functions. Examples of these areelementary functions. Examples of these are

Riemann sumsRiemann sums provide us with a provide us with a good approximationgood approximation of a of a definite integraldefinite integral, but there are better, but there are better techniques techniques and and formulasformulas, called , called quadrature formulasquadrature formulas, that allow a , that allow a more more efficientefficient way of computing approximate values of definite way of computing approximate values of definite integrals.integrals.

2

( ) xf x e2

( ) xf x e 1/2( ) xg x x e 1/2( ) xg x x e 1( )

lnh x

x

1( )

lnh x

x

The Trapezoidal RuleThe Trapezoidal Rule

Consider the problem of finding the area under the curve Consider the problem of finding the area under the curve of of ff((xx)) for the interval for the interval [[aa, , bb]]::

xx

yy

bb

RR

( )f x( )f x

aa


The trapezoidal rule is based on the notion of The trapezoidal rule is based on the notion of dividingdividing the the area to be evaluated into area to be evaluated into trapezoidstrapezoids that that approximateapproximate the the area area under the curve:under the curve:

xx

RR11 RR22 RR33 RR44

bb

( )f x( )f x

aa

yy

RR55 RR66


The The incrementsincrements xx used for each used for each trapezoidtrapezoid are obtained are obtained by dividing the interval into by dividing the interval into nn equal segmentsequal segments (in our example (in our example nn = 6 = 6):):

xx

RR11 RR22 RR33 RR44 RR55 RR66

bb

( )f x( )f x

aa

yy b ax

n

b ax

n

xx


The area of each The area of each trapezoidtrapezoid is calculated by multiplying its is calculated by multiplying its basebase, , xx , by its , by its average heightaverage height::

xxbbxx00 = = aa

( )f x( )f x

xx11

ff((xx00))

0 11

( ) ( )

2

f x f xR x

0 1

1

( ) ( )

2

f x f xR x

yy

xx

RR11 ff((xx11))



xxbbaa

( )f x( )f x

xx11

ff((xx11))

1 22

( ) ( )

2

f x f xR x

1 2

2

( ) ( )

2

f x f xR x

yy

xx

ff((xx22))RR22

xx22



xxbbaa

( )f x( )f x

ff((xx22))

2 33

( ) ( )

2

f x f xR x

2 3

3

( ) ( )

2

f x f xR x

yy

ff((xx33))

xx33

RR33

xx22



xxbbaa

( )f x( )f x

ff((xx33))

3 44

( ) ( )

2

f x f xR x

3 4

4

( ) ( )

2

f x f xR x

yy

xx33

ff((xx44))RR44

xx44



xxbbaa

( )f x( )f x

ff((xx44))

4 55

( ) ( )

2

f x f xR x

4 5

5

( ) ( )

2

f x f xR x

yy

xx55xx44

ff((xx55))RR55



xxb = xb = x66aa

( )f x( )f x

ff((xx55))

5 66

( ) ( )

2

f x f xR x

5 6

6

( ) ( )

2

f x f xR x

yy

xx55

ff((xx66))RR66


Adding the areasAdding the areas RR11 through through RRnn ( (nn = 6 = 6 in this case) of the in this case) of the

trapezoidstrapezoids gives an gives an approximationapproximation of the of the desired areadesired area of of the region the region RR::

xx

RR11 RR22 RR33 RR44 RR55 RR66

bbaa

( )f x( )f x

yy1 2 ... nR R R R 1 2 ... nR R R R


Adding the areasAdding the areas RR11 through through RRnn of the of the trapezoidstrapezoids yields the yields the

following rule:following rule:

✦ Trapezoidal RueTrapezoidal Rue

0 1 2 1( ) [ ( ) 2 ( ) 2 ( ) ... 2 ( ) ( )]2

b

n na

xf x dx f x f x f x f x f x

0 1 2 1( ) [ ( ) 2 ( ) 2 ( ) ... 2 ( ) ( )]

2

b

n na


where .b a

xn

where .

b ax

n

ExampleExample

Approximate the value ofApproximate the value of using the using the trapezoidal ruletrapezoidal rule with with nn = 10 = 10..

Compare this result with the Compare this result with the exact valueexact value of the integral. of the integral.

SolutionSolution Here, Here, aa = 1 = 1, , bb = 2 = 2, an , an nn = 10 = 10, so, so

and and xx00 = 1 = 1, , xx11 = 1.1 = 1.1, , xx22 = 1.2 = 1.2, , xx33 = 1.3 = 1.3, … , , … , xx99 = 1.9 = 1.9, , xx1010 = 1.10 = 1.10..

