INTEGRALS. Chapter Introduction

Size: px
Start display at page:

Download "INTEGRALS. Chapter Introduction"

Transcription

1 INTEGRALS 87 hpter 7 INTEGRALS Just s mountineer clims mountin ecuse it is there, so goo mthemtics stuent stuies new mteril ecuse it is there. JAMES B. BRISTOL 7. Introuction Differentil lculus is centre on the concept of the erivtive. The originl motivtion for the erivtive ws the prolem of efining tngent lines to the grphs of functions n clculting the slope of such lines. Integrl lculus is motivte y the prolem of efining n clculting the re of the region oune y the grph of the functions. If function f is ifferentile in n intervl I, i.e., its erivtive f eists t ech point of I, then nturl question rises tht given f t ech point of I, cn we etermine the function? The functions tht coul possily hve given function s erivtive re clle nti erivtives (or primitive) of the function. Further, the formul tht gives G.W. Leinitz (66-76) ll these nti erivtives is clle the inefinite integrl of the function n such process of fining nti erivtives is clle integrtion. Such type of prolems rise in mny prcticl situtions. For instnce, if we know the instntneous velocity of n oject t ny instnt, then there rises nturl question, i.e., cn we etermine the position of the oject t ny instnt? There re severl such prcticl n theoreticl situtions where the process of integrtion is involve. The evelopment of integrl clculus rises out of the efforts of solving the prolems of the following types: () the prolem of fining function whenever its erivtive is given, () the prolem of fining the re oune y the grph of function uner certin conitions. These two prolems le to the two forms of the integrls, e.g., inefinite n efinite integrls, which together constitute the Integrl lculus.

2 88 MATHEMATIS There is connection, known s the Funmentl Theorem of lculus, etween inefinite integrl n efinite integrl which mkes the efinite integrl s prcticl tool for science n engineering. The efinite integrl is lso use to solve mny interesting prolems from vrious isciplines like economics, finnce n proility. In this hpter, we shll confine ourselves to the stuy of inefinite n efinite integrls n their elementry properties incluing some techniques of integrtion. 7. Integrtion s n Inverse Process of Differentition Integrtion is the inverse process of ifferentition. Inste of ifferentiting function, we re given the erivtive of function n ske to fin its primitive, i.e., the originl function. Such process is clle integrtion or nti ifferentition. Let us consier the following emples: We know tht (sin ) cos... () ( )... () n ( e ) e... () We oserve tht in (), the function cos is the erive function of sin. We sy tht sin is n nti erivtive (or n integrl) of cos. Similrly, in () n (), n e re the nti erivtives (or integrls) of n e, respectively. Agin, we note tht for ny rel numer, trete s constnt function, its erivtive is zero n hence, we cn write (), () n () s follows : (sin + ) cos, ( + ) n ( e + ) e Thus, nti erivtives (or integrls) of the ove cite functions re not unique. Actully, there eist infinitely mny nti erivtives of ech of these functions which cn e otine y choosing ritrrily from the set of rel numers. For this reson is customrily referre to s ritrry constnt. In fct, is the prmeter y vrying which one gets ifferent nti erivtives (or integrls) of the given function. More generlly, if there is function F such tht F() f (), I (intervl), then for ny ritrry rel numer, (lso clle constnt of integrtion) [ F( )+] f (), I

3 INTEGRALS 89 Thus, {F +, R} enotes fmily of nti erivtives of f. Remrk Functions with sme erivtives iffer y constnt. To show this, let g n h e two functions hving the sme erivtives on n intervl I. onsier the function f g h efine y f () g() h(), I f Then f g h giving f () g () h () I or f (), I y hypothesis, i.e., the rte of chnge of f with respect to is zero on I n hence f is constnt. In view of the ove remrk, it is justifie to infer tht the fmily {F +, R} provies ll possile nti erivtives of f. We introuce new symol, nmely, f () which will represent the entire clss of nti erivtives re s the inefinite integrl of f with respect to. Symoliclly, we write f ( ) F( ) +. y Nottion Given tht f (), we write y f (). For the ske of convenience, we mention elow the following symols/terms/phrses with their menings s given in the Tle (7.). Symols/Terms/Phrses f () f () in f () in f () Integrte An integrl of f Integrtion onstnt of Integrtion Tle 7. Mening Integrl of f with respect to Integrn Vrile of integrtion Fin the integrl A function F such tht F () f () The process of fining the integrl Any rel numer, consiere s constnt function

4 9 MATHEMATIS We lrey know the formule for the erivtives of mny importnt functions. From these formule, we cn write own immeitely the corresponing formule (referre to s stnr formule) for the integrls of these functions, s liste elow which will e use to fin integrls of other functions. Derivtives (i) n+ Integrls (Anti erivtives) n+ n n ; n+ +, n n+ Prticulrly, we note tht () ; + sin cos ; cos sin + cos sin ; sin cos + tn sec ; sec tn + cot cosec ; cosec cot + sec sec tn ; sec tn sec + cosec cosec cot ; cosec cot cosec + sin ; sin + (ii) ( ) (iii) ( ) (iv) ( ) (v) ( ) (vi) ( ) (vii) ( ) (viii) ( ) (i) ( cos ) () ( tn ) (i) ( cot ) ; + + ; ; cos + tn + + cot + +

5 INTEGRALS 9 (ii) ( sec ) (iii) ( cosec ) ; ; sec + cosec + (iv) ( e ) e ; e e + (v) ( log ) ; log + (vi) ; log + log Note In prctice, we normlly o not mention the intervl over which the vrious functions re efine. However, in ny specific prolem one hs to keep it in min. 7.. Geometricl interprettion of inefinite integrl Let f (). Then f () +. For ifferent vlues of, we get ifferent integrls. But these integrls re very similr geometriclly. Thus, y +, where is ritrry constnt, represents fmily of integrls. By ssigning ifferent vlues to, we get ifferent memers of the fmily. These together constitute the inefinite integrl. In this cse, ech integrl represents prol with its is long y-is. lerly, for, we otin y, prol with its verte on the origin. The curve y + for is otine y shifting the prol y one unit long y-is in positive irection. For, y is otine y shifting the prol y one unit long y-is in the negtive irection. Thus, for ech positive vlue of, ech prol of the fmily hs its verte on the positive sie of the y-is n for negtive vlues of, ech hs its verte long the negtive sie of the y-is. Some of these hve een shown in the Fig 7.. Let us consier the intersection of ll these prols y line. In the Fig 7., we hve tken >. The sme is true when <. If the line intersects the prols y, y +, y +, y, y t P, P, P, P, P etc., respectively, then y t these points equls. This inictes tht the tngents to the curves t these points re prllel. Thus, + F () (sy), implies tht

6 9 MATHEMATIS Fig 7. the tngents to ll the curves y F (), R, t the points of intersection of the curves y the line, ( R), re prllel. Further, the following eqution (sttement) f ( ) F ( ) + y (sy), represents fmily of curves. The ifferent vlues of will correspon to ifferent memers of this fmily n these memers cn e otine y shifting ny one of the curves prllel to itself. This is the geometricl interprettion of inefinite integrl. 7.. Some properties of inefinite integrl In this su section, we shll erive some properties of inefinite integrls. (I) The process of ifferentition n integrtion re inverses of ech other in the sense of the following results : f () f () n f () f () +, where is ny ritrry constnt.

7 INTEGRALS 9 Proof Let F e ny nti erivtive of f, i.e., F( ) f () Then f () F() + Therefore f () ( F( )+) Similrly, we note tht F( ) f( ) f () f () (II) n hence f () f () + where is ritrry constnt clle constnt of integrtion. Two inefinite integrls with the sme erivtive le to the sme fmily of curves n so they re equivlent. Proof Let f n g e two functions such tht f () () g or f () g (), where is ny rel numer (Why?) Hence f () g () or f () g () + So the fmilies of curves { f ( ) +, R} n { g ( ) +, R} re ienticl. re equivlent. Hence, in this sense, f() n g()

8 9 MATHEMATIS Note The equivlence of the fmilies { f ( ) + }, { g( ) +, } f R n R is customrily epresse y writing () g() without mentioning the prmeter. (III) [ ] f ()+ g() f () + g() Proof By Property (I), we hve, [ f ()+ g()] f () + g ()... () On the otherhn, we fin tht f () + () f () + g() g f () + g ()... () Thus, in view of Property (II), it follows y () n () tht ( () + ()) f g f () + g(). (IV) For ny rel numer k, k f () k f () (V) Proof By the Property (I), k f() k f(). Also k f () k f() k f() Therefore, using the Property (II), we hve k f () k f (). Properties (III) n (IV) cn e generlise to finite numer of functions f, f,..., f n n the rel numers, k, k,..., k n giving [ kf() + kf() knfn() ] k f() + k f () kn fn(). To fin n nti erivtive of given function, we serch intuitively for function whose erivtive is the given function. The serch for the requisite function for fining n nti erivtive is known s integrtion y the metho of inspection. We illustrte it through some emples.

