MAT 266 Calculus for Engineers II Notes on Chapter 6 Professor: John Quigg Semester: spring 2017

Transcription

1 MAT 66 Clculus for Engineers II Noes on Chper 6 Professor: John Quigg Semeser: spring 7 Secion 6.: Inegrion by prs The Produc Rule is d d f()g() = f()g () + f ()g() Tking indefinie inegrls gives [f()g () + f ()g()] d = f()g() + C. Rerrnging gives: Inegrion by Prs: f()g () d = f()g() f ()g() d Leing u = f() nd v = g(), he formul becomes u dv = uv v du Emple: Consider e d. Le Then so u = nd dv = e d du = d nd v = e e d = e e d = e e + C Emple: Someimes we hve o inegre by prs more hn once: Consider e d. Le u = nd dv = e d Then du = d nd v = e so e d = e e ( d) = e e d = e (e e ) + C (preceding emple) = e e + e + C Clerly we could do somehing similr wih ny inegrl of he form n e d for posiive ineger n. In fc, i suffices o do i once nd recognize he pern, hen pply his pern n imes, reducing he power of by one ech ime. This sregy lso works for oher inegrls, like n sin d, n cos d, sin n d, cos n d, nd so on. The pern is clled reducion formul.

2 Emple: Someimes he choices of u nd dv re no obvious: Consider ln d. Le Then so u = ln nd dv = d du = d nd v = ln d = ln = ln ( ) d d = ln + C Emple: Someimes we hve o inegre by prs wice nd solve for he inegrl: Consider e sin d. Le Then so Now le Then so Thus u = e nd dv = sin d du = e d nd v = e sin d = e ( = e cos ) ( cos + cos ) cos (e d) e cos d u = e nd dv = cos d du = e d nd v = e cos d = e sin e sin d = e cos = e cos Solving for he inegrl gives ( + 4 ) 9 sin e sin d + ( e sin + e sin 4 e sin d 9 9 e sin d = e cos + e sin 9 ) e sin d + C

3 nd so e sin d = 9 ( ) e cos + e sin + C 9 = e cos + e sin + C Noe h s soon s we hd n indefinie inegrl one side bu no indefinie inegrl on he oher side, we hd o sring dding he consn of inegrion o he side wihou ny indefinie inegrl. Prs nd definie inegrls. b f()g () d = f()g() Emple: Consider / n d. Le Then so / ] b b u = n nd dv = d du = ] / / n d = n d + nd v = d + = ( ) n n = ( π ) ] 4/ 6 ln w (ln 4 ) ln = π 6 4/ f ()g() d dw w (subsiuing w = + ) = π 6 ln 4 Noe: when using prs on definie inegrl, i s usully bes o keep he limis of inegrion s you go, so you cn do ny compuions h rise long he wy (rher hn doing ll of he indefinie inegrion firs nd puing he limis bck he end). Secion 6.: Trig inegrls nd subsiuions Trig inegrls. Mny inegrls involving rig funcions cn be evlued by mnipuling he inegrnd using rig ideniies. Emple: Consider sin 5 d. We hve sin 5 = sin 4 sin = (sin ) sin = ( cos ) sin.

4 4 Le u = cos. Then du = sin d, so sin 5 d = ( cos ) sin d = ( u ) du = ( u + u 4 ) du = ( + u u 4 ) du = u + u u5 5 + C = cos + cos 5 cos5 + C Somehing similr could be done wih ny posiive odd power sin k+, or more generlly sin k+ cos n for ny ineger n. Moreover, he roles of sin nd cos could be reversed. Emple: Even powers of sin nd cos re bi more work: Consider cos d. By he hlf-ngle formul, so cos = ( + cos ), cos d = = = ( + cos ) d ( ) sin + + C sin C Emple: Consider cos 4 d. We hve ( ) cos 4 = (cos ) = ( + cos ) = 4 ( + cos + cos ) so = 4 + cos + 4 cos cos 4 d = sin + 4 cos d We hve o do he ls inegrl using he sme sregy s in he preceding emple: cos d = ( + cos 4) d = + sin 4 4 = + sin 4 8

