math 122 integral calculus

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MTH 122 Integral Calculus

NOUN 2

MTH 122: INTEGRAL CALCULUS

COURSE GUIDE

NATIONAL OPEN UNIVERSITY OF NIGERIA




Acknowlede!en": I acknowledge previous writers on their field (soe of whichare listed for further reading! whose ideas and works have ade it possi"le for e to

present # point$

TA#LE OF CONTENT

INT%O&UCTION'HT )OU 'I** *+%N IN THI, COU%,+COU%,+ IMCOU%,+ O-.+CTI/+,'O%0IN TH%OUH THI, COU%,+COU%,+ MT+%I*,,TU&) UNIT,T+T-OO0,,,+,,M+NTTUTO% M%0+& ,,INM+NTCOU%,+ O/+%/I+'

INTRODUCTION

This course is the second of the two calculus courses that #ou are re3uired to stud#$In the introduction of calculus4 #ou were infored a"out the iportance of thecourse calculus to huanit#$ lso #ou were infored a"out those categories ofstudents which need calculus as a working tool for their respective prograes inthe National Open Universit# of Nigeria$ Therefore4 in this introduction4 #ou ight

not "e told what #ou are alread# failiar with a"out the su"5ect calculus$ However4for ephasis4 the second course of calculus deals with the "ranch of calculus knownas 6Integral Calculus7 or 6anti8differential calculus7$ Interestingl#4 this second coursein calculus happens to "e the first that was known to the earl# atheaticians in thesense that the stud# of coputation of areas under a curve is older than the stud# ofdifferential calculus$ If #ou know the distance function of a oving o"5ect #ou canfind the velocit# "# differentiating the distance functions$ In this place #ou will "egiven the velocit# and will "e re3uired to find the distance a "od# travels within aspecific interval of tie$ In other words4 integral calculus deals with the inverseoperation of differentiation$ It e9tends the concept of addition to ena"le #ou to findthe su of a continuousl# changing 3uantities then integration continues where

finding the su of changing 3uantities "# discrete ethod fails$ Thus integralcalculus is widel# used in ost situation that re3uires finding the su of 3uantitieslike coputing the areas under curves4 volues generated "# rotating certain solidsalong their a9is of s#etr#4 the length of arc4 work done "# oving a "od# along astraight line etc$

NOUN :




In this course4 #ou will stud# one of the a5or applications of integral calculus4 learnthe coputation of areas under one or ore curves$ In unit 1 #ou will "e introducedto the siple nuerical ethod of coputing area under a curve$ Then in unit ; theuse of integration will "e introduced in finding the areas "ounded "# curves$ Therelationship "etween the area under a curve and the definition of definite integration

was what lead to the s#"ol < of integration to "e an elongated ,4 where , stands fora su$ s was done in calculus4 I ephasis has "een placed on the techni3ue ofintegration so that #ou can narrow #our guess work apprecia"l#$ The solvede9aples provided throughout in the stud# units will ena"le #ou work through thecourse effortlessl#$

$HAT YOU $ILL LEARN IN THE COURSE%

In this course4 #ou will learn how to copute areas under a curve$ )ou learntheore (without proof! that connects integration and differentiation and called theinverse operations to each other i$e$ integration is the inverse operation ofdifferentiation$ )ou will use this theore which is known as fundaental theoreof integral calculus to develop techni3ues of indefinite integration$ )ou will learnhow to use integration to copute the areas "ounded "# two curves4 the voluegenerated "# solid of revolution4 the distance covered "# a oving "od# with aknown constant velocit# etc$

COURSE AIM

This course ai at developing #our skills in the art of integration with a little effort

on #our part$ This is easil# achieva"le "# recalling previous knowledge gained frocalculus I$ Thus in this course4 special techni3ues are introduced that will akeintegration ore of a routine than a guess work$

COURSE O#&ECTIVES

On successful copletion of this course4 #ou should "e a"le to

i$ copute nuericall# the area under a curveii$ evaluate definite integralsiii$ evaluate indefinite integrals

iv$ evaluate integrals involving trigonoetric functions such as sin94cos94 tan 94 col9 cosec9 and sec9$

v$ +valuate integrals involving rational functions involve a2=u24 √ a2=u2

etc$vi$ evaluate integrals involving rational functions of sin94 cos94 tan9 etc$vii$ evaluate integrals involving product functions such as >auCu2 "u4 >au

?18"u >9n etc$

NOUN @




viii$ o"tain reduction forular for certain categories of functions$i9$ copute area "ounded "# two intersecting curves$9$ copute the volues of solid of revolution9i$ find the distance traveled "# a oving o"5ect with a constant velocit#$9ii$ copute the work done "# copressing or stretching a spring$

COURSE MATERIALS

Herewith list of aterials that #ou will need in order to successfull# coplete thiscourseA

: Course uide for MTH 111@ ,tud# Units for MTH 111B %ecoended list of "ooks ssignent fileD &ates of tutorials4 ssessent and +9aination

COURSE OVERVIE$

There are 2 odules in this course$ +ach containing B units$

Mod'le I:

Unit 1A Coputation of reas "# CalculusUnit 2A &efinite Calculus

Unit :A Indefinite IntegralUnit @A Integration of Transcendental functionsUnit BA Integration of Eowers of Trigonoetric functions

Mod'le II:

Unit A Further Techni3ues of Integration IUnit DA Further Techni3ues of Integration IIUnit GA Further Techni3ues of Integration IIIUnit ;A pplication of Integration IUnit 1A pplication of Integration II

SET TE(T#OO)S

The following are recoended te9t"ooks #ou could "orrow or purchase theA

1$ odan 4 Tal"ert .$F$ dditional Matheatics Eure andpplied in ,$I$ *ongan

NOUN B




(viii! Thoas $-$ and Calculus and nal#tic eoetr# Bth Finne# %$*$ (1;G2! +d$ ddison J 'esle# Eu"lishing Co$

'orld student series +dition4 *ondon4,#dne#4 Tok#o4 Manila %eading$

(i9! ,atrino *, K +inar H$ Calculus 2nd +dition .ohn 'ile# K ,ons1$ New )ork *ondon4 ,#dne#4 Toronto$

@$ Osisiogu U$4 NwoLuC$%$ +ssential Matheatics for pplied andet al (21! Manageent ,ciences$ -estsoft

+ducational -ook4 Nigeria$

B$ Osisiogu U$$ (+d! (21! Fundaental of Matheatical nal#sis/ol$ I4 -estsoft +ducational -ooks4

Nigeria$

$ Osisiogu U$$ (+d! (21! Fundaental of Matheatical nal#sis/ol$ II 4 -estsoft +ducational -ooks4

Nigeria$ASSIGNMENT FILE AND TUTOR MAR)ED ASSIGNMENT *TMA+

There are at lest : e9ercise at the end of each section of ever# unit$ These e9ercisesare there as reinforceent to each topic studied$ )ou are to do it all alone$ Theanswers are supplied for self evaluation$

There are two categories of assignent given throughout this course$ The first are

e9ercises given at the end of each section of ever# unit$ These e9ercises are given toreinforce #our understanding of the concept studied$ )ou are re3uired to do it allalone$ The answers to the e9ercises are given in the unit for self evaluation$ Thiswill ake #ou onitor #our progress$

The second are e9ercises given at the end of each unit$ There are at least ten ofthe$ These are assignent that #ou ust do and su"it to #our tutor at #our stud#centre$ These assignents will "e supplied to #ou in #our assignent files$

E(AMINATION ,nd MAR)ING

The final e9aination for the course MT 12@ will "e for a duration of 2 hours$

STRATEGIES FOR STUDYING THE COURSE

Integration involves finding a function whose denuative is the integrand$ s such itinvolves certain level of guess work$ In view of this4 #ou are re3uired to spend uchtie reviewing #our differential calculus and trigonoetr# functions$ Thus

NOUN




differential calculus is a necessar# tool #ou need to develop #our skills in techni3uesof integration$ Therefore4 #ou are re3uired to spend @ hours or ore stud#ing eachunit$ -ecause of the tie re3uired to stud# a unit4 this course has "een designed tocontain onl# 1 units instead of 2 units$ )ou are re3uired to spend a"out one houror ore on the solved e9aples of each unit$ -# doing this4 #ou would have

developed enough skills that ost of #our guesses will "e accurate$ 'hile readingthrough this course4 ake sure that #ou check up an# topic #ou are referred to in an#

previous unit #ou have studied$

SUMMARY

In this course4 #ou have studied how toA

1$ Copute area under a curve as a su of areas of rectangles inscri"ed orcircuscri"ed under a curve within a given interval$

2$ ppl# the fundaental theore of integral calculus in evaluating the definitean indefinite integrals$

:$ To evaluate integrals using "oth standard notation an properties of indefiniteintegrals$

@$ &erive standard integration forulas such as ?sinudu8Cosu=c4 ?1duInlul=Cetc$

B$ +valuate integrals of odd and even powers of cos9 and sin9$$ O"tain reduction forula for ?>a9 Cu8"v d94 ?>99nd9 ?Cosn9 du etc$D$ +valuate integrals of this t#pe

? du ? du

a9

=u2

4 √ a2

=u2

etcG$ Copute integrals "# the following ethods

(1G! partial fractions(1;! copleting the s3uare(2! half angles forula and(21! ethod of integration "# parts$