9 INTEGRALS 95 Emple Write n nti erivtive for ech of the following functions using the metho of inspection: (i) cos (ii) + (iii) Solution (i), We look for function whose erivtive is cos. Recll tht sin cos or cos (sin ) sin (ii) (iii) Therefore, n nti erivtive of cos is sin. We look for function whose erivtive is +. Note tht ( ) + +. Therefore, n nti erivtive of + is +. We know tht (log ), > n [log ( )] ( ),< omining ove, we get ( ) log, Therefore, log is one of the nti erivtives of. Emple Fin the following integrls: (i) (ii) ( + ) (iii) ( + e ) Solution (i) We hve (y Property V)

10 96 MATHEMATIS ;, re constnts of integrtion , where is nother constnt of integrtion. Note From now onwrs, we shll write only one constnt of integrtion in the finl nswer. (ii) We hve ( + ) + (iii) We hve ( + e ) + e + + e log e log + 5 Emple Fin the following integrls: (i) (sin + cos ) (ii) cosec (cosec + cot ) (iii) sin cos Solution (i) We hve (sin + cos ) sin + cos cos + sin + 5

11 INTEGRALS 97 (ii) (iii) We hve + (cosec (cosec + cot ) cosec cosec cot We hve sin sin cos cos cos cot cosec + sec tn sec tn sec + Emple Fin the nti erivtive F of f efine y f () 6, where F () Solution One nti erivtive of f () is 6 since ( 6 ) 6 Therefore, the nti erivtive F is given y F() 6 +, where is constnt. Given tht F(), which gives, 6 + or Hence, the require nti erivtive is the unique function F efine y Remrks (i) (ii) F() 6 +. We see tht if F is n nti erivtive of f, then so is F +, where is ny constnt. Thus, if we know one nti erivtive F of function f, we cn write own n infinite numer of nti erivtives of f y ing ny constnt to F epresse y F() +, R. In pplictions, it is often necessry to stisfy n itionl conition which then etermines specific vlue of giving unique nti erivtive of the given function. Sometimes, F is not epressile in terms of elementry functions viz., polynomil, logrithmic, eponentil, trigonometric functions n their inverses etc. We re therefore locke for fining f (). For emple, it is not possile to fin e y inspection since we cn not fin function whose erivtive is e

12 98 MATHEMATIS (iii) When the vrile of integrtion is enote y vrile other thn, the integrl formule re moifie ccoringly. For instnce + y 5 y y + y omprison etween ifferentition n integrtion. Both re opertions on functions.. Both stisfy the property of linerity, i.e., k f () + k f () k f () + k f () (i) [ ] (ii) [ + ] + k f () k f () k f () k f () Here k n k re constnts.. We hve lrey seen tht ll functions re not ifferentile. Similrly, ll functions re not integrle. We will lern more out nonifferentile functions n nonintegrle functions in higher clsses.. The erivtive of function, when it eists, is unique function. The integrl of function is not so. However, they re unique upto n itive constnt, i.e., ny two integrls of function iffer y constnt. 5. When polynomil function P is ifferentite, the result is polynomil whose egree is less thn the egree of P. When polynomil function P is integrte, the result is polynomil whose egree is more thn tht of P. 6. We cn spek of the erivtive t point. We never spek of the integrl t point, we spek of the integrl of function over n intervl on which the integrl is efine s will e seen in Section The erivtive of function hs geometricl mening, nmely, the slope of the tngent to the corresponing curve t point. Similrly, the inefinite integrl of function represents geometriclly, fmily of curves plce prllel to ech other hving prllel tngents t the points of intersection of the curves of the fmily with the lines orthogonl (perpeniculr) to the is representing the vrile of integrtion. 8. The erivtive is use for fining some physicl quntities like the velocity of moving prticle, when the istnce trverse t ny time t is known. Similrly, the integrl is use in clculting the istnce trverse when the velocity t time t is known. 9. Differentition is process involving limits. So is integrtion, s will e seen in Section 7.7.

13 INTEGRALS 99. The process of ifferentition n integrtion re inverses of ech other s iscusse in Section 7.. (i). EXERISE 7. Fin n nti erivtive (or integrl) of the following functions y the metho of inspection.. sin. cos. e. ( + ) 5. sin e Fin the following integrls in Eercises 6 to : 6. ( e + ) 7. ( ) 8. ( + + c) ( + e ) ( ) ( + + ) 6. ( cos + e ) ( sin + 5 ) 8. sec (sec + tn ) sec. sin 9. cosec. cos hoose the correct nswer in Eercises n.. The nti erivtive of + equls (A) + + (B) + + () + + (D) + +. If f ( ) such tht f (). Then f () is 9 (A) + 8 (B) () (D) 9 + 8

14 MATHEMATIS 7. Methos of Integrtion In previous section, we iscusse integrls of those functions which were reily otinle from erivtives of some functions. It ws se on inspection, i.e., on the serch of function F whose erivtive is f which le us to the integrl of f. However, this metho, which epens on inspection, is not very suitle for mny functions. Hence, we nee to evelop itionl techniques or methos for fining the integrls y reucing them into stnr forms. Prominent mong them re methos se on:. Integrtion y Sustitution. Integrtion using Prtil Frctions. Integrtion y Prts 7.. Integrtion y sustitution In this section, we consier the metho of integrtion y sustitution. The given integrl f () cn e trnsforme into nother form y chnging the inepenent vrile to t y sustituting g (t). onsier I f () Put g(t) so tht g (t). t We write g (t) t Thus I f ( ) f ( g( t)) g ( t) t This chnge of vrile formul is one of the importnt tools ville to us in the nme of integrtion y sustitution. It is often importnt to guess wht will e the useful sustitution. Usully, we mke sustitution for function whose erivtive lso occurs in the integrn s illustrte in the following emples. Emple 5 Integrte the following functions w.r.t. : (i) sin m (ii) sin ( + ) (iii) tn sec (iv) sin (tn ) + Solution (i) We know tht erivtive of m is m. Thus, we mke the sustitution m t so tht m t. Therefore, sin m sin t t m m cos t + cos m + m

15 (ii) INTEGRALS Derivtive of + is. Thus, we use the sustitution + t so tht t. Therefore, sin ( + ) sin t t cos t + cos ( + ) + (iii) Derivtive of is. Thus, we use the sustitution t so tht t giving t t. tn sec ttn t sec t t Thus, tn t sec t t t Agin, we mke nother sustitution tn t u so tht sec t t u 5 u Therefore, tn t sec t t u u + 5 tn 5 5 t+ (since u tn t) (iv) Hence, tn sec tn 5 + (since t ) 5 tn Alterntively, mke the sustitution tn t Derivtive of tn. Thus, we use the sustitution + tn t so tht t. + sin t t sin (tn ) Therefore, + cos t + cos(tn ) + Now, we iscuss some importnt integrls involving trigonometric functions n their stnr integrls using sustitution technique. These will e use lter without reference. (i) tn log sec + We hve sin tn cos

16 MATHEMATIS (ii) Put cos t so tht sin t Then tn t log t + log cos + t or tn log sec + cot log sin + cos We hve cot sin Put sin t so tht cos t t Then cot log t + log sin + t (iii) sec log sec + tn + We hve sec (sec + tn ) sec sec + tn Put sec + tn t so tht sec (tn + sec ) t Therefore, sec t log t + log sec + tn + t (iv) cosec log cosec cot + We hve cosec (cosec + cot ) cosec (cosec + cot ) Put cosec + cot t so tht cosec (cosec + cot ) t So cosec t log t log cosec + cot + t Emple 6 Fin the following integrls: (i) sin cos cosec cot log + cosec cot log cosec cot + (ii) sin (iii) sin ( + ) + tn

17 INTEGRALS Solution (i) We hve sin cos sin cos (sin ) ( cos ) cos (sin ) Put t cos so tht t sin Therefore, sin cos (sin ) ( t ) t t (ii) Put + t. Then t. Therefore sin sin ( t ) t sin ( + ) sin t 5 t t ( t t ) t cos + cos + 5 sin tcos cos tsin t sin t cos t sin cot t t (cos ) t (sin ) log sin t+ (cos ) ( + ) (sin ) log sin ( + ) cos cos (sin ) log sin ( ) sin (iii) sin Hence, sin ( + ) cos sin log sin ( + ) +, where, sin + cos, is nother ritrry constnt. cos + tn cos + sin (cos + sin + cos sin ) cos + sin

18 MATHEMATIS cos sin + cos + sin cos sin () cos + sin Now, consier I cos sin cos + sin Put cos + sin t so tht (cos sin ) t Therefore I t log t + log cos + sin + t Putting it in (), we get + + log cos sin + tn log cos + sin log cos + sin +, + EXERISE 7. Integrte the functions in Eercises to 7: ( log ) sin sin (cos ) 5. sin( + )cos( + ) + log (+ ) ( ) (+ ).. e + 7. e 6. +, > (log ) m, >, m

19 INTEGRALS 5 8. e tn + 9. e e +. e e e + e. tn ( ). sec (7 ). sin. cos sin 6cos + sin 7. sin cos cos ( tn ) cos + sin cos cot log sin. sin + cos. sin ( + cos ). + cot. tn ( ). tn sin cos ( ) 5. ( + log ) 6. ( + ) + log sin tn hoose the correct nswer in Eercises 8 n loge 8. equls + (A) + (B) + + () ( ) + (D) log ( + ) + 9. equls sin cos (A) tn + cot + (B) tn cot + () tn cot + (D) tn cot Integrtion using trigonometric ientities When the integrn involves some trigonometric functions, we use some known ientities to fin the integrl s illustrte through the following emple. Emple 7 Fin (i) cos (ii) sin cos (iii) sin

20 6 MATHEMATIS Solution (i) Recll the ientity cos cos, which gives cos + cos Therefore, cos ( + cos ) cos + + sin + (ii) Recll the ientity sin cos y [sin ( + y) + sin ( y)] (Why?) Then sin cos sin 5 sin (iii) cos 5 cos cos 5+ cos + From the ientity sin sin sin, we fin tht Therefore, Alterntively, sin sin sin sin sin sin cos + cos + sin sin sin Put cos t so tht sin t Therefore, sin ( t ) t ( cos ) sin t t+ t t t+ + cos + cos + Remrk It cn e shown using trigonometric ientities tht oth nswers re equivlent.