5 5 Thus cos 4 d = sin + 4 ( + ) sin 4 + C 8 = sin + sin 4 + C We could hve done ll of he power reducions, repeedly using he hlf-ngle formul, before sring o inegre. Furher, ny even power cos k could be hndled similrly: wrie cos k = ( cos ) k = ( ( + cos ) ) k, hen muliply ou his kh power, nd coninue pplying he hlf-ngle formul nd muliplying ou unil no even powers re lef. Also, we could do somehing similr wih sin k, using he oher hlf-ngle formul sin = ( cos ). If we hve produc sink cos m, we cn wrie sin k cos m = (sin ) k cos m = ( cos ) k cos m or symmericlly we could hve rewrien he cos m in erms of sin. Alernively, jus pply boh hlf-ngle formuls o he powers of sin nd cos. Producs of he form sin s cos, sin s sin, nd cos s cos cn be rewrien using he ddiion nd subrcion formuls. For emple: sin( + 4) + sin( 4) sin 6 sin sin cos 4 = = Emple: We cn hndle some powers of n nd sec similrly: n sec 8 = n sec 6 sec so we could subsiue u = n, du = sec d. Alernively, we could hve done: = n (sec ) sec = n ( + n ) sec, n sec 8 = n sec 7 n sec = (sec ) sec n, nd hen subsiue u = sec, du = sec n d. The sme powers of co nd csc cn be hndled similrly, using + co = csc nd he differeniion formuls for co nd csc. Emple: Some powers involving n nd sec re no hndled so esily, however. Firs of ll, we cn inegre n by iself, bu wh bou sec? This urns ou o be ricky one: ( ) sec + n sec d = sec d sec + n sec + sec n = d sec + n

6 6 Now le u = sec + n, so du = (sec + sec n ) d, nd we ge du sec d = u = ln u + C = ln sec + n + C This one is probbly bes jus regrded s noher formul o memorize. Emple: Similr echniques show h csc d = ln csc co + C Emple: Consider sec d = sec sec d. Inegre by prs wih We ge u = sec du = sec n d dv = sec d v = n sec d = sec n n sec n d = sec n sec n d = sec n sec (sec ) d = sec n sec d + sec d = sec n + ln sec + n sec d Solving for he inegrl gives sec sec n + ln sec + n d = + C csc d is hndled similrly. Inegrls of his ype cn occur in oher similr siuions. Trig subsiuions. Emple: In d, subsiue = sin θ. Then d = cos θ dθ, bu more impornly = sin θ = cos θ = cos θ We don wn he bsolue vlue, so we wn cos θ. We need o hink bou how his subsiuion is working: we cn hink of our subsiuion s θ = sin nd hen we need o hink bou how he inverse rig funcions re defined: sin hs domin [, ] (which is fine becuse is only defined for ) nd rnge [ π, π ]. Thus we ll hve π θ π,

7 7 nd hence cos θ, s we wned. Thus we hve = sin θ = cos θ = cos θ nd hence d = cos θ cos θ dθ = cos θ dθ = ( + cos θ) dθ = θ + sin θ + C 4 Now of course we hve o subsiue bck in erms of. The firs erm is esy: θ = sin. For he oher erm, we firs use he double-ngle formul: sin θ = sin θ cos θ We know h cos θ =, nd our originl subsiuion ws sin θ =. Thus d = sin C Of course we could hndle ny oher rig funcions of θ h migh pper, since we know sin θ nd cos θ. If we hd been given somehing like 9 5, we would hve subsiued 5 = sin θ, since hen we would hve 9 5 = 9 9 sin θ = sin θ = cos θ. Emple: Noe h if we hve definie inegrl hen we don subsiue bck: Consider / d. Agin subsiue = sin θ. Remembering h his mens θ = sin, he new inegrl is / d = = = π/6 π/6 π/6 sin θ(cos θ)(cos θ dθ) sin θ cos dθ sin θ( cos θ) cos dθ

