gts 213-ch5 integrals

8/15/2019 GTS 213-Ch5 Integrals

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GTS 213

Calculus for Technologist II

CHAPTER 5INTEGRALS

1

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Food for Thought

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2


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Topical Outline3

Topic Content

Integrals(Ch. and !"

Review of Calculus for Technologist I• Concepts (re#ie$"• Techni%ue of Integration (re#ie$"

Calculus for Technologist II starts here• I proper Integrals (Section .1'"• pplication of Integrals (Chapter !"

GTS 213, S2 & S3


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Integrals)

*hat does +integrate ean- *hat is athe atical integral-

GTS 213, S2 & S3


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rea ro/le

rea of a rectangle, a triangle, or a pol0gon

a pol0gon the area is found /0 di#iding the pol0gon intotriangles and adding the areas

GTS 213, S2 & S3


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ut $hat if the region has cur#ed sides-!

o$ do $e find an area under the cur#e-For e4a ple, area under 0 5 4 2 fro ' to 1

GTS 213, S2 & S3


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rea 6nder Continuous Functions7

The sa e concept /rea8 it do$n to s all pieces andadd the

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eight of rectangle

GTS 213, S2 & S3

=

Function y = x 2 fro ' to 1Three $a0s to "easure the height of four rectangles

#sing right points

9ight points 1>) 1>2 3>) 1eight ()" 2 (1"2

#sing left points

?eft points

eight #sing "i!!le points

@iddle points


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rea of rectangles under y = x 2

GTS 213, S2 & S3

1'

Rectangle $ Rectangle % Rectangle & Rectangle '

9ight points 1>) 1>2 3>) 1

eight ()" 2 (1"2

The area of these appro4i ating

rectangles

9) 5 (1>)"(


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dding strips> left endpoints & right endpoints11

ppro4i ating S $ith : rectanglesDo$ di#ide ', 1 into : inter#als? : 5 HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH 5 '.273)379 : 5 HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH 5 '.3=:)37Do$, 0ou see that L ) ( A ( R ) ('.273)37 E E '.3=:)37 "

?o$er esti ation 6pper esti ationGTS 213, S2 & S3


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S aller 9ectangles, etter ;sti ation12

Increasing the nu /er of strips, or increasing n

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The greater nu /er of strips, the /etter appro4i ation13

Area under x 2 from 0 to 1.o 0ou notice so ething fro the ta/le-

GTS 213, S2 & S3

Increase n (n J"


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Ke0 Concept1)

Increase n (n J"To use a li it of appro4i ations to arri#e at the truearea under the cur#e.

*eLfirst appro4i ate the region S /0 rectangles and thenta*e the li"it of the areas of these rectangles as $e increasethe nu /er of rectangles.

GTS 213, S2 & S3


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;4a ple to sho$ a li it of appro4i ation1

For region S in the preceding e4a ple, sho$ that thesu of the areas of the upper appro4i atingrectangles approaches , that is,⅓

Solution Rn is the su of

the areas of the n rectanglesas sho$n on the right

→∞= 1lim

3nnR

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;4a ple to sho$ a li it of appro4i ation (contMd"1!

;ach rectangle has *idth 1> n andThe heights are the #alues of the function f ( x " 5 x 2 at the points1>n, 2>n, 3>n,L, n>n, leading to

GTS 213, S2 & S3


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;4a ple to sho$ a li it of appro4i ation (contMd"1:

This gi#es

GTS 213, S2 & S3


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efinition of Integral1=

?i8e$ise the lo$er appro4i ating su s alsoapproach , that is,⅓

s n increases, /oth Ln and Rn /eco e /etter and /etter appro4i ations to the area of S

→∞= 1lim

3nnL

GTS 213, S2 & S3


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efinition of Integral (contMd"2'

Therefore $e define the area A to /e the li it of thesu s of the areas of the appro4i ating rectangles

Concept di#ide an area into s all areas and ta8eli it of the su of the s all areas ppl0 this idea to ore general region.

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ppl0ing the Idea to @ore General 9egion21

*e start /0 su/di#iding S into n strips S 1, S 2,L, S n of e%ual $idth.The $idth of the inter#al is b N a .The $idth of su/inter#al ( x " is (b N a)/n

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ppl0ing the Idea to @ore General 9egion (contMd"22

Then the area of the i th rectangle is f ( x i " x .