21 INTEGRALS 7 EXERISE 7. Fin the integrls of the functions in Eercises to :. sin ( + 5). sin cos. cos cos cos 6. sin ( + ) 5. sin cos 6. sin sin sin 7. sin sin 8 cos cos cos + cos. sin. cos.. cos cos α cos cosα. 6. tn sin cos. cos sin + sin sin + cos sin cos cos ( cos + sin ). cos ( ) cos ( ) hoose the correct nswer in Eercises n.. sin cos is equl to sin cos (A) tn + cot + () tn + cot + e ( + ). equls cos ( e) (A) cot (e ) + () tn (e ) + sin + cos 5. tn sec 8. cos + sin cos. sin (cos ) (B) tn + cosec + (D) tn + sec + (B) tn (e ) + (D) cot (e ) + 7. Integrls of Some Prticulr Functions In this section, we mention elow some importnt formule of integrls n pply them for integrting mny other relte stnr integrls: () log + +

22 8 MATHEMATIS + () log + () + tn + () log + + (5) sin + (6) + log We now prove the ove results: () We hve ( )( + ) ( + ) ( ) ( )( + ) + Therefore, + () In view of () ove, we hve log ( ) log ( + ) + [ ] log + + ( + ) + ( ) ( + )( ) + +

23 INTEGRALS 9 Therefore, + + [ log log ] log + Note The technique use in () will e epline in Section 7.5. () Put tn θ. Then sec θ θ. sec θθ Therefore, + tn θ+ tn θ + + θ () Let secθ. Then secθ tnθ θ. Therefore, secθ tnθθ secθ secθθ log sec θ + tn θ + log + + log + log + log (5) Let sinθ. Then cosθ θ. + +, where log Therefore, cosθθ sin θ θθ +sin + (6) Let tnθ. Then sec θ θ. Therefore, + sec θθ tn θ+ sec θθ log (secθ+ tn θ) +

24 MATHEMATIS log log + + log + log + + +, where log Applying these stnr formule, we now otin some more formule which re useful from pplictions point of view n cn e pplie irectly to evlute other integrls. (7) To fin the integrl, we write + + c + + c c c Now, put + tso tht t n writing c k ±. We fin the t c integrl reuce to the form epening upon the sign of t ± k n hence cn e evlute. (8) To fin the integrl of the type, proceeing s in (7), we + + c otin the integrl using the stnr formule. p+ q (9) To fin the integrl of the type, where p, q,,, c re + + c constnts, we re to fin rel numers A, B such tht p + q A ( + + c)+ba( + )+B To etermine A n B, we equte from oth sies the coefficients of n the constnt terms. A n B re thus otine n hence the integrl is reuce to one of the known forms.

25 INTEGRALS ( p+ q) () For the evlution of the integrl of the type + + c s in (9) n trnsform the integrl into known stnr forms. Let us illustrte the ove methos y some emples. Emple 8 Fin the following integrls: (i) (ii) 6 Solution, we procee (i) We hve 6 log [y 7. ()] (ii) ( ) Put t. Then t. Therefore, t t sin ( t ) + [y 7. (5)] sin ( ) + Emple 9 Fin the following integrls : (i) 6 + (ii) + (iii) 5 Solution (i) We hve ( ) + So, Let Therefore, ( ) 6 + t. Then t t tn t + [y 7. ()] t + tn +

26 MATHEMATIS (ii) The given integrl is of the form 7. (7). We write the enomintor of the integrn, (completing the squre) Thus Put + t. Then t. 6 Therefore, + t 7 t 6 7 t log t+ 66 [y 7. (i)] 7 + log log log + + log log +, where + log

27 INTEGRALS (iii) We hve Put t. Then t. 5 Therefore, 5 (completing the squre) t 5 5 t t log t log [y 7. ()] Emple Fin the following integrls: (i) + + (ii) Solution (i) Using the formul 7. (9), we epress + ( ) A B A (+ 6) + B Equting the coefficients of n the constnt terms from oth sies, we get A n 6A + B or A n B. Therefore, I+ I (sy)... ()

28 MATHEMATIS In I, put t, so tht ( + 6) t t Therefore, I log t + t log () n I Put + t, so tht t, we get t I t + tn t+ [y 7. ()] tn + + Using () n () in (), we get ( ) tn () + log tn ( ++ ) (ii) where, + This integrl is of the form given in 7. (). Let us epress + A (5 ) + B A ( ) + B Equting the coefficients of n the constnt terms from oth sies, we get A n A + B, i.e., A n B

29 INTEGRALS 5 Therefore, + 5 ( ) In I, put 5 t, so tht ( ) t. Therefore, I Now consier I Put + t, so tht t. Therefore, I Sustituting () n () in (), we otin I + I... () ( ) t 5 t t () 5 9 ( + ) t sin + [y 7. (5)] t t sin () 5 + sin +, where. EXERISE 7. Integrte the functions in Eercises to ( ) sec tn +

30 6 MATHEMATIS ( )( ) ( )( ) ( 5)( ) hoose the correct nswer in Eercises n 5.. equls + + (A) tn ( + ) + (B) tn ( + ) + () ( + ) tn + (D) tn + 5. equls 9 (A) 9 8 sin (B) 8 9 sin + 9 () 9 8 sin Integrtion y Prtil Frctions (D) 9 8 sin + 9 Recll tht rtionl function is efine s the rtio of two polynomils in the form P( ), where P () n Q() re polynomils in n Q(). If the egree of P() Q( ) is less thn the egree of Q(), then the rtionl function is clle proper, otherwise, it is clle improper. The improper rtionl functions cn e reuce to the proper rtionl

31 INTEGRALS 7 functions y long ivision process. Thus, if P( ) Q( ) is improper, then P( ) P( ) T( ) +, Q( ) Q( ) where T() is polynomil in n P( ) is proper rtionl function. As we know Q( ) how to integrte polynomils, the integrtion of ny rtionl function is reuce to the integrtion of proper rtionl function. The rtionl functions which we shll consier here for integrtion purposes will e those whose enomintors cn e fctorise into P( ) liner n qurtic fctors. Assume tht we wnt to evlute, where P( ) Q( ) Q( ) is proper rtionl function. It is lwys possile to write the integrn s sum of simpler rtionl functions y metho clle prtil frction ecomposition. After this, the integrtion cn e crrie out esily using the lrey known methos. The following Tle 7. inictes the types of simpler prtil frctions tht re to e ssocite with vrious kin of rtionl functions. Tle 7. S.No. Form of the rtionl function Form of the prtil frction. p+ q, ( )( ) A B + p+ q. ( ) A + B ( ). p + q + r A + B + ( )( )( c) c. p + q+ r ( ) ( ) A + B + ( ) 5. p + q+ r A B + + ( )( + + c) + + c, where + + c cnnot e fctorise further In the ove tle, A, B n re rel numers to e etermine suitly.