8 8 We wn o subsiue gin; we could probbly do i in our hed, bu i s much sfer o wrie i ou: le u = cos θ. Then du = sin θ dθ, so we ge / d = = = π/6 / / [ u = u ] / sin θ( cos θ) cos dθ ( u ) du (u ) du ] / ] / = u u ( ) = ( /) = 8 + = 8 Ok, o sum up, we hndle he form u, where is posiive consn nd u is some consn muliple of, by subsiuing u = sin θ, keeping in mind h π/ θ π/. We hve wo oher forms: + u nd u. Emple: Here s simple one involving + u : Consider + d. Le = n θ. Then d = sec θ dθ nd + = + n θ = sec θ = sec θ Agin we wn o ge rid of he bsolue vlue, so we need o know h sec θ, equivlenly cos θ. The subsiuion hs θ = n, so π < θ < π,

9 9 nd hence cos θ, like we wned. Thus + = sec θ, so + d = = = = n θ sec θ sec θ dθ sec θ n θ dθ / cos θ sin θ/ cos θ dθ cos θ sin θ dθ = sin θ + C Bu now we hve o subsiue bck. We know h n θ = nd sec θ = +. Thus sin θ = n θ cos θ = n θ sec θ = +. Thus + + d = + C. Noe h we could hve similrly epressed ny rig funcion of θ in erms of. If we hve n inegrl involving + u, we would ry he rig subsiuionu = n θ, keeping in mind h π/ < θ < π/. Emple: Finlly, we mus fce he remining form u, nd gin we do i he firs ime wih he simples cse : Consider d. Le = sec θ. Then d = sec θ n θ nd = sec θ = n θ = n θ Agin we would like o know h n θ, so h he bsolue vlue goes wy. The subsiuion hs θ = sec, nd we recll h he rnge is [, π ) [ π, π )

10 In boh of hese inervls we hve n θ, s desired. Thus = n θ, so n θ d = (sec θ n θ dθ) sec θ n θ = sec θ dθ sec θ = dθ sec θ = (sec θ cos θ) dθ = ln sec θ + n θ + sin θ + C = ln sec θ + n θ + n θ sec θ + C = ln C Noe how we epressed sin θ in erms of sec θ nd n θ, nd used his o epress i in erms of, similrly o wh we did erlier. We could hve used his echnique o epress ny rig funcion of θ in erms of. Trig subsiuions cn be used even wih epressions of he form c + c + c : firs complee he squre, so h inside he squre roo we ge sum or difference of posiive number nd somehing of he form u, where u = + b, hen chnge vribles from o u, hen mke rig subsiuion. Noe finlly h some inegrls involving he forms u, + u, or u cn be hndled more efficienly using somehing oher hn rig subsiuion. For wo quick emples, we would subsiue u = + in + d, nd is he derivive of sin. Secion 6.: Pril frcions We use he mehod of pril frcions o inegre rionl funcion f() = P () Q() = n n + n n b k k + b k k + + b + b We suppose h n nd b k, so h he polynomils P () nd Q() hve degrees n nd k, respecively. The firs sep is o mke sure h he rionl funcion is proper, h is, he degree of he op is less hn he degree of he boom, h is, n < k. If no, firs divide i ou, using long division of polynomils if necessry, o wrie f() = S() + R() Q() where S(), R(), nd Q() re polynomils nd he rionl funcion R() is proper. Q() So from now on we ssume h he originl rionl funcion f = P is proper. Q The second sep is o fcor he denominor Q() s fr s possible. I urns ou h i is lwys possible o fcor i ino liner fcors nd irreducible qudric fcors, where

