AP Calulus BC Problm Drill 6: Indtrminat Forms, L Hopital s Rul, & Impropr Intrgals Qustion No. of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd.. Evaluat th it (if possibl): sin (A) (B) Qustion # 6 (C) 6 (D) (E) B. Inorrt! Rmmbr you must hk vry tim L Hopital s Rul is usd whthr or not th nw it is an indtrminat form or not bfor using L Hopital s Rul again. Fdbak on Eah Answr Choi B. Inorrt! Th drivativ of os in th sond stp is sin not sin so th sign will b ngativ /6 at th nd. C. Corrt! Good job, you ndd to us L Hopital s Rul thr tims to obtain this rsult. D. Inorrt! Don t forgt to inlud th dnominator and b arful or th signs. Cos, so os. E. Inorrt! Don t forgt to inlud th dnominator whih is 6 whn th final appliation of L Hopital s Rul is applid. To valuat this it w first idntify th it form. W not that as, and sin so w idntify this as a / indtrminat form. Consquntly, w an apply L Hopital s Rul. Hn, d sin d d d ( sin ) ( ) os On again, w noti that as, and os so this is still a / indtrminat form. W an apply L Hopital s Rul again to produ d ( os ) d sin d 6 ( ) d On again, w noti that as, 6 and sin so this is still a / indtrminat form. W an apply L Hopital s Rul again to produ d ( sin ) d os d 6 ( 6 ) d Taking th it yilds os os 6 6 6 sin 6 RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. of Instrution: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd.. Find th it: ( ) π tan ln sin Qustion # (A) (B) (C) (D) os sin (E) Th it dosn t ist A. Inorrt! Don t forgt to us th hain rul whn diffrntiating ln(sin ). Th drivativ should b (/sin ). os. Fdbak on Eah Answr Choi B. Corrt! Ni work! You hav idntifid this is a indtrminat form and thn transformd it to a / indtrminat form so you an us L Hopital s Rul. C. Inorrt! Don t forgt w hav ln(sin ) not just sin. This would bom os whn diffrntiating and thn whn L Hopital s Rul is applid you would obtain th it of sin whih lads to on th it is takn. D. Inorrt! Don t forgt to tak th it aftr applying L Hopital s Rul. E. Inorrt! You would gt this answr if you applid L Hopital s Rul to th produt tan. ln (sin ). Rmmbr L Hopital s Rul only applis to / or / indtrminat forms. You must hang to on of ths forms bfor applying L Hopital s Rul. To valuat this it w first idntify th it form. W not that as π, tan and sin so ln(sin ) so w idntify this as a indtrminat form. Consquntly, w first transform this to a / or / form so w an apply L Hopital s Rul. Hn, using tan /ot th it boms ( ) ln sin ( ) π tan ln sin π ot Now as π, ot and ln(sin ) so this is now a / indtrminat form. Applying L Hopital s Rul produs d ln( sin ) ( ln( sin ) ) os d sin π ot π d s ( ot ) d Simplifying using th fundamntal trigonomtri funtions yilds os sin π π os sin sin Now tak th it. ( π ) ( π ) ( )( ) π os sin os sin : π ( ) tan ln sin RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd.. Evaluat: sin Qustion # (A) (B) 7 (C) (D) (E) A. Inorrt!. Also, b arful whn hking th it aftr using L Hopital s Rul that you obtain / and not just in th numrator. Fdbak on Eah Answr Choi B. Inorrt! B sur whn you us L Hopital s Rul for th final tim that you diffrntiat as not. C. Inorrt! B arful th drivativ of sin is os not os. D. Corrt! Outstanding! You hav sussfully usd th algbrai pross of finding a ommon dnominator to hang this it to a / indtrminat form allowing L Hopital s Rul to b applid. E. Inorrt! Rmmbr as thn not and w don t hav. W first not that ( ) sin and as. Hn, w hav an indtrminat form. Using th ommon dnominator of sin w transform this it to Simplifying produs ( ) sin. sin sin sin sin W now not that as, form. Applying L Hopital s Rul produs, and sin sin 6 os sin sin os whih mans w now hav a / indtrminat 6 os sin os Again, w noti that as, 6 os sin os indtrminat form. Consquntly, w apply L Hopital s Rul again to produ and so w still hav a / 6 sin os os ( sin ) Taking th it yilds 6 sin os sin 6 6. () ()() sin RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. 