;$ Use definite integration in finding

(1! areas under two curves(2! distance traveled "# an o"5ect with a constant velocit#(:! volues of solid of revolution etc$

NOUN D




NOUN G

MTH 122: INTEGRAL CALCULUS

COURSE DEVELOPMENT

Course Developer &r$ U$ $ Osisiogu

+"on#i ,tate Universit#"akaliki

Unit Writer &r$ U$ $ Osisiogu

+"on#i ,tate Universit#"akaliki

Programme Leader

Dr. Makanjuola Oki

Course Coordinator B. Abiola

NATIONAL OPEN UNIVERSITY OF NIGERIA




CALCULUS II

UNIT 1

COMPUTAION OF AREAS #Y CALCULUS

TA#LE OF CONTENT

1$ INT%O&UCTION2$ O-.+CTI/+,:$1 %+ UN&+% CU%/+:$2 E%TITION OF C*O,+& INT+%N*:$: COMEUTTION OF %+ , *IMIT,@$ CONC*U,IONB$ ,UMM%)$ ++%CI,+, J TM,D$ FU%TH+% %+&IN

1%- INTRODUCTION

One of the earl# atheaticians that attepted to find the area under a curvewas a reek naed rchiedes$ He used ingenious ethods to copute thearea "ounded "# a para"ola and a chord$ ,ee Fig (1$1!$

In this unit4 #ou will stud# how to develop necessar# tools of calculus tocopute areas under curve as a ere routine e9ercise$ The area under a

curve gave "irth to the second "ranch of calculus known as integration$ Thetools that will "e developed here will naturall# lead to the definition ofintegration in the ne9t unit J unit 2$ %ecall that the word to integrateconnotes whole of which could "e interpreted to ean find the whole areaof$ This concept is what will "e introduced in this unit and this will "e full#developed in the ne9t unit$

NOUN ;




2%- O#&ECTIVES

fter stud#ing this unit4 #ou should "e a"le to correctl#

i$ appro9iate area under a curve "# the su of areas of rectanglesinscri"ed in the curve$

ii$ appro9iate the area under a curve "# the su of the areas ofrectangles circuscri"ed over the curve$

iii$ define a partition of a closed interval (a4 "!iv$ copute the e9act value of the area under a curve "# the liiting

process$

.%1 AREA UNDER A CURVE

)ou are 3uite failiar with the coputation of the areas of plane figures suchas triangles4 parallelogra trapeLiu4 regular pol#gons etc$ Interestingl#4#ou studied in eleentar# geoetr# that the area of a regular pol#gon can "ecoputed "# cutting it into triangles and su up the areas of the triangles$

)ou are also aware t hat the area of a circle is πr2$ This forula was derived "# the ethod of liit$ )ou could recall that the liit of the areas ofinscri"ed regular pol#gons as the nu"er of sides approaches infinit# is e3ualto the area of the circle$ ,ee Fig$ 1$2 a8c

Inscri"ed pol#gons Inscri"ed pol#gons Inscri"ed pol#gonsof @ sides of sides of G sides

F/% 1%2, F/% 1%20 F/% 1%2c

*et # f(9! "e a continuous function (see the first course on calculus i$e$calculus I unit @ for definition of continuous function! of 9 on a closedinterval a4 "P$ In this case for "etter understanding4 #ou assue that the f(9!is positive in the closed interval i$e$ f(9! Q $ for all + a4 "P$ Then the

pro"le to "e considered is to calculate the area "ounded "# the graph# f(9! and the vertical lines # f(a! and # f("! and "elow "# the 9 J a9isas shown in Fig$ 1$:$

NOUN 1




)

a "

F/% 1%.

)ou can start "# dividing the area into n thin strips of unifor widthR9 ("8a! "# lines perpendicular to the 9 J a9is at the end points 9 a and

n9 " and an# interediate points which can "e nu"ered as 14 24 n81

(see fig 1$@!$

)

f(91! - C F f(92!

f(9n81! & + a 1 2 % % % n82 n81 n "

F/% 1%

The su of the areas of these n rectangular strips gives an appro9iate valuefor the area under the curve$ To put the a"ove ore atheaticall#4 #ou candefine the area of each strip in ters of f(9! and 9$ iven that R9 91 J a

NOUN 11




92891 $$ "89n81$ For e9aple the area of the rectangular strip -C& in Fig$1$@ a"ove is given asA

rea of -C& f(92! $ (9189! f(92! R9

E,!3le:

,uppose f(9! 928: in Fig 1$B with n were a 24 " G4d9 G82 ThereforeA d9 1 i$e$ #ou have rectangular strips$

2

1 1 2 : @ B

2 : @ B D G

F/% 1%4

rea is given as su off(2! R9 1$1 1f(:! R9 $1 f(@! R9 1:$1 1:f(B! R9 22$1 22f(! R9 ::$1 ::f(D! R9 @$1 @

In fig$ 1$@ a"ove the area under the curve is larger than the su of the areasof the inscri"ed rectangular strips nu"ered 1 to i$e$ su of areas of strips

1==1:=22=::=@ 121 which is less than area under curve$

E,!3le:

Using the sae e9aple ) 928: use circuscri"ed rectangular stripsinstead of inscri"ed ones to copute the area under the curve$ ,ee Fig$ 1$

NOUN 12




)

1 1 2 : @ B

2 : @ B D G

rea is given as the su of

f(9! $ R9 $1 f(9! $ R9 1:$1 1:f(9! $ R9 22$1 22f(9! $ R9 ::$1 ::

f(9! $ R9 @$1 @f(9! $ R9 1$1 1

rea =1:=22=::=@=1 1G1s should "e e9pected this area is greater than the area under the curve f(9! 928:$

In the coputation with the circuscri"ed rectangular strips the sides of therectangles are assued in this case to "e the points of the function in theirrespective su"intervals$ In the case of the inscri"ed rectangles4 the sides of

the rectangles are the iniu values of the function in their respectivesu"intervals$

Therefore the area under the curve lies "etween the su of the areas of theinscri"ed rectangles and the su of the areas of the circuscri"e rectangles$This takes to the issue of liit$ Therefore it will "e right to sa# as n SR9 this iplies that the

NOUN 1:




*i (Ma9 rea J Min rea! as R9 $

Fro the foregoing4 #ou can now define the area under curve as the liit ofthe sus of the areas of inscri"ed (circuscri"ed! rectangles as their coon

"ase of length d9 approaches Lero and the nu"er of rectangles increaseswithout "ound$ In s#"ols #ou can write the a"ove liit asA

li f(9! R9 = f(9D! R9 =$$= f(9n81!P R9 n

li f(91! R9 = f(92! f =$$= f(9n81!P R9 n

O% n81 n

li f(9k ! d9 li (f9k ! d9 n 91 01 n 91 01

Ee5c/6e:

%epeat the a"ove e9aple using n 1$ Find the difference "etween the suof areas of the inscri"ed rectangle (i$e$ the iniu area! and the su ofareas of the circuscri"ed rectangles (i$e$ the a9iu area!$

.%2 PARTITION OF A CLOSED INTERVAL

*et a4 "P "e a "ounded closed interval of real nu"ers$ partition of aclosed interval a4 "P is a finite set of points

E Va 94 914 924 $$4 9n814 9n8"W where

a 9 X 91 X 92 X $$ X 9n814 9n "

E,!3le:

E V4 1Y4 1Y:4 Z4 2Y:4 BY4 1W and[ V4 \4 Z4 ]4 1: are "oth partition of V4 1:W

EU[ V4 1Y4 \ 4 1Y:4 Z 4 2Y:4 ]4 BY 1W is a partition of 4 1P

En[ V4 Z4 1W is a partition of 41P

,ee fig$ 1$(a! to (c!

NOUN 1@




E E

1Y 1Y: Z 2Y: BY 1

[

\ Z ] 1

En[

1Y \ 1Y: Z 2Y: DY@ BY

Z 1

partition of E V94 914 $$4 9nW of a4 "P divides a4 "P into n closed su"interval 94 91P4 914 92P4 ^4 9n814 nP

The closed interval

9r814 9r P is called the r th su"interval of the partition$

iven a partition of Ea 94 924 ^4 9n "P the length of the su"interval s arethe sae and it is denoted "# R9r 9r 8 9r81

This e3ual to the length of the interval a4 "P divided "# the nu"er ofsu"intervals n

i$e$ R9r " J a n

E,!3le: R9 for p is 1 J 1Y

R9 for [ is 1 J \@

Not in all case #ou will get su"intervals of the sae length$ +9aple is EU[The length of 91 J 1Y J 1YThe length of 92 J 1 \ 8 1Y 1Y12

NOUN 1B




,uch partitions in which the su"intervals are not of the sae length are calledirregular partition$

Ee5c/6e: 'rite down a regular partition for

(1! 24 GP4 n 12(2! 14 GP4 n D

An6:

(i! 24 BY24 Y24 DY24 GY24 ;Y24 1Y24 11Y24 12Y24 1:Y24 1@Y24 1BY21Y2P

(ii! 14 24 :4 @4 B4 4 D4 GP

.%. COMPUTATION OF AREAS AS LIMITS

In this section #ou will co"ine the results of section :$1 and :$2 to coputethe areas under curves using the liiting process$

E,!3le

good starting point is to consider the area under the curve ) (see Fig$1$D!

