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# exam%202%20green - Math 152 — ExamQ(Sp1ing 2007 ’...

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Unformatted text preview: Math 152 — ExamQ.(Sp1ing 2007) - ’ Sections 7.3-7.6, 8.1-8.2 Name I a Q 3 Score /100 Show all work t receive full credit. 1. What trigonometric identity should be used to evaluate I cos2 6xdx ? (3 points) .2 , /+é05/<9>‘ 066 COXV "'73" 2. Evaluate Isin xcos3 xdx.(6 points) ’0 3 3m “9%; @052);- ws : 35m 2.1—sén2x)@osxd£¢ ’"S (erth- _ Slnizk) wsxeéc lO u:)mx 591:6,»st ‘5 03"“- u") do: "7% it, éa’3 + a ’3 ' iI ’- /I- '- ' '7/L Sl/L'SL /3 5L“ x’ 3. What trigonometric identity should be used to evaluateJ‘cos 2x cos 7xdx ? Do not solve the integral. (3 points) (105mm 71 5 j 6125(90 r [Losing] 4. Evaluate JCde. (6 points) I 3/2. uzwig Sxe+3oLJL u; x U:%(X*3),,LOM (12,)(4'3 \$5éuz’3)u§ujj& 0L4}: (1+3) )L: “2’3 I M staid“ :SZ‘Q‘LH'WQOL“ j X L p73 . 3/1 ‘ 3 (mi/W :Jus. (0%274 ‘éKCXﬁB I)?’ 5/ /5 . r '42 A. ’_ 1 . a amt—amid = gnaw evlzwa * - 5 , 5/1 y’all— -; ’g¥[x+3) 2,766) + a 31 ’1 1'6 1 1B = 9590364 "I 5. Evaluate I dx .(6 points) l gsedfvm1l (1(- 5 4 J sealévl x2 x2—4 Kim 6,2501% Y 97.66%)3’9 509% 36:95?“ a ‘W d/lsjseat mtdf 4/5266 aim" 1,. :2 2 y : beg}: 5— :) asf 7i 5429—9—5th 527,1- d05th 3L X [4’ : ‘I’ Q, ~ ’3' L’ + C, a? 7732’ 6. A trigonometric substitution can be used to evaluate J. Jak— . Write the equation for this substitution. (3 x +9 kﬁmf points) 8x+l dx. (6 points) x2 +x—2 Swot/1c 7. Use partial fractions to evaluate I 2 . . 7x +9 8. Use partlal fractlons to evaluate 3 dx. (6 points) x +3x 2 75+? ,4 eye /~_ ’ _— X (134-3) ; X X24 3 ‘ 7f+9:/)(X9+3§ +~9L(8x+0) ’3 _ :gwmmzmsm {5:6, 0.:(3 . 2B P(x) x(x+l)3 A §—+£_+D “"4” 3L >5“ CHIS2 (leB 9. Partial fractions can be used to evaluate dx. Set up these partial fractions. Do not solve. (4 points) 10. Use a table of integrals to evaluate I JgidLs . Record the number of the integral formula you used. (6 x + points) . 4&3 é Q m 3 {32(31'9'5N3y—4—5 w a r 3 - b, 5 :£(3x—/o>1)3x+s + C 037 3dx 11. Use the Trapezoidal Rule with n = 4 to approximate (6 points) , 3" — 32— 7 J— l x l/\ V "Z," ’ H a? 5; (I) +\$7£ 4322'? 4299 +£(3\> i , :achthé +09:;?L+9?'%.+J§,> , ‘ I (O N .v \$Zi;(l+ %+/+ §+§> : éM/J/7 3 . dx . 12. Use the Parabolic Rule with n = 4 to approxnnate (6 pomts) 53% ‘” 72 (LU)49£(V%)+09£(Q\+9(Li)+~l3(3\> 1 A"; _ ' :._é(l+/7l.%+°p.;L+»lf-.