8-integralEvaluation

7/30/2019 8-integralEvaluation

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MTH 252Integral Calculus

Chapter 8 Principles of

Integral Evaluation

Section 8.1 An Overview of

Integration Methods

Copyright 2005 by Ron Wallace, all rights reserved.


2/77

Four Approaches to Integration

1. Recognition Review list on pages 511-512

2. Computer Algebra Systems (CAS) Examples: Mathematica, Maple, Derive

Problems: Cant solve some and others giveunnecessary complicated answers.

3. Tables See inside front & back covers

Problem: Not exhaustive4. Transformation Methods

Change to something recognizable.

Examples: Substitution & Algebraic Equivalence

The topic of this chapter!

Also, for definiteintegrals, numerical

approximationmethods.


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Transformation Methods

Substitution

Integration by Parts

Trigonometric Integrals Powers of Trigonometric Functions

Trigonometric Substitutions

Partial Fractions

All of these attempt to change the integral into

something that can be integrated by recognition.

Many problemswill require theuse of two ormore of thesemethods!


4/77



Integral Evaluation

Section 8.2 Integration by Parts



5/77

Review: The Product Rule

( ) ( ) '( ) ( ) ( ) '( )d f x g x f x g x f x g xdx

Example:

2 tand

x xdx

2 22 tan secx x x x

2 2tan tand d

x x x xdx dx


6/77

Integration Equivalent

( ) ( ) '( ) ( ) ( ) '( )d f x g x f x g x f x g xdx

( ) ( ) '( ) ( ) ( ) '( )f x g x f x g x f x g x dx

( ) ( ) '( ) ( ) ( ) '( )f x g x f x g x dx f x g x dx

( ) '( ) ( ) ( ) '( ) ( )f x g x dx f x g x f x g x dx


7/77

Integration Equivalent

( ) '( ) ( ) ( ) '( ) ( )f x g x dx f x g x f x g x dx

Let ( ) and '( )u f x dv g x dx

d '( ) and '( ) ( )u f x dx v g x dx g x

u dv uv v du


8/77

Integration by Parts

( ) ( )f x g x d vux d u dv

=u dv uv v du

'( )

( )

du f x dx

v g x dx

NOTE:f(x) org(x) can be equal to 1 (but not both).


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Example

2xxe dx

=u dv uv v du

2 xu x dv e dx

212

xdu dx v e

2 21 1

2 2

x xxe e dx 2 21 1

2 4

x xxe e c


10/77

Choosing u & dv

In general u should be such that du is simpler

dvshould (must) be easy to integrate

=u dv uv v du

Previous Example:

2xxe dx2 xu e dv xdx

2 2 2 21

2

x xx e x e dx

2 212

2

xdu e dx v x

This made theproblem MORE

difficult!

d


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RepeatedIntegration by Parts

=u dv uv v du

Sometimes the resulting integral is integrated by the same method!

Example: 2 cosx x dx2 cosu x dv x dx

2 sindu x dx v x

2 sin 2 sinx x x x dx sinu x dv x dx

cosdu dx v x

2 sin 2 cos cosx x x x x dx

2 sin 2 cos 2sinx x x x x c

Ci l R d


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Circular RepeatedIntegration by Parts

=u dv uv v du

Example:cosxe x dx

cosx

u e dv x dx sinxdu e dx v x

sin sin

x x

e x e x dx

sinxu e dv x dx

cosxdu e dx v x

cos dx sin cos cosx x x xe x e x e x e x dx

2 cos dx sin cosx x xe x e x e x

sin cos

cos dx 2

x xx e x e x

e x c


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Integral Evaluation

Section 8.3 Trigonometric

Integrals



14/77

Odd Powers of SIN & COS

sinn x dx1sin sinn x x dx

n > 2 is a positive odd integer

n - 1 is a positive even integer

1

2 21 cos sinn

x x dx

1

2 2sin sinn

x x dx

cossin

u xdu x dx

1

2 21n

u du

Multiply out thepolynomial, integrate,

and substitute back.