E E f("!

f(a! [

E - a 91 92 Z 4 9n "

F/ 1-%7which the interval _ a4 "P let there "e n8regular partition of a4 "P i$e$

R9 " J a n

E a4 914 924 $$4 9n814 9n "P

91 a = 99

NOUN 1




92 a = 2R99: a = :R9

9n81 a = (n81! R9

reas of inscri"ed rectangles are

f(a! $ R9 a$R9f(91! $ R9 (a=R9!$R9f(92! $ R9 (a=2R9!$ R9$$f(9n81! R9 (a=(n81!R9! $ R9

,u of the areas of the rectangles is given as

√ (a $ R9 = (a = R9! = ^ = (a= (n81! R9!$R9 a=((1=2=:=^(n81! d9 n81

na = (4 0! R9 ! R9 01

n81

( 0 (n81!n (The su of an arithetic 1 2 progression with different d1!

, na = (n81!n d9 P d9 2

"ut d9 " J a therefore n

, na = (n81!n " J c! P " J c 2 n n

a = " J a n J 1 P (" J a!2 n

Taking liit as n #ou get

*i , li (a = " J a $ n J 1 (" J a! n n ( 2 n !

(a = " J a! (" J a! lis n J 1 2 n n

(a = (" J a! (" J a ! $ 1

NOUN 1D




2

a = " $ (" J a! 2

In fig$ 1$G4 the area of trapeLiu [E- is the sae as the area under thecurve and as #ou know the area of trapeLiu is given asA

Z "ase 9 su of two parallel sides Z (" J a ! 9 f(a! = f("! Z (" J c! (" = a!

E,!3le

Find the area under the graph ) 9 = 1 ` 9 `

Sol'"/on:

*et n "e a positive integer that there "e a partition of a4 "P into n regular partition$

Therefore R9 " n

91 R9 92 2R9 9: :R9

$$

9n81 (n81! R9 C

- n81 n

1 2 &

O 91 92 9n81 91

F/% 1-%8

rea of (n81! rectangles is given as

f(!$R9 1$R9

NOUN 1G




f(91!$R9 (R9 = $1! R9

su of areas of rectangles is

, R9 = (R9=1!R9= (2R9=1! R9=(:Rr=1!R9

= ^ = (n81! R9 =1! R9 (R9= (R9=1!=(2R9=1! = A (n81!(R9=! R9 R9 = n =! n81 kR9P R9 01

, R9 = (n= (n81!n R9P R9 let R9 l 2 nthen

, "P = (n = (n81! n "!P " nP ( 2 ! n!P n

" = (1 = " $ n 81!n 2 ! $ "

Taking liits as n

*i , li " = " li (1 = " (n81! n 2 n n n J 1 n

= " (1 = "Y2!

li , " = " 2 n 2

Ee5c/6e:

,how that the area of the trapeLiu -C& in Fig$ 1$G is e3ual to "("=2! 2

%- CONCLUSION

In this unit4 #ou have studied how to find an appro9iate value of the areaunder a curve "# coputing the sus of areas of rectangles inscri"ed underthe cure and circuscri"ed over if #ou have defined a partition of a closedinterval$ )ou have studied that as the nu"er of partitions of a closedinterval a4 "P is increased without "ound the value of the su of the areas ofthe rectangles (inscri"ed or circuscri"ed! approaches the e9act value of the

NOUN 1;




area under the curve in the given interval a4 "P that is the liit of the su ofareas of the rectangles is e3ual to the e9act area under the given curve as thenu"er n of partition tends to infinit# or the length d9 of the su"interval ofthe partition tends to Lero$

4%- SUMMARY

In this unit #ou studied how to

• copute the iniu value of area under a curve i$e$ su of area ofrectangles inscri"ed under a curve within an interval

• copute the a9iu value of the area under a curve i$e$ the su ofareas of rectangles circuscri"ed over the curve$

• define a partition of a closed interval a4 "P i$e$ a 91 X 92 X ^ X 92 " Ea4 "P

•

copute the e9act area under a curve in a given interval a4 "P "#taking the liit of the su of the areas of the rectangles (inscri"ed orcircuscri"ed! as the nu"er n of partition of a4 "P is increasedwithout "ound i$e$ li ?n where d9 " J a

n n

9%- E(ERCISE

1$ ,how that the sets

V4 1W V4 Z4 1W4 V4 \4 Z4 1W andV4 1Y@ 4 1Y:4 Z4 BYG4 1W are partition of V4 1W

2$ 'hich of the partition of 4 1P in e9ercise (1! a"ove are regularb

:$ Find the iniu and a9iu values of the area under the curvef(9! 29 for 94 1P and E(4 \4 Z4 1!

2$ Find the iniu value of the area under the curve f(9! 18 9 9e42P E(11Y:4 ]4 14 2!$

:$ Find the area under the curve ) 92 4 "P "# taking appropriateliits$

@$ Find the area under the curve ) 9 a ` 9 ̀ " "# taking appropriateliits$

NOUN 2




B$ ,ketch the graph of ) = 1$ &ivide the internal into n su"intervals with d9 (" J a!Y$ ,ketch the inscri"ed rectangles$

$ %epeat 9 D "ut this tie sketch the circuscri"ed rectangle$

D$ Copute the sus of areas in 9 D and 9 G a"ove$

G$ Find the area under the curve ) 9 = 1 a ` " "# taking appropriateliits of results of e9ercise ; a"ove$

NOUN 21




UNIT 2

DEFINITE INTEGRAL

TA#LE OF CONTENT

1$ INT%O&UCTION2$ O-.+CTI/+,:$ &+FINITION OF TH+ &+FINIT+ INT+%*:$1 TH+ FUN&NM+NT* TH+O%+M OF INT+%* C*CU*U,:$2 +/*UION OF &+FINIT+ INT+%*@$ CONC*U,IONB$ ,UMM%)$ TUTO% M%0+& ,,INM+NTD$ FU%TH+% %+&IN

1%- INTRODUCTION

In unit 14 #ou studied how to copute the area under a curve and showedhow #ou could estiate it "# coputing sus of area of rectangles$ Usingthe a"ove estiate #ou applied the concept of liit to get the e9act value ofthe area under a curve$ These ethods were applied to functions or graphsthat could easil# "e sketched i$e$ not too coplicated functions$ In this unit4#ou will "e introduced to the faous path taken "# *ei"niL and Newton inshowing how e9act areas can "e coputed easil# "# using integral calculus$

It is necessar# #ou refer once ore to unit 1 of this course "efore e"arkingon this one$ It will help #ou have a proper grasp of this unit if #ou do so$

2%- O#&ECTIVES

fter stud#ing this unit #ou should "e a"le toA

&efine the definite integral of a function within an interval a4 "P$ +valuate definite integrals of function$ ,tate the fundaental theore of integral calculus$

.%- DEFINITION OF DEFINITE INTEGRAL

In unit 14 #ou studied that the su of the areas of inscri"ed rectangles gives alower (iniu! appro9iation of the area under the curve of the functionf(9!$ If #ou list all the values of the function f(9! in a given interval a4 "P andtake the least aong these value #ou will have what is known as the infiuof f(9! for all 9 a4 "P

NOUN 22




*(E! ` Z (91=9o! (9189o! = Z (92=91! (92891! = ^= Z (9n=9n81!(9n89r81! ` u(E!

"ut Z (91=9o! (9189o! = Z (92=91! (92891! = $$ = Z (9n=9n81! (n=n81!

Z (92 J 92 = 92 J 92 = ^ = 92 J 92n81! Z (92n J 92o !

1 2 1

2 2

* (E! ` Z (9n J 9o! ` u(E!

* (E! ` Z ("28a2!

? "9 d9 Z ("2 8 a2! a

The following properties of definite integral are here"# stated without their

proofs are "e#ond the scope this courseA

1$ If a X c X " then ?cf(9!d9 = ? "f(9!d9 ? "f(9!d9 a c a

2$ If a X " then J? "f(9!d9 ?af(9!d9 a "

:$ ?af(9! d9 a

E,!3le: iven that

?1f(9!d9 4 ?:f(9!d9 B 1

?Df(9!d9 2 :

Find (i! ?Df(9!d9 (ii! ?:f(9!d9 (iii! ?1f(9!d9 (iv! ?1f(9!d9 1 1 D

Sol'"/on: (i! ?Df(9!d9

?Df(9!d9 ?tf(9!d9 = ?Df(9!d9 9

let t : i$e$ ?Df(9!d9 ?:f(9!d9 = ?7 f (9!d9 B = 2 D :

NOUN 2B




Teo5e! 1: If f(9! is a continuous function on a4 "P4 the function F(9!defined on a4 "P "# setting F(9! ?9f(t! dt a

is a(i! continuous function on a4 "P and (ii! satisfies F1(9! f(9! for all 9 in(a4 "!$

P5oo;: )ou will "egin with 9 a4 "P and show that L/! F(9=h! < F(9! f(9!

h O8 h

In figure 2$1 F(9 =h! area fro a to 9 = h

f(9! f(9=h!

F(9! a 9 h 9=h "

F(9! area fro a to 9 F(9 = h! 8 F(9! area fro 9 to 9 = h (rea "ase 9 height!