4 of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd. 4. Evaluat: Qustion #4 (A) (B) (C) (D) (E) Fdbak on Eah Answr Choi A. Inorrt! B sur you don t missing taking th final stp of dtrmining that atual it and not just th it of th logarithm of th funtion. B. Inorrt! Th it of ( ) / as is qual to so this annot b tru. C. Inorrt! B arful to simplify th final quotint orrtly to and not /. D. Inorrt! B arful that you diffrntiat / auratly as / not / or th sign will b inorrt. E. Corrt! Wll don! You hav mastrd th logarithmi rquirmnts to produ a / indtrminat it form to whih L Hopital s Rul an b applid. W first not that as w larly hav an indtrminat form. W us logarithms to transform this it. To do this w assum th it ists and is qual to y. That is, y W now tak th natural logarithm of both sids. That is, ln Using th algbra of its and ontinuity allows this to b simplifid to ln p Using th logarithm proprty lna plna this boms ln Writing this as a quotint yilds ln Consquntly, th it is now a / indtrminat form. W apply L Hopital s Rul to produ ln Taking th it produs Taking of both sids to find y lads to That is, y ln y y RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. 5 of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd. 5. Find: Qustion #5 Fdbak on Eah Answr Choi (A) -6 (B) 6 (C) 6 (D) (E) A. Corrt! Ni job! You hav orrtly dtrmind th indtrminat form as and auratly applid logarithmi thniqus. B. Inorrt! B arful to orrtly intrprt th find rsult of taking th it and th ponnt laws. Rmmbr, -6 / 6. C. Inorrt! Mak sur you omplt th final stp of finding th it rquird and not th logarithm of th it. D. Inorrt! This is an indtrminat form sin if you attmpt to tak th it dirtly th rsult is, so indtrminat thniqus must b mployd sin th it an not dtrmind dirtly. E. Inorrt! B arful not to misintrprt th it. This is an indtrminat form sin if you attmpt to tak th it dirtly th rsult is, so indtrminat thniqus must b mployd sin th it an not dtrmind dirtly. W first not that as, but / and w larly hav an indtrminat form. W us logarithms to transform this it. To do this w assum th it ists and is qual to y. That is, y W now tak th natural logarithm of both sids. That is, ln Using th algbra of its and ontinuity allows this to b simplifid to ln p Using th logarithm proprty lna plna this boms ln Pulling th onstant out and writing this as a quotint yilds ln( ) Consquntly, th it is now a / indtrminat form. W apply L Hopital s Rul to produ ln( ) ( ) ( ) ( ) 6 ( ) Taking th it produs 6 ( 6)( ) 6 ( ) Taking of both sids to find y lads to ln y 6 y 6 RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. 6 of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd. 6. Evaluat th impropr intgral: ln d Qustion #6 Fdbak on Eah Answr Choi (A) (B) (C) (D) (E) A. Inorrt! Chk your intgration. Also, a ngativ would suggst that th ara undr th urv f () (ln )/ is infinitly ngativ! B. Inorrt! You should noti that th funtion (ln )/ is ontinuous on [, ) whih is an infinit or unboundd intrval. Consquntly, this is an impropr intgral of Typ Cas. Chk your intgration. C. Inorrt! This answr is suggsting that thr is no ara undr th urv f () (ln )/. A skth of th urv should rval this is not tru. Th qustion rally is Is this ara infinit or finit? D. Corrt! Outstanding! You hav orrtly idntifid this as a Typ Cas impropr intgral problm. E. Inorrt! B arful to arry th ngativ sign through th intgration pross. Also, a ngativ would suggst that th ara undr th urv f () (ln )/ is ngativ! To valuat this intgral (if it an b), w first idntify th typ of impropr intgral problm. Sin th intgral is to b dtrmind ovr th infinit (unboundd) intrval [, ) and th intgrand (ln )/ is ontinuous on th intrval [, ) w idntify this as Typ Cas impropr intgral. If th impropr intgral onvrgs th it will ist and will b a finit numbr. Sin this intgral is a Typ Cas impropr intgral w hav ln ln d d W nt valuat th indfinit intgral (Omitting th onstant of intgration). ln d ln d W intgrat this intgral using intgration by parts with u ln, du d, dv d, and v. ln d (ln ) d ln ln ( ) d ln Hn ln ln ln d d W valuat this using th Fundamntal Thorm of Calulus ln ln () ln Taking th it produs ln ln Th rmaining it is an / indtrminat form so w apply L Hopital s Rul ln Th intgral is onvrgnt. ln d RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. 