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by: Julia Tang

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# Calculus 2 MidTerm 2 MATH 2924 - 010

Julia Tang
OU
GPA 4.0
Differential and Integral Calculus II

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The test will be over limits, partial fractions, integrals using trig functions These practice Q's came from his previous tests
COURSE
Differential and Integral Calculus II
PROF.
TYPE
Study Guide
PAGES
5
WORDS
KARMA
50 ?

## Popular in Math

This 5 page Study Guide was uploaded by Julia Tang on Monday October 12, 2015. The Study Guide belongs to MATH 2924 - 010 at University of Oklahoma taught by Brady in Summer 2015. Since its upload, it has received 76 views. For similar materials see Differential and Integral Calculus II in Math at University of Oklahoma.

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Date Created: 10/12/15
1211391 I I z ms l 2 Si A Si L z 13032 3511112A z Qms l 1 z 1 23in2 1 3511A I 13633 I 11391 Hyperh li htegi ti by Part3 f d 3 Hi fijdu Invemen Trig 1 1321qu z lial E 1 39 39 RH quot Jun M C d Dm ry it E SEEI 1 E SEEE ta lfm 35233 z E and tanfm E if m 2 1 S d E d d E 3 Tri Sub titu izi m Far Far Far r13 3 me I aiMQ 12 2 1133 m Mum 3 13 me x E E 9 sin 2 11 1 1 cos21 11 31 Dh A sin2rd1 3 2 I see tan 11 Let quotu sec11 tanr and quotu sec1 ta11139 Recall that quotup l and sec1 u quot11 2 and tautr u tr 2 Recall also that sec17d1 duu dvv We get 1 u 13922u 39r22secrdr F u2 2 2 sec1rlr Since 111 1 the square term becomes u 2 39u l and so the integral becomes 1 139 u l 1 1 l 1 m 39uf1 mg f1 39u 1 111 u 39I quot1 C 16 u 16 quotu 64 8 64 Tliuuuiuuu 39 mum Ge memnm A Answer ll since ewpenentials in 1 with base greater then 1 always grew faster than any power of 1 seem in Mel Answer I sinee 1113 always grews mere slewly then any pesitive pewer if 1 again seen in elass Cempute the fellewirrg limit I lirlr1 quotE r w Take legs he begin with i Nli iv 1111 gPemy 39 l lm Ilje By the fem ref l Hepitel e rule we get ml HI l M4 Thus lim 3 slimeshy e 1 EEK PH 339 sin 332 legarithrn e dilferentiatien ln y r r2 lnsin e and se taking derivatives vent e ef heth sides gives it ill y 21 lnsin a 32 i1 E s1nr Thus 2 yquotr 2 sin 31 212 lnsin a 32 set a 39 Lie 2m23 This gave peeple mere treuhle than I expected Yen netiee that there is ne rdzr en the nunteraterT se can t expect a suhs39titutien ef39 the farm a 12 er a 2132 3 tr werk Else you see after a fevv attempts that integratien byquot parts is leading newhere Se it must he semething WE already knew let s seertvhat derivative invelves reeipreeals 33 and asigns vesl the derivative ef tan 1a In fact we remember rent class vvith the merrier jog en the sheet at the end ef the event that d l t 1 1 Il dzr a e erg keg E n r Thus our integral is dual 1 K sEl vimrebels X a equot u l 1s1 iiiF F39 H Ennus nestian Cnnrpnte te lirnit Take legs rst it in lira MET IIFEEE 3 I l 5E2 whieh is lilfI pitalis rule in the II easel te a senses 1 sins as 1 2 quot39 11131 n T 39T39 2 11m 2 I a 211 E I 21 5111 33 sense sine Again Thy lilfI pitalis rule Helease this is in turn equal te i ense esinn enss i sinrr inn 2 11131 tar rt 511 sin s E39s see 1 zz s 11 em a 2s ens s gsin by lilfIepitalis rule IIlllsease this is equal te I ens 1 11m I b SEIfI i EESLHEE 6 Finally lint y e1i as Iv 39 f e23 sint3m is Use integratien by parts with n sin3r and da 2 sharia tn get JEE 1 I u ggm I i r fe m39sin3edc sin3sl 2 EjehS ens 3sjds w Sfe m ens3rdr This last integral is similar tn the nriginal se tr integratien by parts an this integral with n ensi3m and def 2 egi dm to get gt sin3ee2i 3 eET 3 V I a felt s1n3rde stin iiskg d3 Simplifying gives N 21 21 Vv f e23 sin3ndr Emigg 3CDSE E f e tsin sjdzt Near the last integral is identical tn the rst Talking it tn the left side gines 13 sin3se2 3ees3ee2 Em a 39 L n n 2 e s1n3r r 2 1 and an 0 fr39 2E I I Hg m39 Egrsimg dm at s1n3re 3ens3re T3 2 4 Remark Yen could have dinne the integratinn by parts quotwith n eEm and d1 sin3edzr ete Yen should get the same argueE rim Use integratien parts with a 2 and in exhale an that aria Zm ria and n 2 e1 fngexdm 2 age 2 memdm We evalnate the last integral abene by parts again with n E 3 anti in 2 side se its 2 dr and n z This gives V In eldm z 3328 2me jexdar z r e Essex 2e C 2 f a v 9 32 Let r 3sin6 an that dm 2 3ees9rit9 and vquot 9 32 3eest9 We get f 2 dx 5 f 3 sin 6i23ees6dr9 139 32 V We dirl this integral in Qnestien 1 above Thns the eriginal integral is equal to Eh nhnn LHzg f ggc 9 magmas 99 9sin29 39TC2 4 Q paints Find constants 1336 which make the following a true statement 1 a A B39s C 33 9n E a 9 Writing the as a single iraetien and eemparing nmneraters gives Ar29 r2Cr 1 3 By the partial fraetiens algebra above we can rewrite the integral as 1 aria 1 mdm d3 9 n 29 r 29 The rst term is a leg integral the seeend term is a leg integral after making the snhstitntien a z 32 9 and dn2 ndzr the third term is aretan mtegral Thus we get 1 3 1 1 1 3 E l11 1 i 2 9 t V139 C

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