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cosn x dx1cos cosn x x dx

n > 2 is a positive odd integer

n - 1 is a positive even integer

1

2 21 sin cosn

x x dx

1

2 2cos cosn

x x dx

sincos

u xdu x dx

1

2 21n

u du




16/77


sin cosm nn x dx1sin cos cosm nx x x dx

mis a positive integern is a positive odd integer

n - 1 is a positiveeven integer

1

2 2sin 1 sin cosn

m x x x dx

1

2 2sin cos cosn

m x x x dx

sincos

u xdu x dx

1

2 21n

mu u du




17/77


sin cosm nn x dx1sin cos sinm nx x x dx

mis a positive odd integern is a positive integer

m - 1 is a positiveeven integer

1

2 21 cos cos sinm

nx x x dx

1

2 2sin cos sinm

nx x x dx

cossin

u xdu x dx

1

2 21m

nu u du



2 21 i


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Powers of SIN & COS

sin cosm nn x dxNOTE: m &n are non-negative integers.

Since: sin cos & cos sind d

x x x x

dx dx

1. Ifn is odd, put one of the cosines w/ dx, changethe remaining cosines to sines, and let u = sin x.

2. Ifm is odd, put one of the sines w/ dx, changethe remaining sines to cosines, and let u = cos x.

3. All other cases use some other method.

2 21 sin cosx x 2 21 cos sinx x

2 21


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Powers of TAN & SEC

tan secm nn x dxFor what values ofm &n can thesame approach be used?[NOTE: m &n are non-negative integers.]

2Hint: tan sec & sec sec tand d

x x x x x

dx dx

1. Ifm is odd and n > 0, put one of the tangents andone of the secants w/ dx, change the remaining

tangents to secants, and let u = sec x.

2. Ifn is even and n > 0, put two of the secants w/dx, change the remaining secants to tangents, and

let u = tan x.

3. All other cases use some other method.

2 21 tan secx x 2 2sec 1 tanx x


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Even Powers of SIN & COS

sin cosm nn x dx mand n are BOTH non-negative

even integers.

Remember the half-angle formulas:

1 coscos

2 2

1 cossin2 2

1 cos2cos2

xx 1 cos2sin2

xx

Let 2x

2 1 cos2

cos 2

x

x

2 1 cos2

sin 2

x

x

Square both sides.


21/77

1.Use the trigonometric identities

2.Multiply everything out.

3.Integrate each term, one at a time.

Even Powers of SIN & COS

sin cosm nn x dx mand n are BOTH non-negative

even integers.

21

cos 1 cos22

x x 21

sin 1 cos22

x x


22/77

Memorize this one!

Powers of SEC

secx dxsec tan

secsec tan

x xx dx

x x

2sec sec tan

sec tan

x x x

dxx x

Let sec tanu x x 2sec tan secdu x x x

1lndu u c

u ln sec tanx x c

ld


23/77

Powers of SEC

secn x dx 2 2sec secnu x dv x dx

sec ln sec tanx dx x x c

n > 2 and apositive integer

2sec tanx dx x c

Use Integration by Parts

22 sec tan tanndu n x x dx v x

2 2 2sec sec tan ( 2) sec tann n nx dx x x n x x dx

2 2 2sec sec tan ( 2) sec sec 1n n nx dx x x n x x dx 2 2sec sec tan ( 2) sec ( 2) secn n n nx dx x x n x dx n x dx

2 2( 1) sec sec tan ( 2) secn n nn x dx x x n x dx

l td


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Powers of SEC

secn x dx 2 2sec secnu x dv x dx

sec ln sec tanx dx x x c

n > 2 and apositive integer

2sec tanx dx x c

Use Integration by Parts

22 sec tan tanndu n x x dx v x

2 2( 1) sec sec tan ( 2) secn n nn x dx x x n x dx

2 21 2sec sec tan sec

1 1

n n nnx dx x x x dx

n n

This kind of identity is called a Reduction Formula.

No! You do NOT need to memorize this one. Just use it.


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Memorize this one too!