32 8 MATHEMATIS Emple Fin ( + ) ( + ) Solution The integrn is proper rtionl function. Therefore, y using the form of prtil frction [Tle 7. (i)], we write A + B ( + ) ( + ) + + where, rel numers A n B re to e etermine suitly. This gives A ( + ) + B ( + ). Equting the coefficients of n the constnt term, we get A + B n A + B Solving these equtions, we get A n B. Thus, the integrn is given y ( + ) ( + ) () Therefore, ( + ) ( + ) + + log + log ++ log Remrk The eqution () ove is n ientity, i.e. sttement true for ll (permissile) vlues of. Some uthors use the symol to inicte tht the sttement is n ientity n use the symol to inicte tht the sttement is n eqution, i.e., to inicte tht the sttement is true only for certin vlues of. Emple Fin Solution Here the integrn y n fin tht is not proper rtionl function, so we ivie

33 INTEGRALS ( ) ( ) 5 5 A B Let + ( ) ( ) So tht 5 5 A ( ) + B ( ) Equting the coefficients of n constnt terms on oth sies, we get A + B 5 n A + B 5. Solving these equtions, we get A 5 n B Thus, Therefore, Emple Fin ( + ) ( + ) log + log +. Solution The integrn is of the type s given in Tle 7. (). We write ( + ) ( + ) A + B + + ( + ) + So tht A ( + ) ( + ) + B ( + ) + ( + ) A ( + + ) + B ( + ) + ( + + ) ompring coefficient of, n constnt term on oth sies, we get A +, A + B + n A + B +. Solving these equtions, we get 5 A, B n. Thus the integrn is given y ( + ) ( + ) 5 ( + ) ( + ) ( + ) 5 Therefore, ( + ) ( + ) + ( + ) + 5 log + + log + + ( +) + 5 log ( + )

34 MATHEMATIS Emple Fin ( + ) ( + ) Solution onsier Then ( + ) ( + ) ( + ) ( + ) n put y. y ( y+ ) ( y+ ) Write y A + B ( y+ ) ( y+ ) y+ y+ So tht y A (y + ) + B (y + ) ompring coefficients of y n constnt terms on oth sies, we get A + B n A + B, which give A n B Thus, ( + ) ( + ) + ( + ) ( + ) Therefore, + ( + ) ( + ) + + tn + tn + tn + tn + In the ove emple, the sustitution ws me only for the prtil frction prt n not for the integrtion prt. Now, we consier n emple, where the integrtion involves comintion of the sustitution metho n the prtil frction metho. ( sinφ ) cosφ Emple 5 Fin φ 5 cos φ sinφ Solution Let y sinφ Then y cosφ φ

35 INTEGRALS Therefore, ( φ ) sin cosφ φ 5 cos φ sinφ ( y ) y 5 ( y ) y y y y y+ y ( ) y I (sy) Now, we write Therefore, y ( y ) A B + y ( y ) y A (y ) + B [y Tle 7. ()] ompring the coefficients of y n constnt term, we get A n B A, which gives A n B. Therefore, the require integrl is given y y y I [ + ] y + y ( y ) y ( y ) log y + + y log sinφ + + sinφ log ( sin φ+ ) + sin φ (since, sinφ is lwys positive) Emple 6 Fin ++ ( + ) ( + ) Solution The integrn is proper rtionl function. Decompose the rtionl function into prtil frction [Tle.(5)]. Write ++ ( + ) ( + ) A B ( + ) Therefore, + + A ( + ) + (B + ) ( + )

36 MATHEMATIS Equting the coefficients of, n of constnt term of oth sies, we get A + B, B + n A +. Solving these equtions, we get A, B n Thus, the integrn is given y ++ ( + ) ( + ) ( + ) ( + ) 5 + Therefore, ( +) ( + ) log ++ log ++ tn EXERISE 7.5 Integrte the rtionl functions in Eercises to.. ( + ) ( + ). 9. ( )( )( ). ( )( )( ) ( ) 7. ( + ) ( ) 8. ( ) ( + ) ( ) (+ ). ( ) ( + ) ( + ) ( ). ( + ) n [Hint: multiply numertor n enomintor y ( + ) n n put n t ] 7. cos ( sin ) ( sin ) [Hint : Put sin t]

37 INTEGRALS 8. ( + ) ( + ) ( + ) ( + ) 9. ( + ) ( + ). ( ). [Hint : Put e ( e ) t] hoose the correct nswer in ech of the Eercises n... equls ( ) ( ) (A) () ( ) log + log equls ( + ) (A) + log log ( +) + (B) ( ) log + (D) log ( ) ( ) + (B) log + log ( +) + log log ( + +) + () log + log ( +) + (D) 7.6 Integrtion y Prts In this section, we escrie one more metho of integrtion, tht is foun quite useful in integrting proucts of functions. If u n v re ny two ifferentile functions of single vrile (sy). Then, y the prouct rule of ifferentition, we hve v u ( uv ) u + v Integrting oth sies, we get v u uv u + v or v u u uv v... () Let u f () n v g(). Then u f () n v () g

38 MATHEMATIS Therefore, epression () cn e rewritten s f () g() f () g () [ g () ] f () i.e., f() g () f () g () [ f () g () ] If we tke f s the first function n g s the secon function, then this formul my e stte s follows: The integrl of the prouct of two functions (first function) (integrl of the secon function) Integrl of [(ifferentil coefficient of the first function) (integrl of the secon function)] Emple 7 Fin cos Solution Put f () (first function) n g () cos (secon function). Then, integrtion y prts gives Suppose, we tke cos cos [ ( ) cos ] sin sin sin + cos + f () cos n g(). Then cos cos [ (cos ) ] ( cos ) sin + Thus, it shows tht the integrl cos is reuce to the comprtively more complicte integrl hving more power of. Therefore, the proper choice of the first function n the secon function is significnt. Remrks (i) It is worth mentioning tht integrtion y prts is not pplicle to prouct of functions in ll cses. For instnce, the metho oes not work for sin. The reson is tht there oes not eist ny function whose erivtive is sin. (ii) Oserve tht while fining the integrl of the secon function, we i not ny constnt of integrtion. If we write the integrl of the secon function cos

39 INTEGRALS 5 s sin + k, where k is ny constnt, then cos (sin + k ) (sin + k ) (sin + k) (sin k (sin + k) cos k+ sin + cos + This shows tht ing constnt to the integrl of the secon function is superfluous so fr s the finl result is concerne while pplying the metho of integrtion y prts. (iii) Usully, if ny function is power of or polynomil in, then we tke it s the first function. However, in cses where other function is inverse trigonometric function or logrithmic function, then we tke them s first function. Emple 8 Fin log Solution To strt with, we re unle to guess function whose erivtive is log. We tke log s the first function n the constnt function s the secon function. Then, the integrl of the secon function is. Hence, (log.) log [ (log ) ] (log ) log +. Emple 9 Fin e Solution Tke first function s n secon function s e. The integrl of the secon function is e. Therefore, Emple Fin sin e e e e e +. Solution Let first function e sin n secon function e First we fin the integrl of the secon function, i.e., Put t. Then t..

40 6 MATHEMATIS Therefore, Hence, t t t sin ( ) (sin ) ( ) sin ++ sin + Alterntively, this integrl cn lso e worke out y mking sustitution sin θ n then integrting y prts. Emple Fin e sin Solution Tke e s the first function n sin s secon function. Then, integrting y prts, we hve I e sin e ( cos ) + e cos e cos + I (sy)... () Tking e n cos s the first n secon functions, respectively, in I, we get I e sin e sin Sustituting the vlue of I in (), we get I e cos + e sin I or I e (sin cos ) e Hence, I e sin (sin cos ) + Alterntively, ove integrl cn lso e etermine y tking sin s the first function n e the secon function Integrl of the type e [ f ( ) + f ( )] +e f e f... () We hve I e [ f ()+ f ()] e f () + e f () I (),wherei () Tking f () n e s the first function n secon function, respectively, in I n integrting it y prts, we hve I f () e f () e + Sustituting I in (), we get I e f () f () e + e f () + e f () +

41 INTEGRALS 7 Thus, e [ f ( ) + f ( )] e f( ) + Emple Fin (i) Solution (i) We hve I ( +) e e (tn + ) (ii) + ( +) e (tn + ) + onsier f () tn, then f () + Thus, the given integrn is of the form e [ f () + f ()]. Therefore, I (tn + ) + e e tn + (ii) We hve ( + ) e I ( +) ++) e [ ] ( +) e [ ] e + ( + ) ( +) [ + ] + ( +) onsier f(), then f () + ( + ) Thus, the given integrn is of the form e [f () + f ()]. Therefore, + e e + ( + ) + EXERISE 7.6 Integrte the functions in Eercises to.. sin. sin. e. log 5. log 6. log 7. sin 8. tn cos 9. cos. (sin ).. tn. (log ) 5. ( + ) log. sec

42 8 MATHEMATIS e 6. e (sin + cos) 7. ( + ) 8. e + sin + cos 9. e ( ) e. ( ). e sin. sin + hoose the correct nswer in Eercises n.. e equls (A) () e + (B) e + (D). e sec (+ tn ) equls (A) e cos + () e sin Integrls of some more types e + e + (B) e sec + (D) e tn + Here, we iscuss some specil types of stnr integrls se on the technique of integrtion y prts : (i) (ii) + (i) Let I (iii) Tking constnt function s the secon function n integrting y prts, we hve I +

43 INTEGRALS 9 or I I or I log + + Similrly, integrting other two integrls y prts, tking constnt function s the secon function, we get (ii) log (iii) + sin + Alterntively, integrls (i), (ii) n (iii) cn lso e foun y mking trigonometric sustitution secθ in (i), tnθ in (ii) n sinθ in (iii) respectively. Emple Fin Solution Note tht ( + ) + Put + y, so tht y. Then Emple Fin Solution Note tht y + y y y++ log y+ y+ + [using 7.6. (ii)] ( + ) log ( )

44 MATHEMATIS Put + y so tht y. Thus y y Integrte the functions in Eercises to 9. y y y + sin + [using 7.6. (iii)] + ( + ) + sin + EXERISE hoose the correct nswer in Eercises to... + is equl to (A) ( ) (B) (D) + + log ( + ) + () + + log is equl to + 9 ( ) + + (A) (B) () (D) ( ) 8 7 9log ( ) 8 7 9log ( ) 8 7 log ( ) 8+ 7 log