7 of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd. 7. Evaluat (if possibl): d Qustion #7 (A) π (B) π (C) π 4 (D) (E) A. Corrt! Ni job, you hav idntifid th typ of impropr intgral problm as a Typ Cas impropr intgral sin th intgrand ( )/( ) is ontinuous on th intrval (, ). B. Inorrt! B arful of your signs partiularly whn intgrating and thn rombining th two intgrals. Fdbak on Eah Answr Choi C. Inorrt! Rmmbr th original intgral by dfinition onsists of two impropr intgrals. Th valus of ths two intgrals must b addd to dtrmin th valu of th original intgral. D. Inorrt! This suggsts that you think that th dominats but th funtions ar dfind on th infinit intrval (, ) so it annot b dtrmind what th valu is without using impropr intgral mthods. This is a Typ Cas impropr intgral. E. Inorrt! This suggsts that on of th impropr intgrals is divrgnt. You nd to hk your intgration and th valus of thos intgrals whn applying th Fundamntal Thorm of Calulus. Both intgrals ar onvrgnt and hn th original intgral is also onvrgnt. To valuat this intgral (if it an b), w first idntify th typ of impropr intgral problm. Sin th intgral is to b dtrmind ovr th infinit (unboundd) intrval (, ) and th intgrand ( )/( ) is ontinuous on th intrval (, ) w idntify this as Typ Cas impropr intgral. If th impropr intgral onvrgs th it will ist for both th impropr intgrals that form this intgral and will b a finit numbr. Sin this intgral is a Typ Cas impropr intgral w hav using th onvnint hoi a, d d d W nd to first intgrat th indfinit intgral (Omitting th onstant of intgration.) d d W intgrat by substitution and ltting u, so du d. Th intgral now boms du d d u ( ) ( ) Intgrating by using u tan θ with du s θ dθ, and th trigonomtri idntity tan θ s θ yilds du θ θ θ θ θ θ s d θ s d tan s θ d u W now nd to rintrodu u using u tan θ or θ tan - (u). But u allows th rintrodution of. Consquntly, du d ( ) tan u tan u Now valuat ah of th intgrals making up original intgral sparatly π π d tan ( ) tan ( ) tan ( ) 4 4 and b b b π π π d tan ( ) tan ( ) tan ( ) b b b 4 4 Putting th two intgrals togthr shows that th original intgral is onvrgnt and π π π d d 4 4 d π RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. 8 of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd. 8. Dtrmin th onvrgn onditions in trms of p for th following intgral. Qustion #8 Fdbak on Eah Answr Choi (A) Th intgral is onvrgnt for all valus of p. (B) Th intgral is divrgnt for all valus of p. (C) Th intgral is onvrgnt for p and divrgnt for p <. (D) Th intgral is divrgnt for p and onvrgnt for p <. (E) Th intgral is divrgnt for p > and onvrgnt for p <. p d A. Inorrt! B arful to tak into onsidration th fft on th intgral of th ponnt p whn p <, p, and p >. Considr th p as first and you will s that th orrsponding intgral divrgs. B. Inorrt! B arful to tak into onsidration of th fft of th ponnt p whn p <, p, and p >. Considr th simpl as whn p as and you will s that th orrsponding intgral onvrgs. C. Inorrt! B arful that you intrprt th rsulting intgral ass orrtly in trms of onvrgn and divrgn. D. Corrt! Outstanding! You hav rognizd this as a Typ Cas impropr intgral problm. Furthr, you hav disovrd that important fft on th intgral of th ponnt p whn p <, p, and p >. E. Inorrt! B arful not to forgt th important but simpl as of whn p. To dtrmin onvrgn for this intgral w first must dtrmin th typ of impropr intgral problm that this rprsnts. Sin th intgral is to b dtrmind ovr th intrval (, ], and th intgrand / p is ontinuous on th intrval (, ], w idntify this as Typ Cas impropr intgral. In trms of p w an onsidr thr diffrnt snarios that afft th intgral du to p bing th ponnt on. Ths snarios ar whn p, p <, and p >. W will onsidr ah of ths ass sparatly. First, suppos p, th intgral boms Intgrating produs d d d d p ln ln ln ln Consquntly, w s that th intgral divrgs whn p. Now suppos th p, Intgrating produs p d d d p p p p p p d p p p p p p If p <, thn whn th it is takn, / p, and w hav onvrgn. Consquntly, p d p If p >, thn whn th it is takn, / p, and w hav divrgn. Consquntly, p d p d p if p < if p RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. 