So the area of S is appro4i ated /0 the su of the areas of these rectangles

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ppl0ing the Idea to @ore General 9egion (contMd"23

Again+ appro,i"ation see"s to -eco"e -etter an! -etter as n increases

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efinition rea A of the region S 2)

Since f is continuous,

If $e use left endpoints

→∞lim always exists; nn R

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efinition rea A of the region S (contMd"2

if $e ta8e the height of the i th rectangle to /e the #alue of f at an0 nu /er x i * in x i P1, x i .

The nu /ers x 1*, x 2*,…, x n* are the samp e points.

A "ore general e,pression for the area of S GTS 213, S2 & S3


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Sig a Dotation2!

Su of an0 ter s use sig a notation

So the area

can /e $ritten as( )

→∞ =

= ∆∑1

limn

ini

A fx x

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The istance ro/le2:

Suppose the odo eter on our car is /ro8en and $e $ant to esti ate the distance dri#en o#er a 3'Psecond ti e inter#al.Solution *e ta8e speedo eter readings e#er0 fi#eseconds and record the

con#ert the #elocit0 to ft>s, using 1 i>h 5 2:'>3!'' ft>s

Relocit0 of the 1 st fi#e second Relocit0 of the 2nd fi#e secondGTS 213, S2 & S3


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The istance ro/le2=

istance tra#eled during thefirst fi#e seconds

2 ft>s s 5 12 ftistance for second ti einter#al

31 ft>s s 5 1 ft

Relocit0 of the 1 st fi#e second Relocit0 of the 2nd fi#e second

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Relocit0 Function


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The istance ro/le3'

n esti ate for the total distance tra#eled

On the other hand, use #elocit0 at the end of each ti e period. n esti atedtotal distance

Relocit0 of the 1 st fi#e second Relocit0 of the 2nd fi#e second

Relocit0Function

L! =

R! =


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istance ro/le Connection *ith rea (contMd"32

The first ti e inter#althe #elocit0 is appro4i atel0 f (t ' " andso the distance tra#eled is appro4i atel0 f (t ' " t .

Si ilarl0, the distance tra#eled during the secondti e inter#al is a/out f (t 1" t.6se the left en!points , the total distance tra#eledduring the ti e inter#al a , b

The "ore fre.uentl/ we "easure the velocit/+the "ore accurate our esti"ates -eco"e

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istance ro/le Connection *ith rea (contMd"33

6se right en!points instead of left endpoints

istance in li it e4pression

Thus the distance tra#eled is e%ual to the area under thegraph of the #elocit0 function.So the !istance pro-le" an! the area pro-le" arethe sa"e pro-le"0

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The sa e concept is also applied in 33)

To esti ate #olu e, $e add s all #olu es together.

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9e#ie$ N end of Section .13

6sing rectangles to sol#e the area pro/leefinition of area

Sig a notation

The distance pro/le as it relates to the areapro/le

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Food for Thought

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3!


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De4t topics N Section .237

efine definite integral using 9ie ann su segin e#aluating definite integralse#elop properties of the definite integral

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Find a difference /et$een these t$o integrals3:

efinite integral Indefinite integral

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efinite Integral3=

ecause f is continuous, the li it in this definition al$a0s e4istsand gi#es the sa e #alue no atter ho$ $e choose the sa plepoints x i U.

GTS 213, S2 & S3


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efinite integral 9ie ann Su)'

*e could replace x $ith an0 other letter $ithoutchanging the #alue of the integral

9ie ann Su

( ) ( ) ( )= =∫ ∫ ∫

b b b

a a a fxdx ftdt frdr

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efinite integral 9ie ann Su (contMd")1

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efinite Integral for Degati#e & ositi#e Function)2

If f ta8es on /oth positi#e and negati#e #alues,

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efinite Integral for 6ne%ual *idth)3

For une%ual $idth, the definition of a definiteintegral /eco es

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;#aluating Integrals

GTS 213, S2 & S3

))

;#aluating definite integrals /0 definition re%uiresthat $e $or8 $ith su s.On the ne4t slide $e gi#e for ulas forL

su s of po$ers of positi#e integers, and $or8ing $ith sig a notation


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6seful for ulas $hen $or8ing $ith sig a notation)

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roperties of the efinite Integral)!