45 7.7 Definite Integrl INTEGRALS In the previous sections, we hve stuie out the inefinite integrls n iscusse few methos of fining them incluing integrls of some specil functions. In this section, we shll stuy wht is clle efinite integrl of function. The efinite integrl hs unique vlue. A efinite integrl is enote y f (), where is clle the lower limit of the integrl n is clle the upper limit of the integrl. The efinite integrl is introuce either s the limit of sum or if it hs n nti erivtive F in the intervl [, ], then its vlue is the ifference etween the vlues of F t the en points, i.e., F() F(). Here, we shll consier these two cses seprtely s iscusse elow: 7.7. Definite integrl s the limit of sum Let f e continuous function efine on close intervl [, ]. Assume tht ll the vlues tken y the function re non negtive, so the grph of the function is curve ove the -is. The efinite integrl f () is the re oune y the curve y f (), the orintes, n the -is. To evlute this re, consier the region PRSQP etween this curve, -is n the orintes n (Fig 7.). Y S Q ( ) y f M DL X' P A B R O r- r n Y' Fig 7. X Divie the intervl [, ] into n equl suintervls enote y [, ], [, ],..., [ r, r ],..., [ n, n ], where, + h, + h,..., r + rh n n + nh or n. We note tht s n, h. h

46 MATHEMATIS The region PRSQP uner consiertion is the sum of n suregions, where ech suregion is efine on suintervls [ r, r ], r,,,, n. From Fig 7., we hve re of the rectngle (ABL) < re of the region (ABDA) < re of the rectngle (ABDM)... () Eviently s r r, i.e., h ll the three res shown in () ecome nerly equl to ech other. Now we form the following sums. s n h [f( ) + + f ( n - )] n h f( r )... () n n S n h[ f( ) + f( ) + + f( n)] h f( r)... () Here, s n n S n enote the sum of res of ll lower rectngles n upper rectngles rise over suintervls [ r, r ] for r,,,, n, respectively. In view of the inequlity () for n ritrry suintervl [ r, r ], we hve s n < re of the region PRSQP < S n... () As n strips ecome nrrower n nrrower, it is ssume tht the limiting vlues of () n () re the sme in oth cses n the common limiting vlue is the require re uner the curve. Symoliclly, we write lim S n lim sn re of the region PRSQP () n n f... (5) It follows tht this re is lso the limiting vlue of ny re which is etween tht of the rectngles elow the curve n tht of the rectngles ove the curve. For the ske of convenience, we shll tke rectngles with height equl to tht of the curve t the left hn ege of ech suintervl. Thus, we rewrite (5) s h r r f ( ) lim h[ f( ) + f( + h) f ( + ( n ) h] or f ( ) ( ) lim [ f( ) + f( + h) f( + ( n ) h]... (6) n n where h s n n The ove epression (6) is known s the efinition of efinite integrl s the limit of sum. Remrk The vlue of the efinite integrl of function over ny prticulr intervl epens on the function n the intervl, ut not on the vrile of integrtion tht we

47 INTEGRALS choose to represent the inepenent vrile. If the inepenent vrile is enote y t or u inste of, we simply write the integrl s f () t t or ( ) f u u inste of f ( ). Hence, the vrile of integrtion is clle ummy vrile. + s the limit of sum. Emple 5 Fin ( ) Solution By efinition f ( ) ( ) lim [ f( ) + f( + h) f( + ( n ) h], n n where, h n In this emple,,, f () +, Therefore, + ( ) h n n ( n ) lim [ f() + f( ) + f( ) f( )] n n n n n ( n ) lim [ + ( + ) + ( + ) ] n + n n n n lim [( ) + ( ( n ) ] n n n n-terms lim [ n+ ( ( n ) ] n n n ( n ) n( n ) lim [ n+ ] n n n 6 ( n ) ( n ) lim [ n + ] n n n lim [ + ( ) ( )] n n n + [ ]

48 MATHEMATIS Emple 6 Evlute Solution By efinition e s the limit of sum. e ( ) lim e + e + e e n n n n n n Using the sum to n terms of G.P., where, r e n, we hve e n n e lim [ ] n n e n lim e n n e n ( e ) e n lim n n ( e ) e [using lim ] h h h EXERISE 7.8 Evlute the following efinite integrls s limit of sums.. 5. ( + ).. 5. ( ) e 6. ( + e ) 7.8 Funmentl Theorem of lculus 7.8. Are function We hve efine f ( ) s the re of the region oune y the curve y f (), the orintes n n -is. Let e given point in [, ]. Then f ( ) represents the re of the light she region Fig 7.

49 INTEGRALS 5 in Fig 7. [Here it is ssume tht f() > for [, ], the ssertion me elow is eqully true for other functions s well]. The re of this she region epens upon the vlue of. In other wors, the re of this she region is function of. We enote this function of y A(). We cll the function A() s Are function n is given y A () f ( )... () Bse on this efinition, the two sic funmentl theorems hve een given. However, we only stte them s their proofs re eyon the scope of this tet ook First funmentl theorem of integrl clculus Theorem Let f e continuous function on the close intervl [, ] n let A () e the re function. Then A () f (), for ll [, ] Secon funmentl theorem of integrl clculus We stte elow n importnt theorem which enles us to evlute efinite integrls y mking use of nti erivtive. Theorem Let f e continuous function efine on the close intervl [, ] n F e n nti erivtive of f. Then f ( ) [F( )] F () F(). Remrks (i) In wors, the Theorem tells us tht f ( ) (vlue of the nti erivtive F of f t the upper limit vlue of the sme nti erivtive t the lower limit ). (ii) This theorem is very useful, ecuse it gives us metho of clculting the efinite integrl more esily, without clculting the limit of sum. (iii) The crucil opertion in evluting efinite integrl is tht of fining function whose erivtive is equl to the integrn. This strengthens the reltionship etween ifferentition n integrtion. (iv) In f ( ), the function f nees to e well efine n continuous in [, ]. For instnce, the consiertion of efinite integrl ( ) is erroneous since the function f epresse y f() ( ) is not efine in portion < < of the close intervl [, ].

50 6 MATHEMATIS Steps for clculting f ( ). (i) Fin the inefinite integrl f ( ). Let this e F(). There is no nee to keep (ii) integrtion constnt ecuse if we consier F() + inste of F(), we get f ( ) [F ( ) + ] [F( ) + ] [F( ) + ] F( ) F( ). Thus, the ritrry constnt isppers in evluting the vlue of the efinite integrl. Evlute F() F() [F ( )], which is the vlue of f ( ). We now consier some emples Emple 7 Evlute the following integrls: (i) (ii) 9 ( ) (iii) ( + ) ( + ) (iv) sin t cos t t Solution (i) (ii) F( ), Let I. Since Therefore, y the secon funmentl theorem, we get I F() F() Let 9 I. We first fin the nti erivtive of the integrn. ( ) Put t. Then t or t t Thus, t ( ) t F( ) ( )

51 INTEGRALS 7 Therefore, y the secon funmentl theorem of clculus, we hve I F(9) F() ( ) 9 ( 7) (iii) Let I ( + ) ( + ) Using prtil frction, we get + ( + ) ( + ) + + So log + + log + F( ) ( + ) ( + ) Therefore, y the secon funmentl theorem of clculus, we hve I F() F() [ log + log ] [ log + log ] log + log + log log 7 (iv) Let I sin t cos t t. onsier sin t cos t t Put sin t u so tht cos t t u or cos t t u So sin t cos t t u u [ u ] sin t F ( t) sy 8 8 Therefore, y the secon funmentl theorem of integrl clculus I F ( ) F () [sin sin ] 8 8

52 8 MATHEMATIS EXERISE 7.9 Evlute the efinite integrls in Eercises to.. ( + ).. ( ) +. sin 5. cos 6. 5 e 7. tn 8. 6 cosec cos e (sec + + ) 8. (sin cos ) ( sin e + ) hoose the correct nswer in Eercises n.. equls + (A) (B) () 6 (D). equls + 9 (A) 6 (B) 7.9 Evlution of Definite Integrls y Sustitution In the previous sections, we hve iscusse severl methos for fining the inefinite integrl. One of the importnt methos for fining the inefinite integrl is the metho of sustitution. () (D)

53 INTEGRALS 9 To evlute f ( ), y sustitution, the steps coul e s follows:. onsier the integrl without limits n sustitute, y f () or g(y) to reuce the given integrl to known form.. Integrte the new integrn with respect to the new vrile without mentioning the constnt of integrtion.. Resustitute for the new vrile n write the nswer in terms of the originl vrile.. Fin the vlues of nswers otine in () t the given limits of integrl n fin the ifference of the vlues t the upper n lower limits. Note In orer to quicken this metho, we cn procee s follows: After performing steps, n, there is no nee of step. Here, the integrl will e kept in the new vrile itself, n the limits of the integrl will ccoringly e chnge, so tht we cn perform the lst step. Let us illustrte this y emples. Emple 8 Evlute Solution Put t 5 +, then t 5. Therefore, t t t ( 5 ) + Hence, ( + ) ( + ) ( ( ) + ) ( ) Alterntively, first we trnsform the integrl n then evlute the trnsforme integrl with new limits.