Powers of TAN

tanx dxsin

cos

xdx

x

Let cosu x sindu x dx

1lndu u c

u

ln secx c

ln cosx c


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Powers of TAN tan ln secx dx x c

tann x dxn > 1 and a positive integer

1.Ifn is even, change to secants using

2.Ifn is odd,

a. Put sec x tan x w/ dxas follows

b. Convert all tangents (except the one w/ dx) to secants.

c. Use the substitution, u = sec x

2 2tan sec 1x x

1tantan sec tan

sec

nn xx dx x x dx

x


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FYI: More Reduction Formulas

2 21 2sec sec tan sec

1 1

n n nnx dx x x x dx

n n

1 21tan tan tan1

n n nx dx x x dxn

1 21 1sin sin cos sinn n nnx dx x x x dxn n

1 21 1

cos cos sin cosn n nn

x dx x x x dxn n


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One more note

1 1 sinConsider: ln2 1 sin

d x

dx x

2

1 sin cos 1 sin cos1 1 sin

2 1 sin1 sin

x x x xx

xx

1 1 2cos

2 1 sin 1 sin

x

x x

2

cos

1 sin

x

x

sec x

1 1 sinTherefore: sec ln

2 1 sin

xx dx c

x

An equivalent form to the other solution.


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MTH 252

Integral Calculus


Integral Evaluation

Section 8.4 Trigonometric

Substitutions



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Derivative Reminders

1

2

1sin

1

dx

dx x

1

2

1tan

1

dx

dx x

1

2

1sinh

1

dx

dx x

1

2

1cosh

1

dx

dx x

1

2

1tanh

1

dx

dx x


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Equivalent Integrals

1

2

1sin

1dx x c

x

121

tan1

dx x cx

1

2

1sinh

1dx x c

x

1

2

1cosh

1dx x c

x

1

2

1tanh

1dx x c

x


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Radicals in Integrals

2 2a x

2 2a x

2 2x a

If an integral

contains one ofthese expressions,what substitutioncan be used to get

rid of the radical?


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2 2a x

Let: sin , - 2 2x a

2 2 2 2 2sina x a a 2 2 2 21 sina x a 2 2 2 2cosa x a 2 2 cosa x a

cosdx a d

1sinx

a


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Example

3 25x x dx2

1

5sin5cos

5 5 cos

sin 5

xdx d

x

x

3

5 sin 5 cos 5 cos d 3 225 5 sin cos d

cos

sin

u

du d

2 225 5 1 u u du 3 5

25 53 5

u uc

3 5

1 1cos sin cos sin5 5

25 53 5

x x

c

3 5

2 25 525 5

15 5 125 5

x xc

3 5

2 25 55

3 25

x xc

2 2 41 5 50 5 315

x x x c


35/77


2 2a xsin

cos

x a

dx a d

2 2

1

cos

sin

a x ax

a

2 2a x 2tan

sec

x a

dx a d

2 2

1

sec

tan

a x a

x

a

2 2x asec

sec tan

x a

dx a d

2 2

1

tan

sec

x a a

x

a


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2 2a x 2tanh

sech

x a

dx a d

2 2

1

sech

tanh

a x ax

a

2 2a xsinh

cosh

x a

dx a d

2 2

1

cosh

sinh

a x a

x

a

2 2x acosh

sinh

x a

dx a d

2 2

1

sinh

cosh

x a a

x

a

HyperbolicFunction Option


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Integrals w/ Quadratics

Complete the square and then make a simple substitution.

Example:

23 24 1x x

23 8 1x x

2 2 23 8 4 1 3(4)x x

23 4 49x 23 49u

Let 4

u xdu dx

Let 3

3

r u

dr du

2 49r


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MTH 252

Integral Calculus


Integral Evaluation

Section 8.5 Integrating Rational

Functions by

Partial FractionsCopyright 2006 by Ron Wallace, all rights reserved.

Review:


39/77

Review:Addition/Subtraction of Fractions

a c ad bcb d bd

Note: The equation works both ways!

Problem:

Find two fractions whose sum/difference

is equal to a third given fraction.

The product of the denominators of thetwo fractions will be the denominator ofthe given fraction.

d b


40/77

Example:

Find two fractions whosesum or difference is:

1135

Denominator = 35

Possibilities for the other two denominators are:

1 & 35 and 5 & 7

ad bc a c

bd b d

11

35 5 7

A B 11 7 5A B

Many solutions, including:

3 & -2 and 1 & 4/5


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Rational Functions

Any function of the form,

where

are polynomials.

( )( )

( )

P xf x

Q x

( ) and ( )P x Q x


42/77

Fundamental Theorem of Algebra

Every polynomial equation of degree nwith complex coefficients has n roots inthe complex numbers.