The The trapezoidal ruletrapezoidal rule yields yields

2

1

1dx

x2

1

1dx

x

2 1 10.1

10 10

b ax

n

2 1 10.1

10 10

b ax

n

2

1

1 0.1 1 1 1 1 1[1 2 2 2 ... 2 ] 0.693771

2 1.1 1.2 1.3 1.9 2dx

x

2

1

1 0.1 1 1 1 1 1[1 2 2 2 ... 2 ] 0.693771

2 1.1 1.2 1.3 1.9 2dx

x

Example 1, page 500

ExampleExample

Approximate the value ofApproximate the value of using the using the trapezoidal ruletrapezoidal rule with with nn = 10 = 10..

Compare this result with the Compare this result with the exact valueexact value of the integral. of the integral.

SolutionSolution By computing the By computing the actual valueactual value of the of the integralintegral we get we get

Thus the Thus the trapezoidal ruletrapezoidal rule with with nn = 10 = 10 yields a result with yields a result with an an errorerror of of 0.000624 0.000624 to six decimal places.to six decimal places.

2

1

1dx

x2

1

1dx

x

2 2

11

1ln ln 2 ln1 ln 2 0.693147dx x

x

2 2

11

1ln ln 2 ln1 ln 2 0.693147dx x

x

Example 1, page 500

Applied Example:Applied Example: Consumers’ Surplus Consumers’ Surplus

The The demand functiondemand function for a certain brand of perfume is for a certain brand of perfume is given bygiven by

where where pp is the unit is the unit priceprice in dollars and in dollars and xx is the is the quantity quantity demandeddemanded each week, measured in ounces. each week, measured in ounces.

Find the Find the consumers’ surplusconsumers’ surplus if the market price is set at if the market price is set at $60$60 per ounce. per ounce.

2( ) 10,000 0.01p D x x 2( ) 10,000 0.01p D x x



SolutionSolution When When pp = 60 = 60, we have, we have

or or xx = 800 = 800 since since xx must be must be nonnegativenonnegative..

Next, using the Next, using the consumers’ surplusconsumers’ surplus formulaformula with with pp = 60 = 60 andand xx = 800 = 800, we see that the consumers’ surplus is given by, we see that the consumers’ surplus is given by

It is It is not easynot easy to evaluate this to evaluate this definite integraldefinite integral by finding an by finding an antiderivativeantiderivative of the integrand. of the integrand.

But we can, instead, use the But we can, instead, use the trapezoidal ruletrapezoidal rule..

2

2

2

10,000 0.01 60

10,000 0.01 3,600

640,000

x

x

x

2

2

2

10,000 0.01 60

10,000 0.01 3,600

640,000

x

x

x

800 2

010,000 0.01 (60)(800)CS x dx

800 2

010,000 0.01 (60)(800)CS x dx



SolutionSolution We can use the We can use the trapezoidal ruletrapezoidal rule with with aa = 0 = 0, , bb = 800 = 800, ,

and and nn = 10 = 10..

and and xx00 = 0 = 0, , xx11 = 80 = 80, , xx22 = 160 = 160, , xx33 = 240 = 240, … , , … , xx99 = 720 = 720, , xx1010 = 800 = 800..

The The trapezoidal ruletrapezoidal rule yields yields800 2

010,000 0.01 (60)(800)CS x dx

800 2

010,000 0.01 (60)(800)CS x dx

800 0 80080

10 10

b ax

n

800 0 80080

10 10

b ax

n

2 2

2 2

80100 2 10,000 (0.01)(80) 2 10,000 (0.01)(160) ...

2

... 2 10,000 (0.01)(720) 10,000 (0.01)(800)

2 2

2 2

80100 2 10,000 (0.01)(80) 2 10,000 (0.01)(160) ...

2

... 2 10,000 (0.01)(720) 10,000 (0.01)(800)



SolutionSolution The The trapezoidal ruletrapezoidal rule yields yields

Therefore, the Therefore, the consumers’ surplusconsumers’ surplus is approximately is approximately $22,294$22,294. .

800 2

010,000 0.01 (60)(800)CS x dx

800 2

010,000 0.01 (60)(800)CS x dx

40(100 199.3590 197.4234 194.1546 189.4835

183.3030 175.4537 165.6985

153.6750 138.7948 60) (60)(800)

70,293.82 48,000

$22,294.82

40(100 199.3590 197.4234 194.1546 189.4835

183.3030 175.4537 165.6985

153.6750 138.7948 60) (60)(800)