54 MATHEMATIS Let t 5 +. Then t 5. Note tht, when, t n when, t Thus, s vries from to, t vries from to Therefore t t t ( ) tn Emple 9 Evlute + Solution Let t tn, then t. The new limits re, when, t n + when, t. Thus, s vries from to, t vries from to. Therefore tn + t t t 6 EXERISE 7. Evlute the integrls in Eercises to 8 using sustitution sin φ cos φφ. + (Put + t ) hoose the correct nswer in Eercises 9 n. sin + sin + cos e ( ) 9. The vlue of the integrl is (A) 6 (B) () (D). If f () t (A) cos + sin () cos sin t t, then f () is (B) sin (D) sin + cos

55 INTEGRALS 7. Some Properties of Definite Integrls We list elow some importnt properties of efinite integrls. These will e useful in evluting the efinite integrls more esily. P : f ( ) f () t t P : f ( ) f ( ). In prticulr, ( ) c P : f ( ) f ( ) + f ( ) c P : f ( ) f ( + ) P : P 5 : P 6 : P 7 : f ( ) f ( ) (Note tht P is prticulr cse of P ) f ( ) f ( ) + f ( ) f f ( ) f ( ),if f ( ) f ( ) n (i) if f ( ) f () f( ) f( ), if f is n even function, i.e., if f ( ) f (). (ii) f ( ), if f is n o function, i.e., if f ( ) f (). We give the proofs of these properties one y one. Proof of P It follows irectly y mking the sustitution t. Proof of P Let F e nti erivtive of f. Then, y the secon funmentl theorem of clculus, we hve f ( ) F( ) F( ) [F( ) F( )] f ( ) Here, we oserve tht, if, then f ( ). Proof of P Let F e nti erivtive of f. Then f ( ) F() F()... () c f ( ) F(c) F()... () n f ( ) F() F(c) c... ()

56 MATHEMATIS c Aing () n (), we get f ( ) + f ( ) F( ) F( ) f ( ) c This proves the property P. Proof of P Let t +. Then t. When, t n when, t. Therefore f ( ) f ( t ) t + f ( + t ) t (y P ) f ( + ) y P Proof of P Put t. Then t. When, t n when, t. Now procee s in P. Proof of P 5 Using P, we hve Let f ( ) f ( ) f ( ) +. t in the secon integrl on the right hn sie. Then t. When, t n when, t. Also t. Therefore, the secon integrl ecomes Hence f ( ) ( ) f t t f ( t ) t f ( ) f ( ) + f ( ) f ( ) Proof of P 6 Using P 5, we hve f( ) f( ) + f( )... () Now, if n if f ( ) Proof of P 7 Using P, we hve f ( ) f (), then () ecomes f ( ) f( ) + f( ) f( ), f( ) f (), then () ecomes f ( ) f ( ) Let f ( ) f ( ) + f ( ). Then t in the first integrl on the right hn sie. t. When, t n when, t. Also t.

57 INTEGRALS Therefore f ( ) f ( t ) t + f ( ) + f( ) f( ) (y P )... () (i) Now, if f is n even function, then f ( ) f () n so () ecomes f ( ) f ( ) + f ( ) f ( ) (ii) If f is n o function, then f ( ) f () n so () ecomes f ( ) f ( ) + f ( ) Emple Evlute Solution We note tht on [, ] n on [, ] n tht on [, ]. So y P we write + + ( ) ( ) ( ) + + ( ) ( ) ( ) ( ) Emple Evlute sin Solution We oserve tht sin is n even function. Therefore, y P 7 (i), we get sin sin

58 MATHEMATIS ( cos ) ( cos ) sin sin sin Emple Evlute + cos sin Solution Let I. Then, y P + cos, we hve ( )sin( ) I + cos ( ) ( ) sin sin I + cos + cos sin or I + cos sin or I + cos Put cos t so tht sin t. When, t n when, t. Therefore, (y P ) we get I t + t t t + t (y P + t 7, since t is even function) + Emple Evlute tn t tn tn 5 sin cos Solution Let I 5 sin cos. Let f() sin 5 cos. Then f ( ) sin 5 ( ) cos ( ) sin 5 cos f (), i.e., f is n o function. Therefore, y P 7 (ii), I

59 INTEGRALS 5 Emple Evlute sin sin + cos Solution Let I sin sin + cos... () Then, y P sin ( ) I sin ( ) + cos ( ) Aing () n (), we get I Hence I cos... () cos sin cos [ ] sin + + cos + sin Emple 5 Evlute + tn 6 Solution Let I cos + tn cos + sin () Then, y P I 6 cos + 6 cos + + sin Aing () n (), we get sin... () sin + cos 6 6 I [ ]. Hence I 6 6

60 6 MATHEMATIS Emple 6 Evlute log sin Solution Let I log sin Then, y P Aing the two vlues of I, we get I log sin log cos I ( log sin + log cos ) ( log sin cos + log log ) (y ing n sutrcting log ) log sin log (Why?) Put t in the first integrl. Then t, when, t n when t. Therefore I log sin t t log, log sin t t log [y P 6 s sin ( t) sin t) log sin log (y chnging vrile t to ) I log Hence log sin log.

61 INTEGRALS 7 EXERISE 7. By using the properties of efinite integrls, evlute the integrls in Eercises to 9.. cos. sin. sin + cos sin sin + cos. 5 cos sin + cos ( ) n 8. log ( + tn ) 9... (log sin log sin ).. + sin sin cos sin cos 8. 7 sin. log ( + cos ) 7. sin 5 cos + 9. Show tht f( ) g( ) f( ), if f n g re efine s f() f( ) n g() + g( ) hoose the correct nswer in Eercises n. 5. The vlue of ( + cos + tn + ) is (A) (B) () (D). The vlue of (A) + sin log + cos is (B) () (D)

62 8 MATHEMATIS Emple 7 Fin cos 6 + sin 6 Miscellneous Emples Solution Put t + sin 6, so tht t 6 cos 6 Therefore Emple 8 Fin cos 6 + sin 6 t t 6 ( ) 5 ( t) + ( + sin 6) Solution We hve ( ) ( ) 5 Put t, so tht t Therefore ( ) t t t Emple 9 Fin ( ) ( + ) Solution We hve ( ) ( + ) ( + ) + + Now epress ( ) ( + ) ( + ) + ( ) ( + ) A B + + ( ) ( + )... ()... ()

63 INTEGRALS 9 So A ( + ) + (B + ) ( ) (A + B) + ( B) + A Equting coefficients on oth sies, we get A + B, B n A, which give A,B. Sustituting vlues of A, B n in (), we get ( ) ( + ) Agin, sustituting () in (), we hve Therefore ( ) ( + + ) ( ) ( + ) ( + ) ( + ) + ( ) ( + ) ( + ) + + log log ( + ) tn + ( ) ( + + ) Emple Fin log (log ) + (log ) Solution Let I log (log ) + (log ) log (log ) + (log )... () In the first integrl, let us tke s the secon function. Then integrting it y prts, we get Agin, consier I log (log ) + log (log ) log (log ) + log (log )... (), tke s the secon function n integrte it y prts, log we hve log log (log )... ()

64 5 MATHEMATIS Putting () in (), we get I log (log ) + log (log ) (log ) Emple Fin cot tn + log log (log ) + Solution We hve I cot + tn tn ( + cot ) Put tn t, so tht sec t t or t t + t Then t I t + t t ( + t ) + t + t ( t + ) t t t t + t + t + t t Put t y, so tht + t t t y. Then I t y y t tn tn + + y + ( ) Emple Fin t tn tn + tn + t tn sin cos 9 cos ( ) Solution Let sin cos I 9 cos

65 INTEGRALS 5 Put cos () t so tht sin cos t Therefore t t I sin sin cos t Emple Evlute sin ( ) sin for Solution Here f () sin sin for Therefore sin sin + sin Integrting oth integrls on righthn sie, we get sin sin sin cos sin cos sin Emple Evlute cos + sin ( ) Solution Let I cos + sin cos ( ) + sin ( ) (using P ) cos + sin cos + sin I cos + sin Thus I cos + sin

66 5 MATHEMATIS or I cos + sin cos + sin (using P 6 ) cos + sin cos + sin + sec cosec + tn cot + + t ( put tn t n cot u) + t u + u t u tn tn tn + tn Miscellneous Eercise on hpter 7 Integrte the functions in Eercises to [Hint: Put t ]. ( + ) 5. + [Hint: + + 6, put t 6 ] 5 6. ( + ) ( + 9) 7. sin sin ( ) 8. e e 5 log log e e log log 9. cos sin. sin 8 8 cos sin cos. cos ( + ) cos ( + ) e.. 8. ( + e ) ( + e ) ( + ) ( + ) 5. cos e log sin 6. e log ( + ) 7. f ( + ) [f ( + )] n sin cos 8. 9., [, ] sin sin ( +α) sin + cos

67 INTEGRALS sin e + cos. + + ( + ) ( + ) + log ( ) log. tn. + + Evlute the efinite integrls in Eercises 5 to sin e cos sin + cos 9. sin sin cos cos + sin +. sin tn (sin ).. [ + + ] Prove the following (Eercises to 9). + log 5. ( + ) cos 7. tn log 9. e cos cos + sin sin + cos 9 + 6sin tn sec + tn e sin sin. Evlute s limit of sum. hoose the correct nswers in Eercises to.. e + e is equl to (A) tn (e ) + (B) tn (e ) + () log (e e ) + (D) log (e + e ) +. cos (sin + cos ) (A) sin + cos + (B) log sin + cos + () log sin cos + (D) (sin + cos )