Each real root r, gives a factor (x-r).

Each complex root +i has a companion root-i. These give a factor: (ax2+bx+c).

Hence, every polynomial can be written asa product of linear & quadratic factors.

Some Easy Integrals Linear


43/77

Some Easy Integralsof Rational Functions

5

3 dxx 5ln 3x c

5

2 3dx

x Let u=2x-3

du=2dx

5 1 5ln 2 3

2 2du x c

u

4

3

xdx

x 12

4 4 12ln 33

dx x x cx

4 5

3

x dxx

174 4 17ln 33

dx x x cx

LinearDenominator

7

5

(2 3) dxx

7 6 65 5 5 52 2 12 12

7

1(2 3)du u du u c x c

u

Let u=2x-3

du=2dx

Partial ( )P x

where Q(x) is a product


44/77

PartialFractions

( )

( )

P xdx

Q xQ( ) p

of linear factors &

deg(P(x)) < deg(Q(x))

2 5

(3 1)( 4)

x

x x

(3 1) ( 4)

A B

x x

2 5 ( 4) (3 1) ( 3 ) ( 4 )x A x B x A B x A B

2 3

5 4

A B

A B

2 5 1 1 1ln 3 1 ln 4

(3 1)( 4) 3 1 4 3

xdx dx x x c

x x x x

Solve this systemfor A & B.

A=-1, B=1

This method can be extended to anynumber of distinct linear factors.

Partial ( )P x



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PartialFractions

( )

( )

P xdx

Q xQ( ) p

of linear factors &

deg(P(x)) < deg(Q(x))

2

4 1

(2 3)

x

x

(2 3) (2 3)

A B

x x

4 1 (2 3) 2 (3 )x A x B Ax A B

4 2

1 3

A

A B

2 2

4 1 2 7 7ln 2 3

(2 3) 2 3 (2 3) 2(2 3)

xdx dx x c

x x x x

Solve this systemfor A & B.

A=2, B=-7

This method can be extended to any power of the denominatorand can be combined with the previous method.

RepeatedLinear Factors 2(2 3) (2 3)

A B

x x

More Easy Integrals Quadratic


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More Easy Integralsof Rational Functions

2

3

4 7 dxx x

QuadraticDenominator

2

1

3 ( 2) 3 dxx Let u=x-2

du=dx

2

13

3du

u

Finish using trig substitutions.

2

3 11

4 7

xdx

x x

2 23 2 4 17

2 4 7 4 7

xdx dx

x x x x

23

2 ln 4 7x x

Just like theone above!

3 11 _____(2 4) _____ 11x x 32 6

2 4 7d

x xdx

More Easy Integrals Quadratic


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More Easy Integralsof Rational Functions

QuadraticDenominator

Complete the

square & trigsubstitution.

2 5

3 11

( 4 7)

xdx

x x

2 5 2 53 2 4 17

2 ( 4 7) ( 4 7)

x dx dxx x x x

Let u = x2 - 4x + 7du = 2x 4 dx

5 4

2 4

3 3

2 8

3( 4 7)

8

u du u c

x x c

Partial ( )P x

where Q(x) is a product of a


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2( 1) ( 2 3)

A B

x x x 2( 1) ( 2 3)

A Bx C

x x x

PartialFractions

( )

( )

P xdx

Q x linear factor & a quadraticfactor & deg(P(x)) < deg(Q(x))2

2

4 3

( 1)( 2 3)

x x

x x x

2 24 3 ( ) (2 ) (3 )x x A B x A B C x A C

4

1 2

3 3

A B

A B C

A C

2

2 2

4 3 3 1 3

( 1)( 2 3) 1 2 3

x x xdx dx dx

x x x x x x

Solve this systemfor A, B, & C.

A=1, B=3, C=0

This method can be extended to anynumber of distinct linear & quadratic factors.

Partial ( )P x



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PartialFractions

( )

( )

P xdx

Q x of quadratic factors &deg(P(x)) < deg(Q(x))

2 2( )

(3 2)P x

x x

and proceed as before!

All of these methods can be combined and extended to handleany rational function where you can factor the denominator

into a product of linear and quadratic factors.