F(9 = h! J F(9! area fro 9 to 9 = h f(9! if h h h

(note f(9! height of the area under curve in Fig$ 2$1If 9 X 9 = h ̀ " thenF(9=h! J F(9! ?9=hf(t!dt J ?9f(t!dt

a a

(since fro stateent of theore F(9! ? "f(t!dt! a

It follows therefore that

F(9=h! J F(9! ?9=h(f(t!dt = ?af(t!dt a 9

?9=h(f(t!dt 9

*et Mh a9iu value of f(9! on 94 9=hPand h iniu value of f(9! on 94 9=h P

NOUN 2D




since Mh (9=h J 9! Mh$h

and h (9=h J 9! h$h

therefore4 Mh upper su (see UNIT 1!

and h lower su (see UNIT 1!therefore

h$h ` ?9=hf(t!dt ` Mh$h 9

h$h ` F(9=h J F(9! ` Mh$h h

since f(9! is a continuous function on 94 9 = hP therefore

*i h$h f(9! li Mh$h h 8 h 8

thus li F(9=h! J F(9! f(9! $ I h h

In a siilar anner #ou can show that if (a4 "!4 then*i F(9=h! J F(9! f(9! Ih = h

Now if 9 (a4 "! then e3uation (I! and (II! hold

Thus *i F(9=h! J F(9! f(9!

H hand li F(9=h! J F(9! F1 (9!

h h

therefore F1(9! f(9!

since F1(9! e9ists then F(9! ust "e continuous on (a4 "!$ -efore #ou provethe fundaental theore of calculus$ *ook at this definition$

De;/n/"/on:

function F(9! is called an anti8derivative for f(9! on (a$ "! if and onl# if

(i! F(9! is continuous on (a4 "! and(ii! F1(9! f(9! for all (a4 "!

Using the a"ove definition #ou can rewrite theore 1 as

NOUN 2G




a

E,!3le: Find ? "9d9 ,

Sol'"/on

*et F(9! Z92 as an anti8derivative

Then ? "9d9 Z ("2 J a2! a

Find the ? "9nd9 when n is a positive integral the anti8derivative to use is a

F(9! 1 9n=1

n=1

F1(9! 9n ? "9n F("! J F(a! a

1 ("n=1 J aa=1! n=1

No","/on:

? "f(9!d9 F(9!P " F("! J F(a! a a

thus ? "9@d9 1 9@=1P " 1YB ("B J aB!

a

@=1a

E,!3le:

?2(9B829:89!d9

*et F(9! 9 J 29@ J 92

@ 2

then ?2(9B J 29: J 9!d9 9 J 19@ J 92P@

1 2 2 1

:;$

E,!3le:

+valuate the following integrals "# appl#ing the fundaental theore$

(i! ?(981! (982!d9

NOUN :




:

(ii! ?D d9: (982!2

(iii! ?1(9:Y@ = Z 9 Z ! d9

o

(iv! ?;(a2 9 J 9@ !d9 a

(v! ?# $ 9 d9 1 9:

(vi! ?G (√ t J 1!dt 1 t2

(vii! ?: J tdt 1

(viii! ?292(981!d91

(i9! ?%√ 9=1d91

(9! ?1(981!1Dd9

Sol'"/on: To evaluate ?(981! (981!d9 8:

#ou e9pand the function (9 8 1! (9 81!

92

J 29 = 1(i! ?:(981! (981!d9 ?:(92829=1! d9

let F(9! 1Y: 9: 8 92 = 9 serve as anti8derivative

therefore ?: (92829=1! 1Y:9: J 92 = 9P: :

(ii! ?D d9 : (982!

construct a function with derivative as 1 it is not difficult to see that (982!2

d 8 1 1d9 982! (982!2

NOUN :1




d9Fro #our e9perience with differentiation #ou can easil# know that # 92

since d# 29 d9Interestingl#4 it is not onl# # 92 that can "e differentiated to give d# 29$

d9Other function like # 928 14 # 92 = 24 # 92 = a4 # 92 = @ can "edifferentiated to #ield d# 29 d9In general an# function of this for # 9 2 = c4 where C is an# constant will#ield a differential e3uation of this t#pe d# 29 d9)ou are now read# to take this definition$

De;/n/"/on 1: n e3uation such as d# f(9! which specifies the derivative asd9

a function of 9 (or as a function of 9 and #! is called a differential e3uation$For e9aple d# sine

d9 is differential e3uation

De;/n/"/on 2: function # F(9! is called a solution of the differentiale3uation d# f(9! if over doain aX9c" F(9! is differentia"le and d9d F(9! F1(9! f(9!d9

in this case F(9! is called an integral of f(9! with respect to 9$De;/n/"/on .: If F(9! is an integral of the function f(9! with respect to 9 so isthe function F(9! = C an integral of f(9! with respect to 94 where c is anar"itrar# constant$ If d F(9! f(9! so also is F(9! = C i$e$ d F(9! = CP d9 d9d F(9! = C dc df(9! = F1(9! f(9!d9 d9 d9

Fro the a"ove if # F(9! is an# solution of d# f(9! then all other solutionsare contained in the forula # F(9!=C d9 where C is an ar"itrar#

constant this gives rise to the s#"ol$ ?f(9! d9 F(9! = C J (1! where thes#"ol ! is called an integral sign (see unit 2!$ +3uation 1 is read the integralof f(9!d9 is e3ual to F(9! plus C since d# 29 and a t#pical

d9

solution is F(9! 92 = C$ then d F(9! 29 d(92 =C!d9 d9

# 92 = C

NOUN :




dt d9 92

Sol'"/on: d# 928 1d9

d# (92 J 1!d9

?d# ! (9281!d9

"ut d(9 : J 9! (9281!d9 :thereforeA # ?d(9: 89! 9 : J 9 = C : :

(2! d# 1 = 9d9 92

?d# ,(1 = 9!d9

92

d(81 = 9 2 ! (1 = 9! d9 9 2 92

# ?d(9 2 J 1! 92 8 1 = c2 9 2 9

(:! d# 9

d9 #?#d# ?9 d9

d(# 2 ! #d# and d(92 ! 9d9 2 2

thereforeA ?#d# ?d(#2!2

?d(# 2 ! ?d(9 2 ! 2 2

#2 92 = C1

2 2

#2 92 = 2C1

#2 92 = C

NOUN :G




2√ 9 Gt = C4 where C C9 = Ct

(1! d# (@92 J 1!d9 92

?d# ? (@92

J 1!d9 9# ?d(@9 : J 1! @9: J 1 = C

: 9 : 9

Ee5c/6e: +valuate the followingA

(1! ?d9 (2! ? (9 = 1!:d9 9B

(:! ? (a92 = "!d9 (@! ? (9: = 1 ! d9

9

(B! ? (9: J 1 !d9 (! ? (t2 J a! (t2 J "!dt 92

(D! ? (√ 9 J 1!d9 (G! ? (B9!@d991Y: 9B

(;! ?d9 (1! ? (9 J 1!2 = 1 !d9√ 1=9 (9=2!2

An6:

(1! 81@9@=C (2! \ (9 = :!@ = C

(:! 1Y:a9: = "9 = C (@! 81 B9: = 2 = C1 9B

(B! Z (9: = 2! (! 1tB= 1("8a!t: J a"t = C 9 B :

(D! 2Y: 9:Y2 J :Y2 92Y: = C (G! 12B = C :91B

(;! 2√ 9 = 1 = C (1! 1Y:(9 J 1!: J 1 = C (9=1!

7%1 P5o3e5"/e6 o; Inde;/n/"e In"e5,l

NOUN @




then du : du :d9 d9

Therefore ?√ :9=1d9 ?u1Y2 du :

here d9 du :

therefore 1Y:! u1Y2 du 1Y: u 1Y2=1 Z=1

1 2u:Y2 2U:Y2

: : ; 2(:9=1! ;

(@! ?√ 8 1 d9

*et U @9 J 1 du @ d9 du d9 @

then ?√ @981d9 ?u1Y2 du @

\ ?u1Y2 du U 1Y2=1 2(@981!:Y2 = C Z=1 12

E,!3le6: +valuate the following integrals(i! ?√ 1@9 (ii! ?: 1 = 9 d9

(iii! ? B 29=1 (iv! ? @ @9 8 2

(v! ? 9 = @

Sol'"/on:

(i! ? √ 18@9d9 let U 18 @9

then du @4 d9 8du d9 @

therefore ?√ 18@9 d9 ?u1Y2(8du! 8 \ ?u1Y2du 8 1 2u:Y2P 8 1(18@9!:Y2 = C @ : P

NOUN @2




(;! ? & √ @9 = 1 8 √ :9! d9 (1! ? 92 = 92 = 29 d9 92

(11! ? (1 J G9!1YGd9 (12! ? (B9 J 2!1YBd9,olved the differential e3uation at the specified pointsA

(1:! d#Yd9 92 J 1 # 9 1 9@

(1@! d#Yd9 1 # 24 9 1√ 1=D9

(1B! d#Yd9 (1 J @9!1Y@ # 14 9 8:

(1! d#Yd9 c√ 1 J 92!d9 # 4 9 1

(1D! Find the total profit of a product if the arginal profit is given asdpYd9 9@= 92 (√ 189:!where E(!