11 qudric is clled irreducible if i cnno be fcored ino wo liner fcors (using rel numbers). The hird sep is o epress he rionl funcion f() s sum of frcions of he form A ( + b) j or A + B ( + b + c) j, where A, B,, b, c re consns nd j is posiive ineger. The resul is clled he pril frcion decomposiion of f. Ne, we wn o observe h we know how o inegre rionl funcions of he bove wo ypes. The firs ype is of course esy: jus subsiue u = + b. For he second, if necessry complee he squre in he boom, puing i ino he form (r + u ) j for some posiive consn r nd liner funcion u = p + q. Noe h i will no involve u r or r u, becuse we ssume h he qudric + b + c is irreducible. Then wors we cn mke he rig subsiuion u = n θ. In he cse of qudric fcor wih eponen j =, he inegrl is esy: he indefinie inegrl will in generl be sum of wo erms: log nd n inverse n. The new procedure here is finding he pril frcion decomposiion. We collec like fcors ino posiive ineger powers. Ech of hese conribues pr of he decomposiion, nd we hve four ypes: Cse : Nonrepeed liner fcors. Any nonrepeed fcor + b in Q() conribues erm in he pril frcion decomposiion of he form A + b for some consn A. If he boom is produc of nonrepeed liner fcors, mening h Q() = ( + b )( + b ) ( k + b k ) nd none of he fcors i + b i is consn muliple of ny of he ohers, hen here re consns A, A,..., A k such h P () Q() = A + b + A + b + + A k k + b k Finding hese consns A i is prey esy, lhough i cn be edious if here re lo of fcors. Emple: We find he pril frcion decomposiion of Firs we check he degrees: he op hs degree, nd he boom hs degree. Since <, he rionl funcion is proper. Now we fcor he boom: + = ( )( + ) These re nonrepeed liner fcors. Thus we know h here re consns A, B such h = A + B +

12 We hve o find A, B. Muliply boh sides by Q() o cler frcions: + 7 = A ( )( + ) + B ( )( + ) = A( + ) + B( ). + Muliply ou he righ-hnd side nd collec like erms: nd hen eque coefficiens: + 7 = (A + B) + (A B) A + B = A B = 7 We cn esily solve his sysem of equions for he unknowns A, B. For emple, dding he equions gives 4A = 8 A = Then subsiuing ino he firs equion gives Thus he pril frcion decomposiion is + B = B = = + There s noher wy o deermine he coefficiens h les for nonrepeed liner fcors is bou s fs, bu I don wn o push you owrd his mehod, so I ll only do i his one ime: in he equion + 7 = A( + ) + B( ), plug in = nd hen =. Wih = we ge Wih = we ge + 7 = A( + ) + B( ) 4 = 4B B = + 7 = A( + ) + B( ) 8 = 4A A = I emphsize h his mehod of plugging in well-chosen vlues of is no relly ny beer, nd ges worse s he kinds of fcors in he denominor of he rionl funcion become more comple. So I emphsize he mehod of equing like coefficiens of nd solving he resuling sysem of liner equions for he unknowns.

13 Cse : Repeed liner fcors. Any repeed fcor ( + b) r in Q() conribues r erms in he pril frcion decomposiion of he form A + b + A ( + b) + A ( + b) + + A r ( + b) r for some consns A, A, A,..., A r. Emple: Consider Agin we firs compre degrees: on op, on boom, so i s proper. Ne we fcor he boom: + = ( + ) = ( ) We hve nonrepeed liner fcor, nd repeed liner fcor ( ). Ech conribues is erms seprely, so he pril frcion decomposiion is of he form = A + B + C ( ) for some consns A, B, C. Agin we muliply boh sides by Q() o cler frcions, hen muliply ou he righ-hnd side nd collec like erms: hen eque coefficiens: 5 + = A( ) + B( ) + C = A( + ) + B( ) + C = (A + B) + ( A B + C) + A, A + B = 5 A B + C = A = The rd equion lredy gives us A =, nd hen he s equion becomes nd hen he nd equion becomes Thus he pril frcion decomposiion is + B = 5 B = () + C = 6 + C = 8 + C = C = + 8 = = + 5 ( )

14 4 Cse : Nonrepeed qudric fcors. Any nonrepeed irreducible qudric fcor + b + c in Q() conribues erm in he pril frcion decomposiion of he form for some consns A, B. Emple: Consider A + B + b + c ( + )( + ) Compring degrees, we hve on op, on boom, so i s proper. The boom is lredy fcored s fr s possible, becuse he qudric fcor + is irreducible. We hve nonrepeed liner fcor + nd nonrepeed qudric fcor +, nd he fcors conribue erms seprely, so he pril frcion decomposiion is of he form ( + )( + ) = A + + B + C + for some consns A, B, C. Cler frcions: = A( + ) + (B + C)( + ) Muliply ou he righ-hnd side nd collec like erms: Eque coefficiens: Subrcing he firs wo equions gives nd dding his o he rd equion gives Then he s equion becomes nd hen he nd equion becomes = A + A + B + B + C + C Thus he pril frcion decomposiion is = (A + B) + (B + C) + (A + C) A + B = B + C = A + C = 5 A C = 4A = 4 A = + B = B = C = ( + )( + ) = + + +