68 5 MATHEMATIS. If f ( + ) f (), then f ( ) is equl to + + (A) f ( ) (B) f ( ) + + () f ( ) (D) f ( ). The vlue of + tn is (A) (B) () (D) Summry Integrtion is the inverse process of ifferentition. In the ifferentil clculus, we re given function n we hve to fin the erivtive or ifferentil of this function, ut in the integrl clculus, we re to fin function whose ifferentil is given. Thus, integrtion is process which is the inverse of ifferentition. Let F( ) f( ). Then we write f ( ) F( ) +. These integrls re clle inefinite integrls or generl integrls, is clle constnt of integrtion. All these integrls iffer y constnt. From the geometric point of view, n inefinite integrl is collection of fmily of curves, ech of which is otine y trnslting one of the curves prllel to itself upwrs or ownwrs long the y-is. Some properties of inefinite integrls re s follows:. [ f ( ) + g ( )] f ( ) + g ( ). For ny rel numer k, k f( ) k f( ) More generlly, if f, f, f,..., f n re functions n k, k,...,k n re rel numers. Then [ k f ( ) + k f ( ) k f ( )] n n n n k f ( ) + k f ( ) k f ( )

If we have a function f(x) which is well-defined for some a x b, its integral over those two values is defined as

If we have a function f(x) which is well-defined for some a x b, its integral over those two values is defined as Y. D. Chong (26) MH28: Complex Methos for the Sciences 2. Integrls If we hve function f(x) which is well-efine for some x, its integrl over those two vlues is efine s N ( ) f(x) = lim x f(x n ) where x

More information

School of Business. Blank Page

School of Business. Blank Page Integrl Clculus This unit is esigne to introuce the lerners to the sic concepts ssocite with Integrl Clculus. Integrl clculus cn e clssifie n iscusse into two thres. One is Inefinite Integrl n the other

More information

APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line

APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line APPENDIX D Preclculus Review APPENDIX D.1 Rel Numers n the Rel Numer Line Rel Numers n the Rel Numer Line Orer n Inequlities Asolute Vlue n Distnce Rel Numers n the Rel Numer Line Rel numers cn e represente

More information

f a L Most reasonable functions are continuous, as seen in the following theorem:

f a L Most reasonable functions are continuous, as seen in the following theorem: Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f

More information

CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS

CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS LEARNING OBJECTIVES After stuying this chpter, you will be ble to: Unerstn the bsics

More information

Chapter 6 Techniques of Integration

Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

More information

Basic Derivative Properties

Basic Derivative Properties Bsic Derivtive Properties Let s strt this section by remining ourselves tht the erivtive is the slope of function Wht is the slope of constnt function? c FACT 2 Let f () =c, where c is constnt Then f 0

More information

VII. The Integral. 50. Area under a Graph. y = f(x)

VII. The Integral. 50. Area under a Graph. y = f(x) VII. The Integrl In this chpter we efine the integrl of function on some intervl [, b]. The most common interprettion of the integrl is in terms of the re uner the grph of the given function, so tht is

More information

Section 4: Integration ECO4112F 2011

Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

More information

Conservation Law. Chapter Goal. 6.2 Theory

Conservation Law. Chapter Goal. 6.2 Theory Chpter 6 Conservtion Lw 6.1 Gol Our long term gol is to unerstn how mthemticl moels re erive. Here, we will stuy how certin quntity chnges with time in given region (sptil omin). We then first erive the

More information

M 106 Integral Calculus and Applications

M 106 Integral Calculus and Applications M 6 Integrl Clculus n Applictions Contents The Inefinite Integrls.................................................... Antierivtives n Inefinite Integrls.. Antierivtives.............................................................

More information

4.5 THE FUNDAMENTAL THEOREM OF CALCULUS

4.5 THE FUNDAMENTAL THEOREM OF CALCULUS 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere

More information

3.4 Conic sections. In polar coordinates (r, θ) conics are parameterized as. Next we consider the objects resulting from

3.4 Conic sections. In polar coordinates (r, θ) conics are parameterized as. Next we consider the objects resulting from 3.4 Conic sections Net we consier the objects resulting from + by + cy + + ey + f 0. Such type of cures re clle conics, becuse they rise from ifferent slices through cone In polr coorintes r, θ) conics

More information

x dx does exist, what does the answer look like? What does the answer to

x dx does exist, what does the answer look like? What does the answer to Review Guie or MAT Finl Em Prt II. Mony Decemer th 8:.m. 9:5.m. (or the 8:3.m. clss) :.m. :5.m. (or the :3.m. clss) Prt is worth 5% o your Finl Em gre. NO CALCULATORS re llowe on this portion o the Finl

More information

When e = 0 we obtain the case of a circle.

When e = 0 we obtain the case of a circle. 3.4 Conic sections Circles belong to specil clss of cures clle conic sections. Other such cures re the ellipse, prbol, n hyperbol. We will briefly escribe the stnr conics. These re chosen to he simple

More information

7. Indefinite Integrals

7. Indefinite Integrals 7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find

More information

Final Exam Review. Exam 1 Material

Final Exam Review. Exam 1 Material Lessons 2-4: Limits Limit Solving Strtegy for Finl Exm Review Exm 1 Mteril For piecewise functions, you lwys nee to look t the left n right its! If f(x) is not piecewise function, plug c into f(x), i.e.,

More information

Overview of Calculus

Overview of Calculus Overview of Clculus June 6, 2016 1 Limits Clculus begins with the notion of limit. In symbols, lim f(x) = L x c In wors, however close you emn tht the function f evlute t x, f(x), to be to the limit L

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Sturm-Liouville Theory

Sturm-Liouville Theory LECTURE 1 Sturm-Liouville Theory In the two preceing lectures I emonstrte the utility of Fourier series in solving PDE/BVPs. As we ll now see, Fourier series re just the tip of the iceerg of the theory

More information

5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship

5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship 5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is

More information

( x) ( ) takes at the right end of each interval to approximate its value on that

( x) ( ) takes at the right end of each interval to approximate its value on that III. INTEGRATION Economists seem much more intereste in mrginl effects n ifferentition thn in integrtion. Integrtion is importnt for fining the expecte vlue n vrince of rnom vriles, which is use in econometrics

More information

Thomas Whitham Sixth Form

Thomas Whitham Sixth Form Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos

More information

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

More information

Section 6.3 The Fundamental Theorem, Part I

Section 6.3 The Fundamental Theorem, Part I Section 6.3 The Funmentl Theorem, Prt I (3//8) Overview: The Funmentl Theorem of Clculus shows tht ifferentition n integrtion re, in sense, inverse opertions. It is presente in two prts. We previewe Prt

More information

AP Calculus AB First Semester Final Review

AP Calculus AB First Semester Final Review P Clculus B This review is esigne to give the stuent BSIC outline of wht nees to e reviewe for the P Clculus B First Semester Finl m. It is up to the iniviul stuent to etermine how much etr work is require

More information

x ) dx dx x sec x over the interval (, ).

x ) dx dx x sec x over the interval (, ). Curve on 6 For -, () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor

More information

5.3 The Fundamental Theorem of Calculus

5.3 The Fundamental Theorem of Calculus CHAPTER 5. THE DEFINITE INTEGRAL 35 5.3 The Funmentl Theorem of Clculus Emple. Let f(t) t +. () Fin the re of the region below f(t), bove the t-is, n between t n t. (You my wnt to look up the re formul

More information

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ). AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

More information

B Veitch. Calculus I Study Guide

B Veitch. Calculus I Study Guide Clculus I Stuy Guie This stuy guie is in no wy exhustive. As stte in clss, ny type of question from clss, quizzes, exms, n homeworks re fir gme. There s no informtion here bout the wor problems. 1. Some

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

Formulae For. Standard Formulae Of Integrals: x dx k, n 1. log. a dx a k. cosec x.cot xdx cosec. e dx e k. sec. ax dx ax k. 1 1 a x.