70,293.82 48,000

$22,294.82


Simpson’s RuleSimpson’s Rule

We’ve seen that the We’ve seen that the trapezoidal ruletrapezoidal rule approximatesapproximates the the area under the curve by adding the area under the curve by adding the areas of trapezoidsareas of trapezoids under the curve:under the curve:

xx

yy

xx00 xx11 xx22

( )f x( )f x

RR11 RR22


The The Simpson’s ruleSimpson’s rule improves upon the trapezoidal ruleimproves upon the trapezoidal rule by by approximating the area under the curve by the approximating the area under the curve by the area under area under a parabolaa parabola, rather than a straight line:, rather than a straight line:

xx

yy

( )f x( )f x

xx00 xx11 xx22

RR

Simpson’s RuleSimpson’s Rule Given any Given any three nonlinear pointsthree nonlinear points there is a there is a unique parabolaunique parabola

that passes through the given points.that passes through the given points. We can We can approximate the functionapproximate the function ff((xx)) on on [[xx00, , xx22]] with a with a

quadratic functionquadratic function whose graph contain these whose graph contain these three pointsthree points::

((xx00, , ff((xx00))))

xx

yy

( )f x( )f x

xx00 xx11 xx22

((xx11, , ff((xx11))))

((xx22, , ff((xx22))))


Simpson’s ruleSimpson’s rule approximates the area under the curve of approximates the area under the curve of a function a function ff((xx)) using a using a quadratic functionquadratic function::

✦ Simpson’s ruleSimpson’s rule

0 1 2 3 4

1

( ) [ ( ) 4 ( ) 2 ( ) 4 ( ) 2 ( )3

4 ( ) ( )]

b

a

n n


f x f x

0 1 2 3 4

1

( ) [ ( ) 4 ( ) 2 ( ) 4 ( ) 2 ( )3

4 ( ) ( )]

b

a

n n


f x f x

where and is even . b a

x nn

where and is even .

b ax n

n

ExampleExample

Find an approximation ofFind an approximation of using using Simpson’s Simpson’s rulerule with with nn = 10 = 10..

SolutionSolution Here, Here, aa = 1 = 1, , bb = 2 = 2, an , an nn = 10 = 10, so, so

Simpson’s ruleSimpson’s rule yields yields

2

1

1dx

x2

1

1dx

x

2 1 10.1

10 10

b ax

n

2 1 10.1

10 10

b ax

n

0.1 1 1 1 1 1 11 4 2 4 2 4

3 1.1 1.2 1.3 1.4 1.9 2

0.1 1 1 1 1 1 11 4 2 4 2 4

3 1.1 1.2 1.3 1.4 1.9 2

2

1

1 0.1[ (1) 4 (1.1) 2 (1.2) 4 (1.3) 2 (1.4) 4 (1.9) (2)]

3 dx f f f f f f f

x

2

1

1 0.1[ (1) 4 (1.1) 2 (1.2) 4 (1.3) 2 (1.4) 4 (1.9) (2)]

3 dx f f f f f f f

x

0.6931500.693150Example 3, page 503

ExampleExample

Find an approximation ofFind an approximation of using using Simpson’s Simpson’s rulerule with with nn = 10 = 10..

SolutionSolution Recall that the Recall that the trapezoidal ruletrapezoidal rule with with nn = 10 = 10 yielded an yielded an

approximation of approximation of 0.6937710.693771, with an , with an errorerror of of 0.0006240.000624 from the value of from the value of lnln 2 2 ≈ 0.693147≈ 0.693147 to six decimal places. to six decimal places.

Simpson’s ruleSimpson’s rule yields an approximation with an yields an approximation with an errorerror of of 0.0000030.000003 to six decimal places, to six decimal places, a definite improvementa definite improvement over the over the trapezoidal ruletrapezoidal rule..

2

1

1dx

x2

1

1dx

x

Example 3, page 503

Applied Example:Applied Example: Cardiac Output Cardiac Output

One method of measuring One method of measuring cardiac outputcardiac output is to inject is to inject 55 to to 1010 mgmg of a of a dyedye into a vein leading to the heart. into a vein leading to the heart.

After making its way After making its way through the lungsthrough the lungs, the dye , the dye returns to returns to the heartthe heart and is and is pumped into the aortapumped into the aorta, where its , where its concentration concentration is measured atis measured at equal time intervals equal time intervals..


44

33

22

11

2.02.0

0.40.4

0.80.80.50.5

0.20.20.10.1

22 44 66 88 1010 1212 1414 1616 1818 2020 2222 2424 2626

The graph of The graph of cc((tt)) shows the shows the concentrationconcentration of dye in a of dye in a person’s aorta, measured in person’s aorta, measured in 22-second-second intervals after intervals after 55 mg mg of of dyedye have been injected: have been injected:

xx

yy

00

4.04.03.93.9

3.23.2

2.52.5

1.81.8

1.31.3



44

33

22

11

2.02.0

0.40.4

0.80.80.50.5

0.20.20.10.1

22 44 66 88 1010 1212 1414 1616 1818 2020 2222 2424 2626

The person’s The person’s cardiac outputcardiac output, measured in , measured in liters per liters per minute (L/min)minute (L/min) is computed using the formula is computed using the formula

where where DD is the is the quantity ofquantity of dye injecteddye injected..