15 In his emple, noe h Q() ws lredy fcored. I is ypiclly difficul o fcor polynomils of degree or higher, nd I don wn he clculus problems o involve such edium (nd i seems h he book grees wih me on his). Emple: Jus o mke sure here s no confusion, le s do n inegrl wih qudric fcor in he boom: Evlue d Complee he squre in he denominor: = = ( + ) + 9 Subsiue u = + : u d = u + 9 du u = u + 9 du u + 9 du = ln(u + 9) n u + C (no bsolue vlue since u + 9 > ) = ln( ( + ) + 9 ) + n + C (subsiue bck) = ln( ) + n + C (simplify) Emple: Evlue d + This ime when we check degrees we ge on op nd on boom, so he rionl funcion is improper (no proper), so firs we divide i: 5 Thus we cn epress he improper frcion s + ) =

16 6 +7 We lredy found he pril frcion decomposiion of he proper frcion in n + erlier emple, so we cn pu i ll ogeher o ge he pril frcion decomposiion of he rionl funcion in his emple: Now we find he indefinie inegrl: d = + = + + ( + ) d + = + ln ln + + C Cse 4: Repeed qudric fcors. Any repeed irreducible qudric fcor ( + b + c) r in Q() conribues r erms in he pril frcion decomposiion of he form A + B + b + c + A + B ( + b + c) + A + B ( + b + c) + + A r + B r ( + b + c) r for some consns A, B, A, B, A, B,..., A r, B r. Clerly he lgebric mnipulions required o find he consns would quickly (immediely?) ge ou of hnd, so in he emples nd eercises we won do ny more hn jus find he form of he pril frcion decomposiion when repeed qudric fcors occur. Hin: if you re old o find pril frcion decomposiion nd you hink you hve repeed qudric fcor, you should be ble o fcor i ino liner fcors! Wrning, however: he book does find he consns (nd hen does he inegrion) wih liner fcor nd squred qudric fcor, bu h requires 5 consns, which is bi beyond wh I wn o do. Emple: Find he form of he pril frcion decomposiion, bu do no find he consns: ( + + ) ( + + ) Firs we check degrees: on op i s 9, nd on boom i s + =, so he frcion is proper. I migh look like he boom is lredy fcored, bu we mus check he qudric fcors: we hve ++ = (+). The fcor ++ is irreducible since he discriminn is negive: Thus he denominor is b 4c = 4()() = 4 = < Q() = ( + ) 4 ( + + ) wih repeed liner fcor nd repeed qudric fcor. Ech of hese conribues erms seprely. The repeed liner fcor ( + ) 4 conribues 4 erms: A + + A ( + ) + A ( + ) + A 4 ( + ) 4

17 7 nd he repeed qudric fcor ( + + ) conribues erms: B + C B + C ( + + ) + B + C ( + + ) Thus he form of he pril frcion decomposiion is ( + + ) ( + + ) = ( + ) 4 ( + + ) = A + + A ( + ) + A ( + ) + A 4 ( + ) 4 + B + C B + C ( + + ) + B + C ( + + ) Emple: Jus for compleeness, le s see wh s involved in inegring when here s repeed qudric fcor: In ( + 4) d mke he rig subsiuion = n θ: ( + 4) d = = = 6 = 6 = 6 = (4 n θ + 4) sec θ dθ (4 sec θ) sec θ dθ sec θ dθ cos θ dθ ( + cos θ) dθ (θ + ) sin θ + C = θ + sin θ cos θ + C = n + n θ cos θ + C = n + n θ sec θ + C = n C = n + 6( + 4) + C Of course, if insed we hd d, we would jus subsiue u = + 4, nd i would ( +4) be much esier. This brings up useful observion: ofen here will be choice of mehod, nd you migh sve some work by hinking bou h choice.