(22! ,olve d#Yd9 2√ 1=#2 if 9 14 # 1 #

(1;! ,olve d#Yd9 92Y#: if 9 4 # 1

(2! ,olve dsYdt (t2=1!2 when , 4 t

UNIT

INTEGRATION OF TRANSCEDENTAL FUNCTIONS

1$ INT%O&UCTION

2$ O-.+CTI/+,

NOUN @G




In the a"ove u is a differentia"le function of 9 and u> for all values of 9 inthe specified doain$E,!3le: Find ?G9 d9

2981

Sol'"/onA let u 92814 du 29 d9 thendu 9 d9

2therefore

? G9 d9 G ? du9281 2

@?du @YuYu?=C 9

E,!3le: Find ?92 8 d9 1=:9:

let u 1=:9:4 du ;92d9 92d9 duY;?92 d9 ?du 1 ?du 1:9

: ; ; u u

1lnYuY = C 1lnY1=:9:Y=C ; ;

E,!3le: Find

?G9 : J 2 d9 9@89=1

let u 9@ J 9 = 14 du (@9: J 1! dv

"ut (G9: J 2!d9 2(@9:81! d9

thereforeA ? (G9: J 2!d9 ?2(@9: 81!d9 ?2du

9@

89=1 9@

89=1 u

2lnYuY=C

2lnY9@ J 9 = 1Y = C

E,!3le: Find ? (1 J 1 ! d9 (9=1 9=2!

NOUN B




let u 9 = 1 and v 9 = 2du d9 du d9

? (1 J 1 ! d9 ?d9 J ?d9 ?du J ?dv

(9=1 9=2! 9=1 9=2 u v

lnYuY 8 lnYvY = C

lnY9=1Y 8 lnY9=2Y = C

E,!3le: Find ?log(9=1!d9 9 = 1

let u log (9=1! du 1 d9 9=1

thereforeA (9=1! du d9 ?log(9=1!d9 ?u$ $(9=1!du 9 = 1 9=1

?udu Z u2 = C

Z log (9=1!2 = CEe5c/6e: +valuate the following integrals

(1! ?d9 (2! ?: d9:8@9 98B

(:! ?9 d9 (@! ?log9 du 9282 9

&'( ?@9 J 2 d992 J 9=1

An6: (1! 81 lnY(:8@9! Y=C (2! :lnY(98B! = C @

(:! Z lnY9282Y = C (@! Z log92 = C

(B! 2lnY92 J 9 = 1Y

The ethod adopted a"ove is to differentiate the denoinator and check if itis a factor of the nuerator if so with appropriate alge"raic anipulation4 thederivative of the denoinator will "e ade to look like the nuerator$ Thisethod was used in UNIT :$

NOUN B1




i$e$ ?g(9!d9 let u E(9! E(9!

and du E1(9!d9 g(9!d9

then ?g(9! ?du lnYuY = C E(v! u

*nYE(9! = C

.%-%2 THE INTEGRAL ? ed'

%ecall that deu de u $ du eu du d9 du d9 d9

then de u eudu d9 d9

deu eudu

then ?deu ?eudu

therefore ?eudu eu = C

E,!3le: Find ! e89d9$

*et u 894 du 1d9

d9 8du

therefore ?e89d9 ?e(8du! 8?eudu

8eu = C e89 C$

E,!3le: Find ?e29d9$ *et u 29 du 2d9

d9 du 2

therefore ?e29d9 ?eu(du! Z ?eudu 2 Z e29 = C

E,!3le: Find ?e9Y:d9 let u 94 du d9

NOUN B2




: : d9 :du4 ! e9Y: ?e9Y:du ?eu$(:du!

?e9Y:d9 :! eudu :eu C :e9Y: = C

E,!3le: ?@e29d9 *et U e29 du 2e29d9$ ?@e29du 2?2e29d9 2?du 2u = C

2e29 = C

E,!3le: ? (e9 9!2 (e9 1!d9*et u e9 = 9 du (e9 1!d9

? (e9 = 9!2 (e9 = 1! d9 ?u2du U: = C (e 9 = 9!: = C

: :

E,!3leA ?9e92d9 *et u 92 du 29 d9

du 9 d9 2

then ?9e92 d9 Z ! eu du Z eu = C Z e92 = C

Ee5c/6e: +valuate the following integrals(1! ?e:9d9 (2! ?es9

2

(:! ?Ge@9 d9 (@! ? (e9 J 9!2 (e9 J 1! d9

(B! ?:92 e9:

An6: (1! 1Y:e:9 = C (2! 2eB9 = C (:! 2e@9 = CB 2

(B! e9: = C

.%1 INTEGRATION OF TRIGONOMETRIC FUNCTIONS

%ecall fro UNIT G of the first course on calculus that for an# differentia"lefunction U of that

NOUN B:




1arc tan u = C a a 1 arc tan (9 = 2! = C a a

E,!3le: Find ?d9 √ a2 = (981!2

*et u 981 du d9 therefore ?d9 arc sin u = C √ a2=u2 a

arc sin 9 J 1 = C a

Ee5c/6e6: Find the following integralsA

(i! ? d9 (ii! ? d91 = @92 √ ; J @92

(iii! ? d9 (iv! ? d9@; = (9=2!2 √ 2B J ;92

(v! ?B d9 2B=92

An6: (i! \ arc tan 9 (ii! arc sin G9

G :(iii! 1YD art tan 9=2 (iv! arc sin :9 D B

(v! Y2

%- CONCLUSION

In this unit #ou have derived the forula for coon rational functions andhow to find their integrals$ )ou studied how to derive the integrationforula of trigonoetric functions$ +valuation and trigonoetric functions

were treated$ )ou also find the integrals of special functions using theinverse functions of sin 9 and tan 9$ The forulas derived in this unit will "eused to stud# ethods and techni3ues of integration which will "e studied inthe ne9t unit of this course$

4%- SUMMARY:

NOUN BG




1$ d(cot u! 8csc2 u du 1$ ? csc2 u du 8cot u = C

11$ d(sec u! sec u tan u du 11$ ? sec u tan u du sec u = C

12$ d(csc u! 8csc u cot u du 12$ ? csc u cot u du 8csc u = C

1:$ d(sin81 u! du 1:$ ? du Vsin81 u = C √ 1 J u2 and ? 1 J u2 V8cos81 u = C

1@$ d(cos81 u! du 1@$ √ 1 J u2

1B$ d(tan81 u! du 1B$ ? du Vtan81 u = C 1 = u2 and ? 1 = u2 V8cot81 u = C

1$ d(cot81 u! 8du 1$ 1= u2

1D$ d(sec81 u! du 1D$ YuY √ u28 1 and ? du Vsec81YuY = C

? u√ u2 J 1 V 8csc81YuY = C

1G$ d(csc81 u! 8du 1G$ YuY u2 8 1

.%1 INTEGRATION INVOLVING PO$ERS OF TRIGONOMETRICFUNCTIONS

Fro the a"ove "asic forula #ou have thatA

(1! ?un du u n=1 = C for n ≠ –1n = 1

and

(2! ?un du lnYuY=C n 1

This could "e used to evaluate integrals involving powers of trigonoetricfunctions$

E,!3le: Find ? sinn a9 cos a9 d9 *et u sin a9 du acos a9 d9 then du cos a9 d94 un sinn a9

athereforeA ? sinn a9 cos a9 d9 ? un du

a nn = 1 = C

NOUN 2




cosec2 9 cot2 9 = 1

E,!3le: ? cotB 9 cosec@ 9 d9

?cotB 9 cosec2 9 cosec2 9 d9

? cotB 9 (cot2 9 = 1! cosec2 9 d9 ? cotD cosec2 9 d9 = ? cotB 9 cosec2 9 d9(u cot 9 du 8cosec2 9 d9!$ 8cot G 9 8 8cot 9 = C

G

In siilar anner when and n are "oth odd #ou have cot 9 cose9n 9 cot81 9 cosecn81 9 cosec 9 cot 9 and then e9press cot81 9 in ters of cosec2

9 using cot2 9 cosec2 9 J 1

E,!3le: ? cotB 9 cosec: 9 d9

? cot@ 9 cosec2 9 cosec 9 cot 9 d9 ? (cosec2 9 J 1!2 cosec2 9 cosec 9 cot 9 d9 ? (cosec 9 J 2 cosec@ 9 = cosec2 9! cosec 9 cot 9 d9

u cosec 9 du 8 cosec 9 cot 9 d9

? (u J 2u@ = u2! (8du! 8uD = 2uB J u: = C

D B : 8cosecD 9 = 2cosecB 9 J cosec: 9 = C

D B :Ee5c/6e6 Find

1$ ? cot: 9 cosec: 9 d92$ ? cot: 9 cosec2 9 d9:$ ? tanB 9 sec2 9 d9

An6:

1$ 8cosecB 9 = cosec: 9 = C

B :2$ 8cot@ 9 = C @

:$ tan 9 = C

%- CONCLUSION

NOUN D1




In this unit4 #ou have reviewed differential forulas and their correspondingintegrals$ These "asic forulas will "e used throughout the reaining part ofthe course$ )ou have developed techni3ues of finding integrals of powers oftrigonoetric functions "# using the trigonoetric identities