*e can define

If a = b , then x 5 ' and so

( ) >even ifba fxdx a b

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Further roperties)7

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Further roperties (contMd"):

*e illustrate the first t$o properties

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;4ercise)=

6se the properties of integrals to e#aluate

( )+1 2

04 3 .x dx

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Solution'

Solution roperties 2 and 3 gi#e

ropert0 1 gi#es lso, $e found earlier that

GTS 213, S2 & S3

( )= − =1

0 4 41 0 4dx=1 2

0

1 , so3

xdx


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ropert0 1

;4ercise ItMs 8no$ that( )

=100 17 and fxdx

( ) ( )= ∫ 8 100 012, nd . fxdx fxdx

GTS 213, S2 & S3

:


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Co parison roperties2

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ropert0 :3

;4ercise 6se ropert0 : to esti ate− 21

0.xe dx

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Solution

GTS 213, S2 & S3

)

ecausefunction on ', 1 , its

a/solute a4i u #alue is " 5 f ('" 5 1 anda/solute ini u #alue is m 5 f (1" 5 eP1.

So /0 ropert0 :,

( ) −

= 2 is a d ecrea singx fx e


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roperties of the efinite Integral (contMd"

GTS 213, S2 & S3


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9e#ie$ N Section .2!

efinition of definite integral9ie ann su s;#aluating definite integrals

Interpreting definite integrals as areasroperties of the definite integral

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De4t N Section .37

Introduce the ;#aluation Theoreiscuss indefinite integrals

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GTS 213, S2 & S3

:

8e0 disco#er0 in the histor0 of calculus, due toDe$ton and ?ei/niV ($or8ing independentl0", $asthat $e can calculate

antideri#ati#e # of f .This disco#er0, called the ;#aluation Theore , ispart of the Funda ental Theore of Calculus.

( ) if we happen to know an

b

a fxdx


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;#aluation Theore=

*e can calculate antideri#ati#e # of f .

;4a ple an antideri#ati#e of f ( x " 5 x 2 is # ( x " 5 ( "⅓ x 3,so

@uch easier than using 9ie ann Su .

( ) if we happen to know anb

a fxdx

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T$o e4a ples!'

;#aluate

Solution n antideri#ati#e of f ( x " 5 e x is # ( x " 5 e x , so

Find the area under the cosine cur#e fro ' to /, $here ' / W>2.Solution ere f(4" 5 cos 4 and F(4" 5 sin 4

31

.xedx

GTS 213, S2 & S3

= = −3

3 311 .

x xedx e e e

]= = = − =∫ 00 cos sin sin sin0 sin .b b A xdx x b b


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Indefinite Integral!1

ecause of the relation /et$een antideri#ati#es andintegrals, the notation 1 f 2 x 3 dx is traditionall0 usedfor an antideri#ati#e of f and is called an indefiniteinte$ra .

Do nu /ers thereX

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Indefinite Integral (condMt"!2

*e recall that if # is an antideri#ati#e of f on aninter#al % , then the ost general antideri#ati#e of f on % is # ( x " A & , $here & is an ar/itrar0 constant.So, an indefinite integral Y f ( x " dx can representeitherL

a particular antideri#ati#e of f , oran entire fa il0 of antideri#ati#es.

GTS 213, S2 & S3


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Ta/le of Integrals!3

*e restate the antideri#ati#es $e 8no$, in thenotation of indefinite integrals. n0 for ula can /e #erified /0 differentiating the function on theright.9eference age !P1'

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GTS 213, S2 & S3

!)


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;4a ple!

Find the general indefinite integralY (1' x ) N 2sec 2 x " dx

Solution 0 Ta/le 1,

Graph the antideri#ati#e for se#eral #alues of &

GTS 213, S2 & S3


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;4ercise, re#ie$ techni%ue!!

;#aluate

Solution First $e $rite the integrand ore si pl0,

and then $e appl0 our antideri#ati#e for ulas.

+ −2 2921

2 1 .t t t dtt

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9e#ie$ N Section .3!7

The ;#aluation TheoreIndefinite Integrals

GTS 213, S2 & S3


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De4t!:

Techni%ue of IntegrationTa/le of Integrals ?oo8 in the ta/le first. If $e cannot findentr0 that rese /les our gi#en integral, then tr0 the follo$ingtechni%ues

The Su/stitution 9uleIntegration /0 artsTrigono etric Su/stitution

artial Fractions

GTS 213, S2 & S3


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Techni%ue of Integration Ta/le of Integrals!=

9eference age ! P1', on the /ac8 of te4t/oo8.