18 8 Secion 6.4: Tbles I hve mied feelings bou covering bles of inegrls. The whole poin here is lerning he echniques of inegrion, so looking up n inegrl in ble seems off-opic o me. However, i s cerinly imporn for you o recognize he form of he inegrnd in order o decide upon mehod, nd perhps bi of prcice wih bles migh help wih h. My coverge of his opic will be rher limied. The book hs quick emples of using ble, nd I ll do couple here. The book lso discusses compuer lgebr sysems for emple Mple nd Mhemic bu I won cover his. I encourge you o ge some prcice wih CAS, bu I won ssign ny of his for homework or on ems. In priculr, in clss nd on ems you will no be using ny echnology (including clculors!), so when you re doing homework problems you mus ge ino he hbi of working ou, nd eplining, he deils by hnd, showing ll he seps. Finlly, he book menions h no ll coninuous funcions cn be inegred using he elemenry funcions, nmely polynomils, rionl funcions, power funcions ( ), eponenil funcions ( ), logs, rig funcions, inverse rig funcions, nd ll funcions obinble from hese by ddiion, subrcion, muliplicion, division, nd composiion. The bes emple o keep in mind is h we cn evlue e d using elemenry funcions. Of course, we could pproime definie inegrl b e d using Riemnn sums, nd his is wh compuer would do. Eercise 6.4 # : Evlue sin d This doesn seem o be in he ble he bck of he book, so we ry firs o subsiue u =. Then du = d, so d = du = u du. The inegrl becomes sin d = sin u(u du) = u sin u du The pproprie formul for his is # 9: u sin u du = u sin u + u u + C 4 4 Thus sin d = u sin u du = u sin u + u u + C = sin + + C

19 Eercise 6.4 # 8: Evlue 4 e d If we hd o do i by hnd, we would inegre by prs 4 imes, decresing he power of ech ime. We recll h his cn be hndled more efficienly by reducion formul, nd he pproprie one is # 97 in he ble he bck of he book: u n e u du = un e u n u n e u du wih n = 4 nd =. Bu noe h we re doing definie inegrl, so we should keep he limis of inegrion hroughou. Plnning hed, we see h we wn o pply he reducion formul 4 imes. So, le s ry wriing he formul for generl n bu puing in everyhing else: Ok, now le s pply his: n e d = ] n e n = n e ] + n = n e + n e + n = e + n 4 e d = e + 4 e d ( = e + 4 e + Eercise 6.4 # : Evlue = 5e + e d ( = 5e + e + = 7e + 4 e d ( = 7e + 4 e + n e d n e d n e d ) e d ) e d ) e d n e d s ime nd ime simplify bi rd ime simplify bi 4h ime [ ] = 4e + 4 e evlue direcly = 65e + 4 simplify sec θ n θ 9 n θ dθ 9

20 This one is poenilly confusing, becuse i looks like rig subsiuion hs lredy been mde bu no good one, becuse 9 n θ does no collpse. Bu we cn mke n ordinry subsiuion, nmely u = n θ, du = sec θ dθ: sec θ n θ 9 n θ dθ = u 9 u du We recognize h i conins he form u (wih = ), nd he pproprie formul is # 4 in he ble he bck of he book: u du = u 9 u + 9 u 9 u sin + C Now subsiue bck: sec θ n θ 9 n θ n θ dθ = 9 n θ + 9 n θ sin + C Secion 6.5: Approimion In he rel world, when definie inegrls re compued, i s no uncommon for hem o be oo compliced o hndle effecively by hnd using he bsic echniques you re lerning here. And some of hem cn be hndled ll using hese echniques (for emple, b e d). Forunely, using modern echnology i s possible o find pproime vlues o mny definie inegrls. We re no going o go very deeply ino his opic, bu you should be wre of few of he bsic echniques for pproime inegrion. Suppose h f is n inegrble funcion on [, b], nd we wn o pproime b f() d. All he pproimion echniques rely upon Riemnn sums. The ones we ll cover use regulr priions P = {,,..., n } of he inervl [, b], where i = + i b n = + i nd he Riemnn sums re of he form n f( i ) i= where he i re smple poins in he subinervls [ i, i ]. (Look bck Chper 5 o review priions nd Riemnn sums.) When you firs lerned bou definie inegrls, you were old h if f is n inegrble funcion on [, b] hen b f() d = lim n n f( i ) for ny choice of smple poins. A very simple-minded choice is i = i = righ-hnd endpoin of ih inervl [ i, i ] nd his gives he righ endpoin pproimion b n f() d R n = f( i ) = [ f( ) + f( ) + + f( n ) ] i= i=