(i! cos2 9 = sin2 9 1 and(ii! 1 = tan2 9 sec2 9 etc$

)ou have also studied how to evaluate the products of even powers of sinesand cosines functions$ These integrals will "e used when developing othertechni3ues of integration in the ne9t unit of this course$

4%- SUMMARYA

)ou have studied in the unit how to

(i! recall "asic differential forulas and corresponding integrals(ii! use these "asic forulas to develop techni3ues of integration of

powers of trigonoetric function(iii! evaluate the integrals of odd powers of trigonoetric function such as

? sinn 9 d94 ? cosn 9 d9(iv! evaluate the integrals of trigonoetric function such as ? tann 9 d94 ?

cotn 9 d9 where n is odd or even(v! evaluate the integrals of even powers of sec 9 and cosec 9(vi! evaluate the integrals of products of even powers of sin 9 and cos 9

such as ? cosn 9 d94 ? sinn 9 d94 ? cosn 9 d9 sin 9 d9 where n or iseven or "oth are even$

9%- TUTOR MAR)ED ASSIGNMENT

(1! Find ? sin2 9 cos2 9 d9(2! ,how that ? tan a9 d9 1 lnYcosa9Y=C

a(:! Find ? sin: @9 d9(@! Find ? tanB 9 sec: 9 d9(B! ,how that ? sec2n 9 d9 ?(1=u2!n81 du where u tan 9(! Find ? cos2Y: 9 sinB 9 d9(D! Find ? sin2 9 cosB 9 d9(G! Find ? sin @9 cos2 9 d9(;! Find ? tan 9 d9(1! Find ? tanB 9 sec@ 9 d9

UNIT 9

NOUN D2




FURTHER TECHNI?UES OF INTEGRATION I

TA#LE OF CONTENTS

1$ INT%O&UCTION

2$ O-.+CTI/+,:$ INT+%*, IN/O*/IN √ a2±u24 and √ u2≠a2

:$1 INT+%*, -) COMEUTIN TH+ ,[U%+ OF a92 = "9 = C@$ CONC*U,IONB$ ,UMM%)$ TUTO% M%0+& ,,INM+NT,D$ FU%TH+% %+&IN,

1%- INTRODUCTION

In continuation of developent of skills in techni3ues for finding integrals ofspecial functions4 #ou will stud# in the unit how to evaluate integralsinvolving rational function with √ a28u24 √ a2 = u24 √ u28 a2 and a2 J u2 asdenoinators$ The use of inverse trigonoetric functions or trigonoetricidentifies will "e needed$ In this unit the ethod or process used in derivingforulas for integrals of functions in the previous unit will "e adopted$

2%- O#&ECTIVES

In this unit #ou should "e a"le to correctl#

(i! recall "asic differential forulas and corresponding integrals as statedin UNIT B$(ii! Use these "asic forulas to evaluate integrals involving √ a28u24

a2 = u24 u24 8 a24 and a2 J u2

(iii! evaluate integrals of rational functions with a92 = " 9 = C a sdenoinator$

.%- INTEGRALS INVOLVING √ ,2 ≠'2 ,nd ,2 ≠'2

%ecall that d (arc tan u! 1 du

d9 1 = u2 d9

therefore ? d (arc tan u! ? 1 du d 1=u2

therefore arc tan u = C 1 J (I! 1=u2

NOUN D:




E,!3leA Find the integral of 1 i$e$ ? dua2 = u2 a2 = u2

To evaluate the a"ove #ou factor out a2 fro a2 = u2

i$e$ a2 = u2 a2 (1 = (u!2! a

let L u4 adL du a

then a2 = u2 a2(1 = L2!

thereforeA ? du 1 ? adL 1 ? dL a2=u2 a2 1=L2 a 1=L2

1 tan81 L a

thus ? du 1 arc tan u = C J II a2=u2 a a

E,!3le: Find ? du ; = u2

Sol'"/on ? du ? du 1 arc tan u = C

a=u (:!2

=u2

: :)ou have to review soe trigonoetric identities #ou studied in MTH 111$

E,!3le:

(i! 1 J sin2 9 cos2 9(ii! 1 = tan2 9 sec2 9 and(iii! sec2 9 J 1 tan2 9$

let u a sin 9

then u2 a2 sin2 9

Multipl#ing identit# (1! through "# a2 #ou getA

a2(18sin2n! a2 cos2 9$ J (i!

a2 J a2 sin2 9 a2 cos2 9$ J (ii!

NOUN D@




i$e$ ? sec 9 (tan 9 = sec 9! d9 lnYssec 9 = tan 9Y = C1

tan 9 = sec 9

Hence ±? sec 9 d9 lnYsec = tan 9Y = C1

If #ou let 9 arc tan u 9∈(8πY24 πY2!

then sec 9 will "e positive and ? du ? sec 9 d9 √ a2 = u2

lnYsec 9 = tan 9Y = C

recall that #ou let a2 = u2 a2 sec2 9 sec 9 √ a2 =u2

a

and tan 9 u a

Then? du lnY√ a2 =u2 = uY = C1

√ a2=u2 a a

let C C1 J ln a #ou then have that

? du lnY√ a2=u2 = uY = C$ I/

E,!3le: Find ? du√ 1 = u2

Sol'"/on: let a2

1 a @ then "# direct su"stitution into I/ #ou get

? du ln Y√ 1 = u2 = uY = C√ 1=u2

E,!3le: Find ? du luY>a> √ u2 J a2

Sol'"/on: )ou can start "# tr#ing the su"stitution u a sec 9then du a sec 9 tan 9 d9

"ut u2 J a2 a2 sec2 9 J a a2 (sec2 9 J 1! a2 tan2 9

)ou will then have that

? du ? a sec 9 tan 9 d9 ? a sec 9 tan 9 d9√ u28a2 √ a2 tan2 9 a tan 9

NOUN DD




Fro Unit B sec :$: #ou have that

? ; cos2 9 d9 ;Y2 ?(1 = cos 29! d9 ;Y2(9 = sin 29! = C

2

;Y2 9 = ; sin 29 = C @

thereforeA ? √ ; J u2 du ; arc sin u u √ ; J u2 = C 2 : 2

E,!3le: Find ? u2 du√ @ J u2

Sol'"/on: let u 2 sin 9 and J π < < π 2 2

du 2 cos 9 d9

@ J u2 @ J @ sin2 9 @ cos2 9

therefore ? u2 du ? @ sin2 9 2 cos 9 d9 ? @ sin2 9 d9 √ @8u2 √ @ cos2 9

Fro unit B sec :$: #ou have that

@ ? sin2 9 d9 @ ? (1 J cos 29! d9

2 29 J 2 sin 9 cos 9 = C 2(arc sin u J (@ J u2 ! = C

2 @E,!3le: Find ? d9

√ 1 J @92

Sol'"/on: let 29 sin u4 @92 sin2 u

2 d9 cos u du

1 J @92 1 J sin2 u cos2 u

thereforeA √ 18@92 √ cos2 u cos u

hence ? d9 Z ? cos u du Z ? du √ 18@92 cos u

Z u = C "ut u arc sin 29

NOUN D;




(iii! Z lnY√ ;92 = @ J 2 Y = C :9

(iv! 9 √ 2B J @92 = 2B arc sin 29 = C

2 @ B

(v! 89 √ ; J @92 = ; arc sin 29 = CG 1 :

(vi! \ arc sin @9 = C (vii! arc sin (√ 2 sin 9! = C 2

(viii! 9 √ 1 J 92 = 2 arc sin 9 = C G @

(i9! 81 lnYa = √ a2 = 92 Y = C a 9

(9! 8 \ √ 2B J @92

.%1 INTEGRAION #Y COMPLETING THE S?UARE OF ,2 = 0 = C

iven a 3uadratic function of this for f(9! a92 = "9 = C "# copleting thes3uare it can "e reduced to the for a(u2 = !

i$e$ a92 = "9 = C a(92 = "9! = C

a a(92 = "9 = " 2 ! = C J " 2 a @ @a

a(9 = "!2 = @ac J " 2 2a @a

if #ou let u 9t " and @98"2 2a @a2

then a92 = "9 = C a(u2 = !$

'hen the integral involves the s3uare root of a92 = "9 = C then #ou have toconsider onl# the case for which √ a(u2=! will have onl# real roots$

E,!3le: Find ? d9x√ 2=29

NOUN G1




(1! how to evaluate the following t#pes of integrals

(i! ? du (ii! ? dua2=u2 √ a2=v2

(iii! ? du (iv! ? du√ a28u2 √ u2 J a2

(2! how to evaluate integrals such asA

(i! ? d9 (ii! ? d9a92="9=C √ a92="9=C

using the ethod of copleting the s3uare$


+valuate the following integralsA

1$ ? d9 2$ ? 92 d9√ 29892=: √ 2B892

:$ ? √ @ J 92 d9 @$ ? du√ u2 J a2

B$ ? du YaY>YuY YuY>Y1Y>

$ ? du D$ ? d9u√ ;u2=@ 92=29=B

G$ ? d9 ;$ d9√ 928G9=:2 √ :928@9=1

1$ ? :9 = 1 √ 92=29=B

UNIT 7

FURTHER TECHNI?UES OF INTEGRATION II

NOUN G@




@92 J 2@9 = 11 : = 1 8 B(9=2B 9 J :!2 9=2 :8: (98:!2

therefore ? @92 J 2@9 = 11 d9 ? : du = ? d9 8 ? Bd9

(9=2!(98:!2 9=2 98: (98:!2

:lnY9=2Y=lnY98:Y= B = C (982!