xx

yy

00

4.04.03.93.9

3.23.2

2.52.5

1.81.8

1.31.3


28

0

60

( )

DR

c t dt

28

0

60

( )

DR

c t dt

28

0( )c t dt

28

0( )c t dt


44

33

22

11

2.02.0

0.40.4

0.80.80.50.5

0.20.20.10.1

22 44 66 88 1010 1212 1414 1616 1818 2020 2222 2424 2626

Use Use Simpson’s ruleSimpson’s rule with with nn = 14 = 14 to to evaluate the integralevaluate the integral and and determinedetermine the person’s the person’s cardiac outputcardiac output..

xx

yy

00

4.04.03.93.9

3.23.2

2.52.5

1.81.8

1.31.3


28

0

60

( )

DR

c t dt

28

0

60

( )

DR

c t dt

28

0( )c t dt

28

0( )c t dt


SolutionSolution We have We have aa = 0 = 0, , bb = 28 = 28, an , an nn = 14 = 14, and , and tt = 2 = 2, so that, so that

tt00 = 0 = 0, , tt11 = 2 = 2, , tt22 = 4 = 4, , tt33 = 6 = 6, … , , … , tt1414 = 28 = 28. .

Simpson’s ruleSimpson’s rule yields yields


28

0

2( ) [ (0) 4 (2) 2 (4) 4 (6) ... 4 (26) (28)]

3c t dt c c c c c c

28

0

2( ) [ (0) 4 (2) 2 (4) 4 (6) ... 4 (26) (28)]

3c t dt c c c c c c

2[0 4(0) 2(0.4) 4(2.0) 2(4.0)

34(4.4) 2(3.9) 4(3.2) 2(2.5) 4(1.8)

2(1.3) 4(0.8) 2(0.5) 4(0.2) 0.1]

49.9

2[0 4(0) 2(0.4) 4(2.0) 2(4.0)

34(4.4) 2(3.9) 4(3.2) 2(2.5) 4(1.8)

2(1.3) 4(0.8) 2(0.5) 4(0.2) 0.1]

49.9


SolutionSolution Therefore, the person’s Therefore, the person’s cardiac outputcardiac output is is

or approximately or approximately 6.06.0 L/minL/min..


28

0

60 60(5)6.0

49.9( )

DR

c t dt

28

0

60 60(5)6.0

49.9( )

DR

c t dt


7.47.4Improper IntegralsImproper Integrals

x

y

1

– 1

1 2 – 2 – 1

R1

R2

Improper IntegralsImproper Integrals

In many applications we are concerned with integrals that In many applications we are concerned with integrals that have have unbounded intervalsunbounded intervals of integration. of integration.

These are called These are called improper integralsimproper integrals.. We will now discuss problems that involve improper We will now discuss problems that involve improper

integrals.integrals.

Improper Integral of Improper Integral of ff over over [[aa, , ))

✦ Let Let ff be a continuous function on the be a continuous function on the unbounded unbounded intervalinterval [[aa, , )). Then the . Then the improper integralimproper integral of of ff over over [[aa, , )) is defined by is defined by

if the limit exists.if the limit exists.

( ) lim ( )b

a abf x dx f x dx

( ) lim ( )

b

a abf x dx f x dx

ExamplesExamples

EvaluateEvaluate if it converges.if it converges.

SolutionSolution

Since Since lnln bb → → , as , as bb → → the the limit does not existlimit does not exist, and we , and we

conclude that the given conclude that the given improper integralimproper integral is is divergentdivergent..

2

1dx

x

2

1dx

x

2 2

2

1 1lim

limln

lim(ln ln 2)

b

b

b

b

b

dx dxx x

x

b

2 2

2

1 1lim

limln

lim(ln ln 2)

b

b

b

b

b

dx dxx x

x

b

Example 2, page 513

11 22 33

ExamplesExamples

Find the Find the area of the regionarea of the region RR under the curve under the curve y y = = ee––xx/2/2 for for xx 0 0..

SolutionSolution The The required arearequired area is shown in the is shown in the diagram belowdiagram below::

yy

xx

11

RR y y = = ee––xx/2/2

Example 3, page 514

ExamplesExamples

Find the Find the area of the regionarea of the region RR under the curve under the curve y y = = ee––xx/2/2 for for xx 0 0..

SolutionSolution Taking Taking bb > 0 > 0, we compute the , we compute the area of the region area of the region under the under the

curve curve y y = = ee––xx/2/2 from from xx = 0 = 0 to to xx = = bb,,

Then, the Then, the area of the regionarea of the region RR is given by is given by

or or 22 square unitssquare units..