21 Alernively, we could use he lef endpoin pproimion b n f() d L n = f( i ) = [ f( ) + f( ) + + f( n ) ] i= A slighly beer pproimion uses he midpoins of he inervls [ i, i ], giving he Midpoin Rule: b i = ( i + i ) f() d M n = [ f( ) + f( ) + + f( n ) ] Anoher pproimion uses he verge of he lef nd righ endpoin pproimions: Trpezoidl Rule: b f() d T n = [ f( ) + f( ) + f( ) + + f( n ) + f( n ) ] Noe he pern of coefficiens of he vlues of f he priion poins,,..., n. I kes only bi of lgebric mnipulion o verify h T n = (L n + R n ) The reson for he nme Trpezoidl Rule is h, while he lef nd righ endpoin rules cn be regrded s pproiming he re under he curve y = f() by recngles, T n pproimes he re by rpezoids wih slned ops inerpoling he funcion f successive poins,, hen,, nd so on (drw picure). The pproimions L n nd R n re roughly eqully (no very) good, nd M n nd T n re lso roughly eqully good (nd somewh beer hn L n, R n ). The ls pproimion echnique we ll consider is much beer hn ll of hese: Simpson s Rule: b f() d S n = [ f( )+4f( )+f( )+4f( )+ +f( n )+4f( n )+f( n ) ] Here n mus be even, nd he pproimion cn be regrded s using he res under prbols inerpoling he funcion f successive poins,,, hen,, 4, hen 4, 5, 6, nd so on. Agin noe he pern of coefficiens of he f( i ):, 4,, 4,, 4,..., 4,, 4, We re no going o sudy ( ll) he error bounds for hese pproimion rules, bu he mos imporn hing o remember is h Simpson s Rule is much beer hn he ohers. In fc, some clculors or compuer progrms use Simpson s Rule o compue definie inegrls. To cully pply hese pproimion rules, you jus plug ino he formul, nd ypiclly you d compue he pproimion s deciml number wih however mny digis is desired for he priculr pplicion. The compuions quickly ge ou of hnd o do by pencil nd pper (by hnd!), bu hey cn esily be done using spredshee or CAS.

22 I will no require you o do ny cul compuions using hese pproimion echniques, bu I migh sk you quesions on ems h es your fmiliriy wih he properies nd he srucure of he echniques. Secion 6.6: Improper inegrls Improper inegrls of Type : Le f be funcion on he inervl [, ), nd suppose h f is inegrble on [, ] for every. Then he improper inegrl of f from o is f() d = lim f() d provided h his limi eiss, in which cse we sy he improper inegrl f() d converges. Similrly, if f is defined on (, b] nd is inegrble on [, b] for every b, hen b f() d = lim b f() d provided his limi eiss, in which cse he improper inegrl b f() d converges. If n improper inegrl converges, we sy i is convergen. If i is no convergen, we sy i is divergen, or diverges. If boh f() d nd f() d converge, we define f() d = f() d + f() d nd sy h he improper inegrl f() d converges, nd oherwise we sy h i diverges. Noe h for his ls improper inegrl ny number cn be used. Also noe h o invesige f() d we mus (choose some number, hen) consider he improper inegrls f() d nd f() d seprely, nd i does no mer in which order. If he firs one we consider diverges, we cn sop nd sy h f() d diverges. If he firs one converges bu he second one diverges, hen gin f() d diverges. On he oher hnd, boh f() d nd f() d converge, hen we cn sy h f() d converges, nd we ge is vlue by dding. If f is nonnegive we inerpre he improper inegrl f() d s he re of he region {(, y), y f()} Similrly for he oher improper inegrls of Type.