E,!3le: Find ? 9 = 1:9:=12928298:

Sol'"/on: :9: = 1292 J 29 J : (@92 81!(29 = :!$

Then 9 = 1 = - = C(@9281!(29=:! 29 81 29=1 29=:$

9 = 9 (29=1!(29=:! = -(2981!(29 =:! = C(2981!(29=1!

-# e3uations coefficients and solving the resulting siultaneous e3uation#ou get

9 = 1 1 8 1 8 1(@9281!;29=:! 2(2981! 2(29=1! 29 = :

thereforeA ? 9 = 1 d9 Z ? d9 = Z ? d9 8 ? d9

:9:

=1292

8298: 2981 29=1 29=: \ lnY2981Y=1Y \ lnY29=1Y8 Z lnY29=:Y = C$

E,!3le: Find ? :92 d91 = 9:

Sol'"/on: :92 = -9 = C1 = 9: 1 = 9 1 J 9 = 92

Therefore :92 (1 J 9 = 92! = (-9 = C! (1 = 9! +3uations coefficient solving the resulting siultaneous e3uations #ields

:92 1 = 29 J 11=9: 1=9 189=92

thereforeA ? :92 d9 ? 1 d9 = ? 29 J 1 d91=9: 1=9 189=92

E,!3le: Find ? 9 = :

NOUN GD




E,!3le: ? 9e89 d9$

Sol'"/on: Use integration "# parts with u 94 du d94 dv e89 d94 v 8e89

Therefore ? 9c89 d9 89e89 8 ?8e89 d9 89e89 J c89 = c

In a"ove e9aple4 it is possi"le to choose u and v differentl#i$e$ ?9e89 d9 ?udv

u e894 du 8e89 d9 dv 9 d9 v 92 2

then integration "# parts #ou get

? 9e89 d9 92 e89 8 ? 8 89 92 d9 2 2

The a"ove is true "ut the resulting integral on the right is harder than thegiven one on the left$ Therefore4 #ou should "e cautious when factoring theintegrand into u and dv$ 'ith ore e9aples4 #ou get use to this techni3ueof integration "# parts$E,!3le6: Find ? ln 9 d9

Sol'"/on: u ln94 dv d94 du 1 d9 9

therefore ? ln d9 9ln9 8 ? 9$ 1 d9 $ 9

9ln9 J 9$ = c$

E,!3le: Find ? arc cos 9 d9

Sol'"/on: u arc cos 9 du 8 d9√ 1892

dv d94 v 9$

therefore

? arc cos 9 d9 uv 8?vdu 9 arc cos 9 8? 89 d9√ 1892

"ut ? 9 d9 8? u du where #2 1 J 92

√ 1892 #

8? d# 8# = c 8#du = 9 d9√ 1892 = C

NOUN ;




dv e89 d94 u 9: du :92dv v e9$

? 9:e89d9 uv 8 ? vdu 89:e898 ? 8e89 :92 d9

:?e89 92 d9 :uv 8 ?vduP

u 92 dv e89 d94 du 29 d94 v 8e89

:? e89 92 d9 : J92 e89 8: ?8e89 29 d9$

= ?e89 9 d9 uv 8 ? vduP

u 94 du d94 dv e89 d94 v 8e89

?e89 9 d9 89 e89 8 ? 8e89 d9 9e89 = e89 = C

thereforeA ? 9:e89 d9 89: e89 J :92 e89 J 9 e89 J e89 = C e89 (89: 8:92 J 9 8! = C

E,!3le: Find ? e9 sin 9 d9

Sol'"/onA let u e94 du e9 d9 du sin9 d9 v cos 9

*et I ? e9 sin 9 d9

therefore I uv 8 ? v du 8e9 cos 9 8 ? 8 cos9 e9 d9$

? cos 9 e9 d9 integrate "# parts again "# letting u e9 du e9 d94 dvcos9 d9$

v sin 9then ? cos 9 e9 d9 e9 sin 9 8 ? sin 9 e9 d9$

Therefore I 8e9 cos 9 = e9 sin 9 J I (since I ? sin9 e2 d9! 2I 8e9 cos 9 = e9 sin 9$

therefore I e9 (sin 9 J cos 9!

2 ? sin9 e9 d9 e9 (sin 9 J cos 9 ! = C

2

Ee5c/6e: +valuate the following integrals

(1! ? 9e9 d9 (2! ? 9 cos 9 d9

(:! ? 92 e29 d9 (@! ? 92 e89 d9

(B! ? 9e29 d9 (! ? ln (92 = 1! d9

NOUN ;:




In this unit #ou have studied

(i! the techni3ue of integration "# partial fraction$

i$e$ ? f(9! d9 ? 1 d9 = ? 2 d9 = ? n d9 g(9! g1(9! g2(9! gn(9!where g(9! g1(9! g2(9! ^ gn(9! and the integrals on the left and aresipler than the given integral on the left$

(ii! how to integrate product of function such as 9e94 e9 sin 9 9 ln9 etc "#the techni3ue of integration "# parts$

i$e$ ? u dv uv 8 ? v duwhere the integral on the left is sipler than the given integral on theleft$

(iii! how to appl# the ethod of integration "# parts two or ore ties$


Find the following integrals$

1$ ? 1 d9 (2! ? B9 J : d9 92 J @ (9=1! (98:!

:$ ? 8 2 9 = @ (@! ? 292

= : d9(92=1!(981!2 9(981!2

(B! ? 92 = 1 d9 (! ? arc tan 9 d9 9: J@92=9=

(D! ? 9:e9 d9 (G! ? 92 cos a9 d9

(;! ? sin (ln9! d9 (1! ?2 9 sin a9 d9

1

UNIT 8

FURTHER TECHNI?UES OF INTEGRATION III

TA#LE OF CONTENTS

NOUN ;B




. I4 = B sin: 9 cos 9 = : .2P @ @

I = B I2 = : I: = Z .oP @

I1 (sin 9! B cos 9 YIIY2 (sin IIY2 !B cos IIY2 J (sin o!B cos o

I2 4 I:

hence. B (¾!(1Y2 .o! B (¾!( Z !(9!YIIY2

B $ : $ Z $ II $ BII

@ 2 :2Ee5c/6e6: Find (1! ? sinB9 d94 (2! ?II sinD 9 d9

(:! ? 9Be9 d9 ( @! ?1 9De9 d9

(B! ? cos: 9 d9

An6:

(1! e9 (9B J B9@ = 2 9: J 9@ = 12 9 J 12!

(1! 8 1 cos 9 (: sin@ 9 = @ sin2 9 = G! = C 1B

(2! :2:B

(:! e9 (9B J B9@ = 29: J 9@ = 129 J 12! = C

(@! B@ J 1GB@e1

(B! 1 sin 9 (cos2 9 = 2! = C:

E,!3le: O"tain a reduction forula for ? ea9 cos "9 d9$

Sol'"/on: *et U ea9 and dv cos "9 d9Then du aea9 d9 and v 1 sin "9

-

Hence ?ea9 cos "9 d9 ea9 sin "9 8 ? 1 sin "9$aea9 d9

NOUN ;;




Sol'"/on: let a Z " 2Y: then "# the reduction forula

?ea9 cos "9 d9 ea9 (" sin "9 = a cos "9! = C a2 = "2

#ou have that

? e9Y2 cos 2Y: 9 d9 e9Y2 (2Y: 9 = Z cos 2Y: 9! \ = @Y;

: e9Y2 (2Y: sin 2Y:9 = Z cos 2Y:9! = C 2B

.%1 RATIONAL E(PRESSIONS IN SIN ( AND COS (

There are certain class of trigonoetric functions that techni3ues studied inthe previous ight not "e a"le to "e used to integrate the specificall#4rational functions of sin9 and cos 9$ n appropriate su"stitution of u tan 9Y2 ight reduce the pro"le of integrating such class of rational functions ofsin9 and co9 to a pro"le of integrating a rational function of u$ This in turncan "e integrated "# the ethod of partial fraction studied in unit D$

E,!3le: If f(9! is a rational e9pression in sin9 and cos94 then thesu"stitution u tan 9Y2 transfors the integral ? f(9! d9 into the integral of a

rational function of u$Sol'"/on: t#pical wa# to e9plain the a"ove is to start "# e9pressing cos9and sin9 in ters of u$

i$e$ cos 9 2 cos2 9 8 1 2 82 sec2(9!