If $e cannot find an0 entr0 that rese /les our gi#enintegral, then tr0 other ethods. 9e#ie$ Se ester 1Mscontent.

GTS 213, S2 & S3


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Zuestion

GTS 213, S2 & S3

7'

The %uestion arises *ill our /asic integrationfor ulas, together $ith the

Su/stitution 9ule,integration /0 parts,ta/les of integrals, andco puter alge/ra s0ste s

ena/le us to find the integral of ever/ continuousfunction-

$


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The ns$er

GTS 213, S2 & S3

71

The ans$er is No , at least not in ter s of thefunctions $ith $hich $e are fa iliar.

It can /e sho$n, for e,a"ple , that $e $ill ne#er

succeed in e#aluatingter s of functions that $e 8no$.?ater, though, $e $ill see ho$ to e,press such

integrals as infinite series .

2

inx

edx

f


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Food for Thought

GTS 213, S2 & S3

72


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De4t N Section .1'73

Calculus II starts hereI"proper IntegralsGoals

efine i proper integrals of T0pes 1 and 2Stud0 con#ergence and di#ergence of i proper integrals6se co parison to help decide $hether an i proper integralcon#erges

GTS 213, S2 & S3

I d i f I I l


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Introduction of I proper Integrals7)

In defining a !efinite integral $e assu ed thatL

f is defined on a finite inter#al a , b and that f has no infinite discontinuit0.

*e $ant to e4tend this to the cases $herethe interval is infinite+ and>or T0pe 1

f has an infinite !iscontinuit/ T0pe 2

Such integrals are called improper .

GTS 213, S2 & S3

( )b

a fxdx

T0 1 I fi i I # l


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T0pe 1 Infinite Inter#als

GTS 213, S2 & S3

7

?et S /e the infinite region that liesunder the cur#e y 5 1> x 2,a/o#e the x Pa4is, andto the right of the line x 5 1.

Is the area of Sinfinite-

T0 1 ( Md"


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T0pe 1 (contMd"

GTS 213, S2 & S3

7!

The area of the part of S thatlies to the left of the line x 5 t is

Dotice that A(t " E 1 no atter ho$ large t is chosen, and A(t " 1 as t J.

The area of the infinite region S is e%ual to 1.

T0 1 ( Md"


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T0pe 1 (contMd"

GTS 213, S2 & S3

77

efine the integral of f o#er an infinite inter#al as the li itof integrals o#er finite inter#als

Pa/ attentionon a !irection of t

T0pe 1 (contMd" Con#ergence and di#ergence


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T0pe 1 (contMd Con#ergence and di#ergenceof T0pe 1

GTS 213, S2 & S3

7:

For e4a ple, ( )∞

∫ 1 1/ is divergent, since xdx

T0 1 ( tMd" # # di#


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T0pe 1 (contMd" con#ergence #s. di#ergence

GTS 213, S2 & S3

7=

Thus $e ha#e sho$n that

4 l # l t0 xd


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;4a ple ;#aluate

GTS 213, S2 & S3

:'

Solution art (/" of the definition gi#es

*e integrate /0 parts $ith u 5 x , d! 5 e x dx , so that du 5 dx , ! 5 e x

*e 8no$ that e t ' as t NJ[ so /0 lM ospitalMs 9ule $e ha#e

Therefore

−∞.xxedx

4 l


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;4a ple

GTS 213, S2 & S3

:1

For $hat #alues of p is the integralcon#ergent-Solution *e 8no$ fro our first e4a ple that theintegral di#erges if p 5 1[ for p \ 1,

∞

1 1 p dxx

S l ti ( tMd"


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Solution (contMd"

GTS 213, S2 & S3

:2

If p ] 1, then p N 1 ] ', so as t J, t p P1 J and 1> t p P1 '.Therefore

and so the integral con#erges.

ut if p E 1, then p N 1 E ' and so

*e su ariVe this result for future reference

T0pe 2 iscontinuous Integrands


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T0pe 2 iscontinuous Integrands

GTS 213, S2 & S3

:3

Suppose that f isa positi#e continuous function defined on a finite inter#al a , b" /ut hasa #ertical as0 ptote at b.

?et S /e the un/ounded region under the graph of f and a/o#e the x Pa4is /et$een a and b'

T0pe 2 (contMd"


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T0pe 2 (contMd"

GTS 213, S2 & S3

:)

The area of the part of S /et$een a and t is

If it happens that A(t " approaches a nu /er A as t b N , then $e sa0 that the area of the region S is A and $e $rite

*e use this e%uation to define an i proper integralof T0pe 2L

e#en $hen f is not positi#e function,no atter $hat t0pe of discontinuit0 f has at b.