23 Emple: d = lim + = lim n + d ] = lim (n n ) = lim n = π Thus he improper inegrl d converges, nd is vlue is π/. Noe h i ws bi + of nuisnce o crry lim hroughou. Wrning: i would no be ok o jus drop i, nd keep connecing he seps wih equls signs. Do no do his! An lernive would be o do sepre clculion wih f() d, hen when h s been crried s fr s possible ke he limi s goes o infiniy. Emple: A similr nlysis shows h d = π (no surprising, since is n + + even funcion of ). Thus + d = + d + In priculr, d converges. + Emple: Consider e d. For ny, we hve + d = π + π = π. e d = e ] = e e = e Thus e d = lim So gin he improper inegrl is convergen. Emple: For ny we hve e d = lim ( e ) = e d = e ] = e e = e

24 4 Thus e d = lim e d = lim (e ) = since lim e =. Thus e d diverges. For his reson e d lso diverges, even hough e d converges. Emple: Consider Thus Thus d diverges. Emple: Consider nd since p >. Thus d. For, ] d = ln = ln ln = ln d = lim d = lim ln = d for consn p >. We hve p ] d = p p p = p p p, lim p = ( d = lim p p ) p p = p Thus d converges for ll p >. On he oher hnd, similr nlysis shows h he p improper inegrl diverges for ll p <. We sw in he preceding emple h i diverges for p =. Puing i ll ogeher, we see h d converges if nd only if p >. p Improper inegrls of Type : Le f be funcion on he inervl [, b), wih vericl sympoe b, nd suppose h f is inegrble on [, ] whenever < < b. Then he improper inegrl of f from o b is b f() d = lim b f() d

25 provided h his limi eiss, in which cse we sy he improper inegrl b f() d converges or is convergen, nd oherwise we sy i diverges or is divergen. Similrly, if f is defined on (, b], wih vericl sympoe, nd is inegrble on [, b] whenever < < b, hen b f() d = lim + b f() d provided h his limi eiss, in which cse we sy he improper inegrl b f() d converges or is convergen, nd oherwise we sy i diverges or is divergen. If f hs vericl sympoe c, where < c < b, nd if boh improper inegrls c f() d nd b f() d converge, we define c b f() d = c f() d + b c f() d nd sy h b f() d converges, nd oherwise we sy h i diverges. Noe: he book includes he priculr cse h f is coninuous on [, b) bu no b. If here eis m, M such h m f() M for < b, hen i urns ou h we could define f(b) rbirrily nd he resuling funcion would be inegrble on [, b] in he usul sense. Similrly. Thus we only need o consider improper inegrls of Type if f hs vericl sympoe. And his is ll h will occur in he problems. Emple: Consider d. For < < we hve d = ] =. Thus d = lim d + = lim +( ) = since lim + =. Thus he improper inegrl d converges, wih vlue. Emple: Consider Thus d. For < <, d = ln = ln ln = ln ] d = lim + d = lim + ln = 5

26 6 Thus he improper inegrl d diverges. Emple: Consider π/4 π/4 csc d. If we ren creful we migh sy π/4 π/4 csc d = co ] π/4 π/4 = co π 4 co π 4 = = Bu his is wrong, becuse csc hs vericl sympoe. In fc, he resul is bsurd on is fce becuse we re inegring nonnegive funcion. We mus brek he inegrl nd consider he lef nd righ prs seprely. We sr wih π/4 csc d. For < < π, 4 so π/4 π/4 csc d = co ] π/4 csc d = lim + = co co π 4 = co π/4 csc d = lim +(co ) = Thus π/4 sec d diverges, nd herefore so does π/4 π/4 csc d. Comprison Theorem: Le f nd g be coninuous, wih f() g() for. () If f() d converges, hen so does g() d. () If g() d diverges, hen so does f() d. Similrly for oher improper inegrls of Type, nd for improper inegrls of Type. Emple: Consider e d. For we hve, so e e. In n erlier emple we showed h e d converges, nd similr nlysis shows h e d converges. Thus by he Comprison Theorem he improper inegrl e d converges. I follows h e d lso converges, nd e d = e d + e d lhough we do no know he vlue of eiher of hese convergen improper inegrls. Emple: Le p >. Then whenever < we hve p, so. In n p erlier emple we showed h d diverges. Thus by he Comprison Theorem d p diverges for ll p >.