Formulae For. Standard Formulae Of Integrals: x dx k, n 1. log. a dx a k. cosec x.cot xdx cosec. e dx e k. sec. ax dx ax k. 1 1 a x. Forule For Stndrd Forule Of Integrls: u Integrl Clculus By OP Gupt [Indir Awrd Winner, +9-965 35 48] A B C D n n k, n n log k k log e e k k E sin cos k F cos sin G tn log sec k OR log cos k H cot log sin

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into

More information

sec x over the interval (, ). x ) dx dx x 14. Use a graphing utility to generate some representative integral curves of the function Curve on 5

sec x over the interval (, ). x ) dx dx x 14. Use a graphing utility to generate some representative integral curves of the function Curve on 5 Curve on Clcultor eperience Fin n ownlo (or type in) progrm on your clcultor tht will fin the re uner curve using given number of rectngles. Mke sure tht the progrm fins LRAM, RRAM, n MRAM. (You nee to

More information

Calculus AB. For a function f(x), the derivative would be f '(

Calculus AB. For a function f(x), the derivative would be f '( lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:

More information

Chapter Five - Eigenvalues, Eigenfunctions, and All That

Chapter Five - Eigenvalues, Eigenfunctions, and All That Chpter Five - Eigenvlues, Eigenfunctions, n All Tht The prtil ifferentil eqution methos escrie in the previous chpter is specil cse of more generl setting in which we hve n eqution of the form L 1 xux,tl

More information

Chapter 1: Logarithmic functions and indices

Chapter 1: Logarithmic functions and indices Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4

More information

MA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations

MA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

More information

Math 211A Homework. Edward Burkard. = tan (2x + z)

Math 211A Homework. Edward Burkard. = tan (2x + z) Mth A Homework Ewr Burkr Eercises 5-C Eercise 8 Show tht the utonomous system: 5 Plne Autonomous Systems = e sin 3y + sin cos + e z, y = sin ( + 3y, z = tn ( + z hs n unstble criticl point t = y = z =

More information

Mathematics Number: Logarithms

Mathematics Number: Logarithms plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement

More information

Mathematics. Area under Curve.

Mathematics. Area under Curve. Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding

More information

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive

More information

The practical version

The practical version Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

More information

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations. Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

10 Vector Integral Calculus

10 Vector Integral Calculus Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve

More information

Antiderivatives Introduction

Antiderivatives Introduction Antierivtives 0. Introuction So fr much of the term hs been sent fining erivtives or rtes of chnge. But in some circumstnces we lrey know the rte of chnge n we wish to etermine the originl function. For

More information

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function? Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

More information

2.4 Linear Inequalities and Interval Notation

2.4 Linear Inequalities and Interval Notation .4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or

More information

Chapter 8.2: The Integral

Chapter 8.2: The Integral Chpter 8.: The Integrl You cn think of Clculus s doule-wide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in

More information

1B40 Practical Skills

1B40 Practical Skills B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need

More information

Here we consider the matrix transformation for a square matrix from a geometric point of view.

Here we consider the matrix transformation for a square matrix from a geometric point of view. Section. he Mgnifiction Fctor In Section.5 we iscusse the mtri trnsformtion etermine mtri A. For n m n mtri A the function f(c) = Ac provies corresponence etween vectors in R n n R m. Here we consier the

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

The Evaluation Theorem

The Evaluation Theorem These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for in-clss presenttion nd should not

More information

Contents PART II. Foreword

Contents PART II. Foreword Contents PART II Foreword v Preface vii 7. Integrals 87 7. Introduction 88 7. Integration as an Inverse Process of Differentiation 88 7. Methods of Integration 00 7.4 Integrals of some Particular Functions

More information

Chapter 8: Methods of Integration

Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln

More information

Math 142: Final Exam Formulas to Know

Math 142: Final Exam Formulas to Know Mth 4: Finl Exm Formuls to Know This ocument tells you every formul/strtegy tht you shoul know in orer to o well on your finl. Stuy it well! The helpful rules/formuls from the vrious review sheets my be

More information

Using integration tables

Using integration tables Using integrtion tbles Integrtion tbles re inclue in most mth tetbooks, n vilble on the Internet. Using them is nother wy to evlute integrls. Sometimes the use is strightforwr; sometimes it tkes severl

More information

= f (c) f (c) the height of the rectangle guaranteed by the MVT for integrals.

= f (c) f (c) the height of the rectangle guaranteed by the MVT for integrals. Get Rey: Given (t) = 8t n v() = 6, fin the isplcement n istnce of the oject from t= to t= If () = 4, fin the position of the prticle t t= I. Averge Vlue of Function Wht oes represent? Cn we rw rectngle

More information

Chapters Five Notes SN AA U1C5

Chapters Five Notes SN AA U1C5 Chpters Five Notes SN AA U1C5 Nme Period Section 5-: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles

More information

Math Review. b c A = ½ ah P = a + b + c a

Math Review. b c A = ½ ah P = a + b + c a Mth Review Geometry Much of wht you lerne in high school geometry is either pretty intuitive or will not e neee for this clss, ut mny formuls you lerne erlier for re or volume will come up lot. Three two-imensionl

More information

Instantaneous Rate of Change of at a :

Instantaneous Rate of Change of at a : AP Clculus AB Formuls & Justiictions Averge Rte o Chnge o on [, ]:.r.c. = ( ) ( ) (lger slope o Deinition o the Derivtive: y ) (slope o secnt line) ( h) ( ) ( ) ( ) '( ) lim lim h0 h 0 3 ( ) ( ) '( ) lim

More information

RAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties

RAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : 947084408 9546359990 M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

Section - 2 MORE PROPERTIES

Section - 2 MORE PROPERTIES LOCUS Section - MORE PROPERTES n section -, we delt with some sic properties tht definite integrls stisf. This section continues with the development of some more properties tht re not so trivil, nd, when

More information

Linear Inequalities. Work Sheet 1

Linear Inequalities. Work Sheet 1 Work Sheet 1 Liner Inequlities Rent--Hep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

1 nonlinear.mcd Find solution root to nonlinear algebraic equation f(x)=0. Instructor: Nam Sun Wang

1 nonlinear.mcd Find solution root to nonlinear algebraic equation f(x)=0. Instructor: Nam Sun Wang nonlinermc Fin solution root to nonliner lgebric eqution ()= Instructor: Nm Sun Wng Bckgroun In science n engineering, we oten encounter lgebric equtions where we wnt to in root(s) tht stisies given eqution

More information

Review of Gaussian Quadrature method

Review of Gaussian Quadrature method Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

Introduction. Calculus I. Calculus II: The Area Problem

Introduction. Calculus I. Calculus II: The Area Problem Introuction Clculus I Clculus I h s its theme the slope problem How o we mke sense of the notion of slope for curves when we only know wht the slope of line mens? The nswer, of course, ws the to efine

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

Topics Covered AP Calculus AB

Topics Covered AP Calculus AB Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.

More information

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

The Fundamental Theorem of Calculus Part 2, The Evaluation Part

The Fundamental Theorem of Calculus Part 2, The Evaluation Part AP Clculus AB 6.4 Funmentl Theorem of Clculus The Funmentl Theorem of Clculus hs two prts. These two prts tie together the concept of integrtion n ifferentition n is regre by some to by the most importnt

More information

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply

More information

Introduction and Review

Introduction and Review Chpter 6A Notes Pge of Introuction n Review Derivtives y = f(x) y x = f (x) Evlute erivtive t x = : y = x x= f f(+h) f() () = lim h h Geometric Interprettion: see figure slope of the line tngent to f t

More information

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!! Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble

More information

( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.

( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists. AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

Mat 210 Updated on April 28, 2013

Mat 210 Updated on April 28, 2013 Mt Brief Clculus Mt Updted on April 8, Alger: m n / / m n m n / mn n m n m n n ( ) ( )( ) n terms n n n n n n ( )( ) Common denomintor: ( ) ( )( ) ( )( ) ( )( ) ( )( ) Prctice prolems: Simplify using common

More information

5.7 Improper Integrals

5.7 Improper Integrals 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

Introduction. Calculus I. Calculus II: The Area Problem

Introduction. Calculus I. Calculus II: The Area Problem Introuction Clculus I Clculus I h s its theme the slope problem How o we mke sense of the notion of slope for curves when we only know wht the slope of line mens? The nswer, of course, ws the to efine

More information

0.1 THE REAL NUMBER LINE AND ORDER

0.1 THE REAL NUMBER LINE AND ORDER 6000_000.qd //0 :6 AM Pge 0-0- CHAPTER 0 A Preclculus Review 0. THE REAL NUMBER LINE AND ORDER Represent, clssify, nd order rel numers. Use inequlities to represent sets of rel numers. Solve inequlities.

More information

SUPPLEMENTARY NOTES ON THE CONNECTION FORMULAE FOR THE SEMICLASSICAL APPROXIMATION

SUPPLEMENTARY NOTES ON THE CONNECTION FORMULAE FOR THE SEMICLASSICAL APPROXIMATION Physics 8.06 Apr, 2008 SUPPLEMENTARY NOTES ON THE CONNECTION FORMULAE FOR THE SEMICLASSICAL APPROXIMATION c R. L. Jffe 2002 The WKB connection formuls llow one to continue semiclssicl solutions from n

More information

Calculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION

Calculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer..

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

Parse trees, ambiguity, and Chomsky normal form

Parse trees, ambiguity, and Chomsky normal form Prse trees, miguity, nd Chomsky norml form In this lecture we will discuss few importnt notions connected with contextfree grmmrs, including prse trees, miguity, nd specil form for context-free grmmrs

More information

Lecture Solution of a System of Linear Equation

Lecture Solution of a System of Linear Equation ChE Lecture Notes, Dept. of Chemicl Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updted /) Lecture 8- - Solution of System of Liner Eqution 8. Why is it importnt to e le to solve system of liner

More information

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

More information

Chapter 9 Definite Integrals

Chapter 9 Definite Integrals Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished

More information

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

More information

MATH 144: Business Calculus Final Review

MATH 144: Business Calculus Final Review MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

More information