RepeatedQuadratic Factors

2 2 2(3 2) (3 2)

Ax B Cx D

x x x x

Partial ( )P x



50/77

PartialFractions

( )

( )

P xdx

Q x of quadratic factors &deg(P(x)) < deg(Q(x))5 3

4 3 33 2

( 1)( 1)( 1)x x x dx

x x x

Example

( )P x


51/77

( )

( )

P xdx

Q x where deg(P(x)) deg(Q(x))

Simplify using long division of polynomials.

Example:

34 3 5

2 1

x xdx

x

3 22 1 4 0 3 5x x x x

22x

3 24 2x x22 3x x

x

22x x

2 5x

1

2 1x

4

42 1x

2 422 1

x dx xdx dx dxx


52/77



Integral Evaluation

Section 8.7 Numerical Integration;

Simpsons Rule


Reminder:


53/77

Reminder:Definition of a Definite Integral

*

max 01

( ) lim ( )kk

n

bk

a xk

f x dx f x x

where

0 1 2,0

1

{ , , ..., }, nn

k k

P x x x xx a x bx x

*

1

1

[ , ]k k k

k k k

x x x

x x x

Since any sequence of partitions may be used

provided maxxk 0, using regular partitions


54/77

Midpoint Approximation


55/77

Midpoint Approximationof a Definite Integral

where

* 1

2

k

k kk k

b ax

n

x a k x

x xx m

1

12

1k

xa k

m x k

a bm1 m2 m3 m4

b

a

n

k

k xxfdxxf1

* )()( n

k

kxfx1

* )(

Midpoint ApproximationE l !


56/77

dpo t pp o at oof a Definite Integral

5 2

11 4x x dx

Example!

Approximate w/ n = 4

5 11

4x

1 1

2 2

3 3

4 4

1.5 4.75

2.5 4.75

3.5 2.75

4.5 1.25

m f m

m f m

m f m

m f m

11

1 11 11

Trapezoid Approximation

1A h


57/77

p ppof a Definite Integral

11

( ) ( ) ( )2

n

bk k

ak

xf x dx f x f x

where

k

b axn

x a k x

1 21

2TA h B B

1

1

( ) 2 ( ) ( )2

n

kk

xf a f x f b

Trapezoid ApproximationE ample!


58/77

p ppof a Definite Integral

5 2

11 4x x dx

Example!


5 11

4x

1 1

2 2

3 3

1 ( ) 4

2 ( ) 5

3 ( ) 4

4 ( ) 1

5 ( ) 4

a f a

x f x

x f x

x f x

b f b

4+2(10)-4 = 20

1

20 102

10

Si R l


59/77

Simpsons Rule

(x+h, y3)

x x+hx-h

(x-h, y1)

(x, y2)

y1 = f(x-h)

y2 = f(x)

y3 = f(x+h)

Three non-linearpoints determine a

unique parabola.

f(x)

p(x) =ax2+bx+c

Si R l


60/77

Simpsons Rule

x x+hx-h

(x+h, y3)

(x-h, y1)

(x, y2)

y1 = f(x-h)

y2 = f(x)

y3 = f(x+h)


unique parabola.

p(x) =ax2+bx+c

h-h

(0, y2)

(h, y3)

(-h, y1)

Si R l


61/77

3

22

21

2

00ycbhahycba

ycbhah

32

22

12

2

yybhah

yc

yybhah

Simpsons Rule

y1 = f(x-h)

y2 = f(x)

y3 = f(x+h)


unique parabola.

p(x) =ax2+bx+c

(0, y2)

x h-h

(h, y3)

(-h, y1) cah

hdxxp

h

h 623)(2

2

312

2

22

ycyyyah

2

2

321

2

2

ych

yyya

321 43

)( yyyh

dxxph

h

Si R l


62/77

Simpsons Rule

(x3, y3)

x2 x3x1

(x1, y1)

(x2

, y2

)f(x)

p(x) =ax2+bx+c

)()(4)(3)( 3213

1 xfxfxf

h

dxxf

x

x

Si R l


63/77

Simpsons Rule

][

)()(4)(2

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

12

3210

nnn

b

a

xfxfxf

xfxfxfxf

x

dxxf

)(xf

a b1x 2x 2nx 1nx

xkax

nabx

evenn

xbxa

k

n

is

0

Simpsons RuleExample!