( ) ( )=∫ ta A t fxdx

( ) ( )−→= ∫ limb ta at b fxdx fxdx

T0pe 2 (contMd"


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T0pe 2 (contMd

GTS 213, S2 & S3

:

T0pe 2 (contMd" Con#ergence and di#ergence


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of T0pe 2

GTS 213, S2 & S3

:!

So 0ou need to 8no$ $here the discontinuit0 is, or $here a

#ertical as0 ptote is.

;4a ple Find5 1 dx


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;4a ple Find

GTS 213, S2 & S3

:7

Solution First, this integral is i proper /ecause as0 ptote x 5 2Since the infinite discontinuit0 occurs at the leftendpoint of 2, , $e use part (/" of the definition

Thus, the integral con#erges.

−2 .2dx

x

( ) = −1/ 2 has the vertical fx x

Solution (contMd"


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Solution (contMd

GTS 213, S2 & S3

::

−5

21 .

2dx

x

;4a ple ;#aluate3

ifpossibledx


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;4a ple ;#aluate

GTS 213, S2 & S3

:=

Solution Dote that the line x 5 1 is a #erticalas0 ptote of the integrand.Since it occurs in the iddle of ', 3 , $e need part(c" of the definition $ith ( 5 1

−0 if possible.1x

Solution (contMd"


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Solution (contMd

GTS 213, S2 & S3

='

/ecause 1 N t ' A as t 1N .Thus the integral di#erges (e#en though

]

( )

( )

− −

−

−

→ →

→

→

= −− −

= − − −

= − = −∞

∫ ∫ 1 00 01 11

1

Now =lim limln 11 1

lim ln 1 ln 1

limln1

t t

t t

t

t

dx dx xx x

t

t

−3

1 con verges).

1dx

x

di#ergei#erge con#erge

Su ar0 I proper Integrals occur $hen


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Su ar0 I proper Integrals occur $hen=1

The do ain ofintegration, fro a to b,is not finite

T0pe 1 Infinite Inter#als

The range of theintegrand is not finiteon this do ain

T0pe 2 iscontinuousIntegrands

GTS 213, S2 & S3

Food for Thought


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Food for Thought

GTS 213, S2 & S3

=2

Co parison Test


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Co parison Test

GTS 213, S2 & S3

=3

oes a gi#en i proper integral con#erge-

So eti es it is i possi/le to find the e4act #alue ofan i proper integralLLand 0et it is i portant to 8no$ $hether it iscon#ergent or di#ergent.In such cases a co parison test such as the follo$ing

theore can /e #er0 useful. *e state the test for T0pe 1 integrals[ a si ilar #ersion holds for T0pe 2

Co parison Test (contMd"


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Co parison Test (contMd

GTS 213, S2 & S3

=)

The idea /ehind the theore is that if the area under the graph of f is finite, then so is the area under

the graph of $[the area under the graph of $ is infinite, then so is the areaunder the graph of f .

;4a ple


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;4a ple

GTS 213, S2 & S3

=

Sho$ thatSolution It is not possi/le to e#aluate the integraldirectl0 since +ordinar0 antideri#ati#e.

Instead $e $rite

∞ − 2

0 is convergent.xe

− 2 has noxe

n ordinar0definite integral

6se Co parison Test

Solution (contMd"


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Solution (contMd

GTS 213, S2 & S3

=!

In the second integral $e use the fact that for x 1 $e ha#e x 2 x , so N x 2 N x and therefore

right

nd the integral of e N x is eas0 to e#aluate.

− −≤2 , as shown on thex xe e

Solution (contMd"


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Solution (contMd

GTS 213, S2 & S3

=7

Thus, ta8ing f ( x " 5 e N x and

in the Co parison Theore sho$s thatIt follo$s that

( ) −=2x gx e

∞−

2

1 is convergent.xe dx∞ − 20

is convergent.xe dx

n ordinar0definite integral

Con#ergent

6se Co parison Test

Con#ergent

Convergent

9e#ie$ N Section 1'


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9e#ie$ N Section .1=:

T$o t0pes of i proper integralT0pe 1 (infinite inter#als"T0pe 2 (discontinuous integrands"

Co parison Theore

gts 213-ch5 integrals

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