/2/2

1( ) lim(2 2 ) 2 2lim 2b

bb bI b e

e

/2

/2

1( ) lim(2 2 ) 2 2lim 2b

bb bI b e

e

/2 /2 /2

00( ) 2 2 2

b bx x bI b e dx e e /2 /2 /2

00( ) 2 2 2

b bx x bI b e dx e e

Example 3, page 514

Improper Integral of Improper Integral of ff over over (–(– , , bb]]

✦ Let Let ff be a continuous function on the be a continuous function on the unbounded unbounded intervalinterval (–(– , , bb]]. Then the . Then the improper integralimproper integral of of ff over over (–(– , , bb]] is defined by is defined by

if the limit exists.if the limit exists.

( ) lim ( )b b

aaf x dx f x dx

( ) lim ( )

b b

aaf x dx f x dx

ExampleExample

Find the Find the areaarea of the of the regionregion R R bounded above by the bounded above by the xx-axis-axis, , below by below by yy = – = – ee22xx, and on the right, by the line , and on the right, by the line xx = 1 = 1..

SolutionSolution The graph of The graph of regionregion RR is: is:

– – 1 1 11

yy

xx

RR

y y = = ee22xx

x x = 1= 1

11

–– 11

– – 33

– – 77

Example 4, page 514

ExampleExample

Find the Find the areaarea of the of the regionregion R R bounded above by the bounded above by the xx-axis-axis, , below by below by yy = – = – ee22xx, and on the right, by the line , and on the right, by the line xx = 1 = 1..

SolutionSolution Taking Taking aa < 1 < 1, compute , compute

Then, the Then, the areaarea under the required under the required regionregion RR is given by is given by

1 2( ) [0 ( )]x

aI a e dx

1 2( ) [0 ( )]x

aI a e dx

1 2 x

ae dx

1 2 x

ae dx

12 2 21 1 1

2 2 2x a

a

e e e 1

2 2 21 1 1

2 2 2x a

a

e e e

2 21 1lim ( ) lim

2 2a

a aI a e e

2 21 1lim ( ) lim

2 2a

a aI a e e

2 21 1lim

2 2a

ae e

2 21 1

lim2 2

a

ae e

21

2e 21

2e

Example 4, page 514

Improper Integral Unbounded on Both SidesImproper Integral Unbounded on Both Sides

Improper IntegralImproper Integral of of ff over over (–(– , , )) Let Let ff be a be a continuous functioncontinuous function over the over the

unbounded intervalunbounded interval (–(– , , )). . Let Let cc be be any real numberany real number and suppose both the and suppose both the

improper integralsimproper integrals

are are convergentconvergent.. Then, the Then, the improper integralimproper integral of of ff over over (–(– , , )) is is

defined bydefined by

( ) ( ) an d c

cf x dx f x dx

( ) ( ) an d c

cf x dx f x dx

( ) ( ) ( )c

cf x dx f x dx f x dx

( ) ( ) ( )

c

cf x dx f x dx f x dx

ExamplesExamples

EvaluateEvaluate the the improper integralimproper integral

and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution Take the number Take the number cc to be to be zerozero and and evaluate evaluate firstfirst for the for the

intervalinterval (–(– , 0), 0)::

2xxe dx

2xxe dx

2 2

2

2

0 0

0

lim

1lim

2

1 1 1lim

2 2 2

x x

aa

x

aa

a

a

xe dx xe dx

e

e

2 2

2

2

0 0

0

lim

1lim

2

1 1 1lim

2 2 2

x x

aa

x

aa

a

a

xe dx xe dx

e

e

Example 5, page 515

ExamplesExamples


and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution Now Now evaluate evaluate for the intervalfor the interval (0, (0, ))::

2xxe dx

2xxe dx

2 2

2

2

0 0

0

lim

1lim

2

1 1 1lim

2 2 2

bx x

b

b

x

b

b

b

xe dx xe dx

e

e

2 2

2

2

0 0

0

lim

1lim

2

1 1 1lim

2 2 2

bx x

b

b

x

b

b

b

xe dx xe dx

e

e

Example 5, page 515

ExamplesExamples


and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution Therefore, Therefore,

2xxe dx

2xxe dx

2 2 20

0

1 10

2 2x x xxe dx xe dx xe dx

2 2 20

0

1 10

2 2x x xxe dx xe dx xe dx

Example 5, page 515

ExamplesExamples


and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution Below is the Below is the graphgraph of of yy = = xexe––xx22

, showing the regions of , showing the regions of interest interest RR11 and and RR22::

2xxe dx

2xxe dx

xx

yy

11

–– 11

11 2 2 –– 22 – – 11

RR11

RR22

Example 5, page 515

ExamplesExamples


and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution Region Region RR11 lies lies belowbelow the the xx-axis-axis, so its area is , so its area is negative negative

((RR11 = –= – ½½):):

2xxe dx

2xxe dx

x

y

1

– 1

1 2 – 2 – 1

R1

R2

Example 5, page 515

ExamplesExamples


and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution While the While the symmetrically identicalsymmetrically identical region region RR22 lies lies aboveabove the the

xx-axis-axis, so its area is , so its area is positive positive ((RR22 = ½= ½):):