9 of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd. 9. Evaluat: d Qustion #9 Fdbak on Eah Answr Choi (A) / (B) (C) / (D) (E) A. Inorrt! Two impropr intgrals mak up this impropr intgral. Mak sur you add th rsults of ah of ths intgrals to dtrmin th final rsult. This answr is th answr to on of th impropr intgrals. B. Corrt! Wll don, you orrtly idntifid this as a Typ Cas impropr intgral. An impropr intgral an onvrg to. C. Inorrt! Two impropr intgrals mak up this impropr intgral. Mak sur you add th rsults of ah of ths intgrals to dtrmin th final rsult. This answr is th answr to on of th impropr intgrals. D. Inorrt! This suggsts that on of th intgrals that this intgral onsists of is divrgnt. You nd to hk your intgration and valuations whn using th Fundamntal Thorm of Calulus. E. Inorrt! Chk your intgration that you hav intgratd orrtly an not omittd th onstant that oms from dividing by th ponnt. First, w obsrv that th intgrand in th intrval [, ] has a disontinuity at. Thrfor, w idntify this as a Typ Cas impropr intgral problm. Consquntly, and th intgral boms d d d or d t d d t r W now onsidr th following indfinit intgral (Omitting th onstant of intgration). Ltting u and du d w hav d du u du u Intgrating and rplaing u with u produs u u du ( ) ( ) W now valuat ah of th impropr intgrals using th Fundamntal Thorm of Calulus, that is t t d ( ) ( t ) ( ) ( ) ( ) t t t and d ( ) ( ) ( r ) () ( ) r r r r r W now rombin th intgrals. Consquntly, d d d d d t r t Hn, sin both its ist (both intgrals onvrg) th impropr intgral onvrgs. d RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd
Qustion No. of Instrutions: () Rad th problm and answr hois arfully () Work th problms on papr as ndd () Pik th answr (4) Go bak to rviw th or onpt tutorial as ndd.. By prforming a suitabl onvrgn tst what an b said about th onvrgn or divrgn of: π d sin Qustion # (A) Convrgs for all ral. (B) Divrgs for all ral. (C) Convrgs for all in th intrval [, π ]. (D) Divrgs for all in th intrval [, π ]. (E) Convrgs whn is in th intrval [, π 4 ] and divrgs whn is in th intrval [ π 4, π ]. Fdbak on Eah Answr Choi A. Inorrt! B arful to idntify th intrval that you ar applying th onvrgn tst to as it annot b applid unlss all th onditions of th tst ar satisfid. Idntify ths onditions first. B. Inorrt! B arful to idntify th intrval that you ar applying th onvrgn tst to as it annot b applid unlss all th onditions ar satisfid. Also, hk th onvrgn rsult for th intgral you ar using as a omparison. C. Corrt! Wll don, you idntifid th Dirt Comparison Tst to us and th intgral of th funtion you nd to us as a omparison. D. Inorrt! Chk th onvrgn rsult for th intgral you ar using as a omparison. E. Inorrt! Idntify th onvrgn tsts you should us as th Dirt Comparison Tst and th orrt funtion that nds to b intgratd for th omparison. This should b g ( ). To tst for onvrgn w an us th Dirt Convrgn Tst whih stats that if two funtions f and g ar ontinuous on a losd intrval [a, b] and f() g() for all in [a, b], thn For this as a and b b a f( ) d onvrgs if g( ) d onvrgs b a π. Also sin, sin whn π thn w hoos g ( ) and f ( ) sin and s that f() g() whn π. Consquntly, th Dirt Comparison Tst an b applid. W first hk th onvrgn th Typ Cas impropr intgral d d d d π π π π ( ) ( ) ( ) ( ) Intgrating using th Fundamntal Thorm of Calulus produs π π ( ) d π Taking th it yilds W onlud that π d π π onvrgs So by th Dirt Comparison Tst sin d d onvrgs thn sin π π onvrgs π d sin onvrgs RapidLarningCntr.om Rapid Larning In. All Rights Rsrvd