%+éx :am %+I+%+é) ,1; w, H 2 13. List three strategies for integration that have been discussed in this class. (4 points) ﬁdesﬁltobok we or. ideade [Meicjrdﬁon L} par+3 CID/Wag aggém, ayﬂm/aﬂméaér H‘xaonomgﬂa 5M4¢5470K5 65%?me 64,635 #ﬁfgw or Patiiwo pal/Ml)” a 3B 14. Select one of your strategies from the previous problem and explain when it should be used. (4 points) 2 15. prossible, determine hit/121008, x . (5 points) 1"” " 8111 x , 53 L )1. (~ 5: n x) z A = 39%”:— Q ’ 77W ’ﬂszQS‘” :0? / I y n x m ...— w m ‘7’ 5% 1w”? x2 16. If possible, determine lim . (5 points) x->0* x —— sin x v 1‘2 \$22.24...— : +00 v. _,——---—-—- : ‘ . 34 am zasmx 3L 0; [’wsx Mag} 5‘ rap” ” 2 17. If possible, determine lim COS x .(5 points) H0 l—sin x ‘ .2 m lvsmx //52n0 /’ )C’IO . . 3 + 7 18. If possible, deterrnme 11m x x x—>oo ex -1 . 1 L“ (9‘ L ; _ gem Vag-f 7x e}; O ’S/ X300 ‘ 55-470 lav” )l‘990 . (5 points) 4B 19. If possible, determine lim x2111 x. (5 points) x—rO" .1— 73 9:2- _ , ﬂ/n Q . “'2’,— "T b; ' j“— ”" gm 95? x: "T, /&m 6713 7‘ ' m “9 27 0 way v “*0 )5 M7 ’4 20. If pow-321e, determine x2”. (6hpoints) é. am M e d” 5’ S) 5 ,z/ MQ’) : m 2- yes a0 344790 p”\ 2 l . y 559,00 L\¢ﬁ,op;_ a W a?“ 23> o ‘ﬁﬂmx Vw lzmﬁﬂ: :be / V” 1; MISCELLANEOUS ALGEBRAIC FORMS 92 /u(au + by! d” 2% _%lnlau + 1,] + C 93 /.u(au + b)‘2 du = (ILZPHAIIIU. + b[ + a”: b] + C u _ Ll(tll( + b)"+1 (nu + b)"+2 .. 94 /u(au+b)du—W—m+C 1fn¢—l,—2 du. 1 u ' (In 95 “=———- ——-—+2‘—3 —-—- 'f.¢1 /wiw MMﬁKWiWﬂ(n M( )” ~1 a2 2!: Liz)” 96 u Vau + b du = (31m - 2b)(au + bf”2 + C a 97 /Ll" Vtm + b (in = “(Ea—3) (u"(au + b)”2 — nb Lz""1 V all. + b du) u (in 2 u" (in. 2 ' it""l du. 98 —-———-,_=-- —2 V .+b+C 9 ———=———-— "V — a (m + b 3a2 (au b) cm 9 .__au + I] “(2"- + 1) (u all + b nb Tl + b) (In 1 Van + — Vb . du 2 nu + b 1003 h =——ln -———~——— + C Ifb > O 1001) ———— = tan~1 + C ifb < O - u Vau b b Vau + b + Vb u'V 1sz + b V—b ‘17 101 (In Van. + b (2" " 3)" / du "*Tﬂﬁ" ”m[‘+ b b(n —-1)u.""' (2n. — 2)b ” _ 2 _ .. 102 /\/2au — Ltzdu = ” ‘7 Vzau — u? + ism" ” a + C 103 / £— = Sin—l ” ” + C u”" 21m - 112 m 2H + 1 a 104 u" V 2am. ~ L12 (in = ~ ( ) + (———) M" V Zau — 112 (la n+2 n+2 105 =__‘L_l 2a“ _. “z + (2’7 — j)" u""' (In 106 V2011 - £12 - V2rm - uz ’7 I7 -\ QM _ “2 u 107 / ng, — u? l (2011 — L12)3/2 n — 3 V 21m — u2 ~lﬁ ( M .____ (lu = V20u — £12 + asin" u _ a + C (I = + ——_.__ u' u (3 — 2n.)1m" (2n — 3)ll . u (in _ V21”! ~ (I2 + 1'! — J / (lu. - 1.1"V21m — U2 "(1 _ 2”)”, (2’7 — 1)” . 11"" V2014 — u2 . __ I 2 109 / ('V Zuu - LIZ)“ (in = u + (2m: - L12)“2 + H“: 1 /('V 21111 — uz}"_2 du . I n . I ‘ _ '1 —n '— I l 110 /‘dFLT=#L:(V2(IlI~lI”]Z +—"—3—— (In "—1 108 ...
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exam%202%20green - Math 152 — ExamQ(Sp1ing 2007 ’...

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