2xxe dx

2xxe dx

x

y

1

– 1

1 2 – 2 – 1

R1

R2

Example 5, page 515

ExamplesExamples


and give a and give a geometric interpretationgeometric interpretation of the result. of the result.SolutionSolution Thus, Thus, adding the areasadding the areas of the two regions of the two regions yieldsyields zerozero::

2xxe dx

2xxe dx

1 2

1 10

2 2R R R 1 2

1 10

2 2R R R

x

y

1

– 1

1 2 – 2 – 1

R1

R2

Example 5, page 515

7.57.5Applications of Calculus to ProbabilityApplications of Calculus to Probability

x

y

R1

1 ( )R P a x b 1 ( )R P a x b

a b

( ) 1R P x ( ) 1R P x

Probability Density FunctionsProbability Density Functions

A A probability density functionprobability density function of a of a random variablerandom variable xx in in an interval an interval II, where , where II may be bounded or unbounded, is a may be bounded or unbounded, is a nonnegative functionnonnegative function ff having the following having the following propertiesproperties..1.1. The The total areatotal area RR of the region under the graph of of the region under the graph of ff is is

equal toequal to 11::

xx

yy

RR

Probability Density FunctionsProbability Density Functions

xx

yy

RR11

1 ( )R P a x b 1 ( )R P a x b

( ) ( )b

aP a x b f x dx ( ) ( )

b

aP a x b f x dx

A A probability density functionprobability density function of a of a random variablerandom variable xx in in an interval an interval II, where , where II may be bounded or unbounded, is a may be bounded or unbounded, is a nonnegative functionnonnegative function ff having the following having the following propertiesproperties..2.2. The The probabilityprobability that an that an observed valueobserved value of the of the random random

variablevariable xx lies in the lies in the intervalinterval [[aa, , bb]] is given by is given by

aa bb

ExamplesExamples Show that the function Show that the function

satisfies the satisfies the nonnegativitynonnegativity conditioncondition of of Property 1Property 1 of of probability density functions.probability density functions.

SolutionSolution Since the factors Since the factors xx and and ((xx – 1) – 1) are both are both nonnegativenonnegative, we see , we see

that that ff((xx) ) 0 0 on on [1, 4][1, 4].. Next, we compute Next, we compute

2( ) ( 1) (1 4)

27 f x x x x

2( ) ( 1) (1 4)

27 f x x x x

4

1

2( 1)

27x x dx

4

1

2( 1)

27x x dx

4

3 2

1

2 1 1

27 3 2x x

4

3 2

1

2 1 1

27 3 2x x

4 2

1

2( )

27x x dx

4 2

1

2( )

27x x dx

2 64 1 18

27 3 3 2

2 64 1 18

27 3 3 2

2 271

27 2

2 271

27 2

Example 1, page 522

ExamplesExamples Show that the function Show that the function

satisfies the satisfies the nonnegativitynonnegativity conditioncondition of of Property 1Property 1 of of probability density functions.probability density functions.

SolutionSolution First, First, ff((xx)) 0 0 for all values of for all values of xx in in [0, [0, )). . Next, we compute Next, we compute

( 1/3)1( ) (0 )

3 xf x e x ( 1/3)1

( ) (0 )3

xf x e x

( 1/3)

0

1

3xe dx

( 1/3)

0

1

3xe dx

( 1/3)

0lim

bx

be

( 1/3)

0lim

bx

be

( 1/3)

0

1lim

3

b x

be dx

( 1/3)

0

1lim

3

b x

be dx

( 1/3)lim 1

1

b

be

( 1/3)lim 1

1

b

be

Example 1, page 522

ExamplesExamples Determine the Determine the value of the constantvalue of the constant kk so that the function so that the function

is a is a probability density functionprobability density function on the on the intervalinterval [0, 5][0, 5]..SolutionSolution We computeWe compute

Since this value must be Since this value must be equal to oneequal to one, we find that , we find that

2( )f x kx 2( )f x kx

5 2

0kx dx

5 2

0kx dx

53

03

kx

53

03

kx

5 2

0k x dx

5 2

0k x dx

125

3k

125

3k

1251

33

125

k

k

1251

33

125

k

k

Example 2, page 522

ExamplesExamples If If xx is a is a continuous random variablecontinuous random variable for the function for the function

computecompute the the probabilityprobability that that xx will assume a value will assume a value between between xx = 1= 1 and and xx = 2= 2..