64/77

pApproximation

5 2

11 4x x dx

Example!


5 11

4x

1 1

2 2

3 3

1 ( ) 4

2 ( ) 5

3 ( ) 4

4 ( ) 1

5 ( ) 4

a f a

x f x

x f x

x f x

b f b

4+4(6)+2(4)-4 = 32

6

3

2

10323

1

E E ti ti


65/77

Error Estimation

],[over)(max

],[over)(''max

)4(

4

2

baxfK

baxfK

Let

2

22

2

3

)(24

1

24

)(xKab

n

KabE

M

Midpoint Method:

2222

3

)(

12

1

12

)(xKab

n

KabET

Trapezoid Method:

4444

5

180

1

180

)(xKab

n

KabES

Simpsons Rule:

What happens whenyou double thenumber of intervals?

E E ti ti Example!


66/77

Error Estimation Example!

52

1 1 4x x dx Approximated w/ n = 4

3

1

)4(24

24

24

)(2

3

2

2

3

n

KabEMMidpoint Method 11

0)4(180

04

180

)(4

5

44

5

n

KabESSimpsons Rule 10-2/3

00)(

22)(''

4

)4(

2

Kxf

Kxf

Trapezoid Method 103

2

)4(12

24

12

)(2

3

22

3

n

KabET

Determining the


67/77

gNumber of Intervals (n)

52

1 1 4x x dx Find n so that the midpointmethod will have an errorless than 10 -4.

22

3

24)(n

KabEM 4

2

3

1024

2)4( n

4

2

013

32

n

42

0132

3

n

7.666,1063

320,00012 n 26.53n


68/77



Integral Evaluation

Section 8.8 Improper Integrals


Definite Integ als


69/77

Definite Integrals

)()()(

aFbFdxxfb

a where )()(' xfxF

Assumptions?

f(x) is continuous over [a,b]But what if

f(x) is only continuous over (a,b]

f(x) is only continuous over [a,b)

f(x) is only continuous over (a,b)

f(x) is not continuous at c(a,b)

a =

b =

ImproperIntegrals

Improper Integrals Examples


70/77

Improper Integrals Examples

1

0ln dxx

3

0 2

1

2dx

x

x

1

1dx

x

Not continuous at x=0.

Not continuous at x=1.

Upper limit is infinite.

Improper Integral


71/77

over an Open Interval

a b

b

adxxf )(

k

adxxf )(

k

abk

b

adxxfdxxf )(lim)(

f(x) is notcontinuous

at b.

Iff(x) is not continuous at athen

Iff(x) is not continuous at a &b

and a < c < bthen

b

kak

b

adxxfdxxf )(lim)(

c

a

b

c

b

a dxxfdxxfdxxf )()()(

Improper Integrals Example


72/77


1

0ln dxx Not continuous at x=0.

1

k0

lnlim dxxk

)ln()1(lim0

kkkk

Integration by Parts!

k

k

k 1

lnlim1

0

1lim10

kk

LHpitals Rule

Improper Integral


73/77

over an Infinite Interval

adxxf )(

k

akdxxf )(lim

k

b

kk

b

dxxfdxxf )(lim)(

c

c

dxxfdxxfdxxf )()()(



74/77


11 dx

xUpper limit is infinite.

k

kdx

x11lim

][ 22lim

kk

k

kx

1

21

2lim

Divergent!

Improper Integrals


75/77

with a Point of Discontinuity

b

adxxf )(

b

c

c

adxxfdxxf )()(



76/77


3

0 2

1

2dx

x

xNot continuous at x=1.

3

1 2

1

0 2

1

2

1

2dx

x

xdx

x

xcxxdx

x

x

1ln2

11ln

2

3

1

22

1ln2

11ln

2

3lim4ln

2

12ln

2

3

01ln2

11ln2

3lim

1

1

kk

kk

k

k Divergent!

Improper IntegralsWarning!


77/77


3

0 2

1

2dx

x

xNot continuous at x=1.

cxxdxx

x

1ln2

11ln

2

3

1

22

1ln

2

11ln

2

34ln

2

12ln

2

3

2ln2

1 This is not correct why?

8-integralEvaluation

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