2

2 8 1 2 8 1 = tan2(9! 1 = u

2 1 8 u2

1 = u2

thereforeA cos 9 1 8 u2 4 u tan 9 1 = u2 2

and sin 9 2 sin 9 cos 9 2 sin 9Y2 $ cos2 9

NOUN 11




G 12 2

Ee5c/6e6: +valuate the following integralsA

(i! ?II sin 29 sin B9 d9 (ii! ? cos B9 cos 9 d9II

(iii! ?II cos B9 sin 9 d9o

.%. OTHER RATIONALISING SU#STITUTIONS

)ou will stud# how to integrate rational functions f(9! involving fractional powers of 9$ This #ou will carr# out "# using the su"stitution 9 un$

E,!3le: Find ? d91 = 9

Sol'"/on: set u2 9 2u du d9 where u √ 9 then? d9 ? 2 u du 8 2? u du1 = √ 9 1 = u 1 = u

2 ?(1 8 1 ! du 1=u

2(u J ln 11 = uY = C

2u J 2lnY1 = uY = C

2√ 9 J 2lnY1 = √ 9Y = C

E,!3le: Find ? d 91 = @

x√

Sol'"/on: ,et u@ 9 @u: du d9

Then ? d9 ? @u: du @ ? u: du

1 = @9 1 = u 1 = u

@ u: J 2u2 = @u J @lnY1 = uY = C :

@ 9:Y@ J 291Y2 = @91Y@ J @Y1 = 91Y@Y = C

NOUN 1




81 9

: 82

F/ B%.

,tep 2A Eoint of intersection is given as 92 J 2 29 = 1 92 J 29 J :

(9 J :!(9 = 1! 9 : or J 1

,tep :A The line # 29 =1 is a"ove the para"ola # 92 J 2 "# an aount29 = 1 J (92 J 2! 29 = : J 92

Therefore the area is given as ?:(29 = : J 92! d9 :2 s3 units 81 :

)ou will now e9tend the a"ove to finding areas "etween curves that are crossed$Consider two f(9! and g(9! shown in Fig ;$@

#

f(9! " 9 a 91 92

g(9!

F/ B%

In fig ;$@ neither f(9! or g(9! reains positive4 i$e$ f(9! > g(9! a1 91P and 924 "P while g(9! > f(9! for 914 92P$ Then the area is given as ?9

1(f(9! Ja

g(9!! d9 = ?92(g(9! J f(9!! d9 = ? "(f(9! J g(9! d9 ?9Yf(9! J g(9!Yd9 = ?9

2Yf(9!8 g(9!Yd991 9

2 a 91

NOUN 112




81

F/ B%7

1

The area is given "# the definite integral ? (8:#2 = @ J #2! d# 81

1

? (@ J @#2! d# 1 81 :

E,!3le: Find the area "etween

9 \ #: # 824 1P and9 Z #: = \ #2 J Z # # 824 1P

Sol'"/on: The region "ounded "# the two curves is disploa#ed in Fig$ ;$G$

# 2

1 9 \ #2

# 1 #

82 1 2 @

NOUN 11B




9 Z #: = \ #2 8 Z #

82

# 82

F/ B%8

The two curves eet at where \ #: Z #: = \ #2 J Z # 9 @

#: 2#: = #2 8 2# #: = #2 J 2# # (#2 = # J 2! # (# J 1!(# = 2! # 4 # 14 # 82

therefore area is given as 1

? Y Z #: = \ #2 J Z # J (1Y@ #:!Y d92

? Y Z #: = \ #2 J Z # J (1Y@ #:!Y d9 82

1

= ? Y Z #:

= \ #2

J Z 8 ( \ #:

!Y d9

?( \ #: J \ # #2 = Z #! d9 82

1 = ? (8 \ #: J \ #2 = Z #! d9

1

2 = B :D : @G @G

Ee5c/6e6: ,ketch the region that is "ounded "# the following curves and

find the area$

1$ # 92 4 # @9 8 :

2$ Z #2 94 9 @ = #

:$ 9 = #2 J @ 4 9 J 2#

NOUN 11




,tep 2A *et 9 denote the distance of a ring fro the centre as shown in Fig1$1 here ≤9 ≤ r$

,tep :A The length of an# ring is given as 29 and the width is d9$ Thus thearea of a t#pical ring is given as 2m9 d9$ The area of the circle is therefore

given as the su of areas of the concentric rings$ r

i$e$ ? 2m9 d9$

92Yr r 2 which is the area of a circle with radius r$

Vol'!e o; Reol'"/on: The volues of an# solids can "e found "# theethod of slicing descri"d a"ove$ -efore #ou continue it is necessar# todefine what is eant "# solid of revolution solid which has a central a9isof s#etric is called a solid of revolution$ Most solid that will "e considerin this unit are solid of revolution$ For e9aple4 a cone4 a c#linder4 a "ucketetc$ To find the volue of such solid displa#ed in fig$ 1$1 #ou irst considerthe area under the region - of the curve # f(9! revolved a"out the 9 8 a9is

y = f(x)

d v

a b

A

x0

y

F/ 1-%1

through :$ +ach point on the curve represents a circle with centre on the98a9is$ solid of revolution as displa#ed in Fig 1$1 hs two circular planeends cutting the 98a9is at 9 a at 9 "$ s ws done in the e9aple of thearea of a circle$ *et v "e the volue of the solid of revolution fro 9 a upto an# value (a4 "! see Fig 1$2

#

# = d#

NOUN 12@




9 9 = d9 9 a

d9

F/ 1-%2 ,ow/n , c5o66 6ec"/on o; 6ol/d o; 5eol'"/on

,uppose thee is an increent in 9 i$e$ d9 with a corresponding increent in #4d# and increent in /4 dv$ In fig 1$2 he volue of the slice with thicknessd9 is given "# dv and is enclosed "etween two c#linders of outer radius # d# and inner radius #$

Thus #2 s9 sv ≤ (# = s#!2 d9

#2 ≤ sv ≤ (# = s#!2 (! s9

hence as s9 4 s9 and sv dv s9 d9

thus (! "ecoes

#2 ≤ dv ≤ #2 d9

dv #2 integrating "oth sides #ou have

/ ? #2 d9

where # f(9! a continuous function and / is the volue of solid generatedwhen the curve # f(9! is rotated through : around the 98a9i9 "etween theliits 9 a and 9 "$

E,!3le: The region in the 98# plane "ounded "# the curve # 924 the line 9 and 9 : and the 9 a9is is revolved a"out 9 J a9is$ 'hat is the

resulting volueb

Sol'"/on: ,ee Fig (1$:!

/ ?: #2 d9

# 92

NOUN 12B




Sol'"/on: The a9is of the cone is a line that passes through the verte9 of thecone and the centre of the "ase$

) [

E# #

9 t N 9

F/% 1-%4 h

/ ? #2 d9

To get #2 #ou consider Fig$ 1$B where a cross section gives #ou a picture oftwo siilar triangles [N and ∆ OET$ Thus

# r # r9 #2 r2 92 9 h h h2

h h

therefore / ? r2 92 d9 r2 9: P 1 r 2h h2 h2 :which is te volue a circular cone$

E,!3le: Find the volue of a sphere "# rotating the circle 92 t#2 r 2 a"outthe 98a9is$

Sol'"/on: Using the forula "

/ ? m#2 d9 $ )

since the centre of the circle is at origin # √ r 2 J 92 8r ≤ 9 r

then a r 4 " r r

/ ? (r 2 J 92! d9 r 292 8 9: Pr @ r :

8r : 8r :

Ee5c/6e6: ,ketch the graph and find the volue generated "# revolving theregion "elow it a"out he 98a9is$

NOUN 12D




1 a@ a: a @ @

E,!3le: Find the average value of f(9! √ r 2 J 92 8r4 rP

Sol'"/on: a r4 " r4 f(9! √ r 2 J 92

r

ThereforeA verage 1 ? (r 2 J 92! Z d9 r

Z r Z 9 √ r 2 J 92 = Z r 2 arc sin 9 P r

Z r Z r 2 P r @

%- CONCLUSION

In this unit #ou have studied how to find the work done when a force isapplied on an o"5ect along a straight line$ )ou have studied how to coputethe work done when a spring is copressed or stretched$ )ou have studiedhow to copute volues of a solid generated "# revolving a region along thea9is of s#etr# of the solid$ )ou have also studied how to find the averagevalue of a set of continuous data in a gien interval$

4%- SUMMARY: In this unit #ou have studied how toA

(i! copute the work done when a spring is copressed or stretched "

i$e$ ' ? f(9! d9 a

(ii! copute the volues of solid of revolution9" #"

i$e$ / ? #2 d9 or / ? 92 d# 9a #a

(iii! copute the average value of a function f(2! i$e$

"

verage f(2! 1 ? f(9! d9 a ≤9≤ " " J a a


NOUN 12;




1$ certain spring e9erts a force of $BN when stretched $:: "e#ondits natural length$ 'hat is the work done in stretching the spring$1 "e#ond its natural lengthb 'hat is the work done in stretchingit an additional $1b

2$ heispherical oil tank of radius 1 is "eing puped out$ Find thework done in lowering the oil level fro 2 "elow the top of the tankto @ "elow the top of the tank$ iven that the pup is placed righton top of the tank$ Take the weight of water wkg$

:$ The "ase of a solid is the region "etween the curves √ 9 = √ # 1 and # 1 J 9$ ,ketch the graphs and find the volue ofthe solid generated "# revolving the region a"out the 98a9is$

@$ Find the volue generated when the plane figure "ounded "# # Bsin 294 the 98a9is and the ordinates 9 and 9 mY@ rotates a"out the98a9is through a coplete revolution$

B$ ,uppose a superarket receives a consignent of 1@ satchets of pure water ever# : da#s$ The pure water is sold to retailers at astead# rate and 9 da#s after the consignent arrives4 the inventor#I(9! of satchets still on hand is I(9! 1@ J 1@ 9$ Find the averagedail# inventor#$

math 122 integral calculus

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