SolutionSolution The required probability is given byThe required probability is given by

2( )f x kx 2( )f x kx

2 2

1

3(1 2)

125P x x dx

2 2

1

3(1 2)

125P x x dx

1 7(8 1)

125 125

1 7(8 1)

125 125

23

1

1

125x

23

1

1

125x

Example 2, page 522

0.60.6

0.40.4

0.20.2

11 22 33 44 55

ExamplesExamples If If xx is a is a continuous random variablecontinuous random variable for the function for the function

computecompute the the probabilityprobability that that xx will assume a value will assume a value between between xx = 1= 1 and and xx = 2= 2..

SolutionSolution The The graphgraph of of ff showing showing PP(1 (1 xx 2) 2) is: is:

2( )f x kx 2( )f x kx

xx

yy

PP(1 (1 xx 2) = 7/125 2) = 7/125

23

125y x 23

125y x

Example 2, page 522

Applied Example:Applied Example: Life Span of Light Bulbs Life Span of Light Bulbs

TKK Inc. manufactures a TKK Inc. manufactures a 200-watt electric light bulb200-watt electric light bulb.. Laboratory tests show that the Laboratory tests show that the life spanslife spans of these light of these light

bulbs have a bulbs have a distributiondistribution described by the described by the probability probability density functiondensity function

where where xx denotes the denotes the life spanlife span of a light bulb. of a light bulb. DetermineDetermine the the probability probability that a light bulb will have a that a light bulb will have a

life spanlife span of of a.a. 500500 hours or less. hours or less.b.b. More than More than 500500 hours. hours.c.c. More than More than 10001000 hours but less than hours but less than 15001500 hours. hours.

0.001( ) 0.001 (0 )xf x e x 0.001( ) 0.001 (0 )xf x e x



SolutionSolution

a.a. The The probabilityprobability that a light bulb will have a that a light bulb will have a life spanlife span of of 500500 hours or lesshours or less is given by is given by

500 0.001

0(0 500) 0.001 xP x e dx

500 0.001

0(0 500) 0.001 xP x e dx

5000.001

0

xe5000.001

0

xe

0.5 1

0.3935

e

0.5 1

0.3935

e



SolutionSolution

b.b. The The probabilityprobability that a light bulb will have a that a light bulb will have a life spanlife span of of more thanmore than 500500 hours hours is given byis given by

0.001

500( 500) 0.001 xP x e dx

0.001

500( 500) 0.001 xP x e dx

0.001

500lim

bx

be

0.001

500lim

bx

be

0.001

500lim 0.001

b x

be dx

0.001

500lim 0.001

b x

be dx

0.001 0.5

0.5

lim

0.6065

b

be e

e

0.001 0.5

0.5

lim

0.6065

b

be e

e



SolutionSolution

c.c. The The probabilityprobability that a light bulb will have a that a light bulb will have a life spanlife span of of more thanmore than 10001000 hours hours butbut less than less than 15001500 hours hours is given byis given by

1500 0.001

1000(1000 1500) 0.001 xP x e dx

1500 0.001

1000(1000 1500) 0.001 xP x e dx

15000.001

1000

xe15000.001

1000

xe

1.5 1

0.2231 0.3679

0.1448

e e

1.5 1

0.2231 0.3679

0.1448

e e


Exponential Density FunctionExponential Density Function The example we just saw involved a function of the formThe example we just saw involved a function of the form

ff((xx) = ) = keke–kx–kx

where where xx 0 0 and and kk is a positive constant, with a graph: is a positive constant, with a graph:

This probability function is called an This probability function is called an exponential density exponential density functionfunction, and the , and the random variablerandom variable associated with it is associated with it is said to be said to be exponentially distributedexponentially distributed..

Such variables are used to represent the Such variables are used to represent the lifelife span ofspan of electric componentselectric components, the , the duration of telephone callsduration of telephone calls, the , the waiting time in a doctor’s officewaiting time in a doctor’s office, etc., etc.

xx

yy

kk


Expected ValueExpected Value

Expected Value of a Continuous Random VariableExpected Value of a Continuous Random Variable Suppose the function Suppose the function f f defined on the interval defined on the interval [[aa, , bb]]

is the is the probability density functionprobability density function associated with a associated with a continuous random variablecontinuous random variable xx..

Then, the Then, the expected valueexpected value of of xx is is

( ) ( )b

aE x xf x dx( ) ( )

b

aE x xf x dx


Show that if a Show that if a continuous random variablecontinuous random variable xx is is exponentially distributedexponentially distributed with the with the probability density probability density functionfunction

ff((xx) = ) = keke–kx–kx (0 (0 xx < < ))

then the then the expected valueexpected value EE((xx)) is equal to is equal to 1/1/kk.. Using this result and Using this result and continuingcontinuing with our with our last examplelast example, ,

determine the determine the average life spanaverage life span of the of the 200-watt light bulb200-watt light bulb manufactured by manufactured by TKK IncTKK Inc..



SolutionSolution We computeWe compute

Integrating by partsIntegrating by parts with with uu = = xx and and dvdv = = ee–kx–kxdxdx

so thatso that We haveWe have

0 0( ) ( ) kxE x xf x dx kxe dx

0 0( ) ( ) kxE x xf x dx kxe dx

0lim

b kx

bk xe dx

0

limb kx

bk xe dx

1 and kxdu dx v e

k

1 and kxdu dx v e

k

0 0

1 1( ) lim

bbkx kx

bE x k xe xe dx

k k

0 0

1 1( ) lim

bbkx kx

bE x k xe xe dx

k k




(0)2 0

1 1 1lim (0)

bkb k kx

bk be e e

k k k

(0)

2 0

1 1 1lim (0)

bkb k kx

bk be e e

k k k

2 2

1 1 1lim kb kb

bk be e

k k k

2 2

1 1 1lim kb kb

bk be e

k k k

1 1 1lim lim lim1

kb kbb b b

b

e k e k

1 1 1lim lim lim1

kb kbb b b

b

e k e k

0 0

1 1( ) lim

bbkx kx

bE x k xe xe dx

k k

0 0

1 1( ) lim

bbkx kx

bE x k xe xe dx

k k



SolutionSolution Now, by taking a Now, by taking a sequence of valuessequence of values of of bb that that approaches approaches

infinityinfinity (such as (such as bb = 10, 100, 1000, 10,000, … = 10, 100, 1000, 10,000, … ) we see that, ) we see that, for a fixed for a fixed kk,,


Finally, since Finally, since kk = 0.001 = 0.001, we see that the , we see that the average life spanaverage life span of of the the TKK light bulbsTKK light bulbs is is 1/(0.001) = 10001/(0.001) = 1000 hourshours..

lim 0kbb

b

e lim 0

kbb

b

e

1 1 1( ) lim lim lim1

1 10 0 (1)

1

kb kbb b b

bE x

e k e k

k k

k

1 1 1( ) lim lim lim1

1 10 0 (1)

1

kb kbb b b

bE x

e k e k

k k

k


Expected Value of an Exponential Density FunctionExpected Value of an Exponential Density Function

If a If a continuous random variablecontinuous random variable xx is is exponentially exponentially distributeddistributed with with probability density functionprobability density function

ff((xx) = ) = keke–kx–kx (0 (0 x x < < ))

then the then the expected (average) valueexpected (average) value of of xx is given by is given by

1( )E x

k

1( )E x

k

Applied Example:Applied Example: Airport Traffic Airport Traffic

On a typical Monday morning, the On a typical Monday morning, the time between time between successive arrivals of planessuccessive arrivals of planes at Jackson International at Jackson International Airport is an Airport is an exponentially distributed random variableexponentially distributed random variable xx with with expected valueexpected value of of 10 10 minutes.minutes.

a.a. Find the Find the probability density functionprobability density function associated with associated with xx..

b.b. What is the What is the probabilityprobability that that betweenbetween 66 andand 88 minutes will minutes will elapse elapse betweenbetween successive arrivalssuccessive arrivals of planes. of planes.

c.c. What is the What is the probabilityprobability that the that the time between successive time between successive arrivalsarrivals of planes will be of planes will be more thanmore than 1515 minutes? minutes?



SolutionSolutiona.a. Since Since xx is is exponentially distributedexponentially distributed, the associated , the associated

probability density functionprobability density function has the form has the form


✦ The The expected valueexpected value of of xx is is 1010, so, so

✦ Thus, the required Thus, the required probability density functionprobability density function is is

ff((xx) = 0.1) = 0.1ee––0.10.1xx

1( ) 10

1

100.1

E xk

k

1( ) 10

1

100.1

E xk

k



SolutionSolutionb.b. The The probabilityprobability that that between between 66 and and 88 minutes will elapse minutes will elapse

between successive arrivalsbetween successive arrivals is given by is given by

8 0.1

6(6 8) 0.1 xP x e dx

8 0.1

6(6 8) 0.1 xP x e dx

80.1

6

xe80.1

6

xe

0.8 0.6

0.10

e e

0.8 0.6

0.10

e e



SolutionSolutionc.c. The The probabilityprobability that the that the time between successive arrivalstime between successive arrivals

will be will be more thanmore than 1515 minutes is given minutes is given

0.1

15( 15) 0.1 xP x e dx

0.1

15( 15) 0.1 xP x e dx

0.1

15lim 0.1

b x

be dx

0.1

15lim 0.1

b x

be dx

0.1

15lim

bx

be

0.1

15lim

bx

be

0.1 1.5lim( )b

be e

0.1 1.5lim( )b

be e

1.5

0.22

e

1.5

0.22

e


End of End of Chapter Chapter

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7 Integration by Parts Integration Using Tables of Integrals Numerical Integration Improper Integrals Applications of Calculus to Probability