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06pf-solution - SOLUTIONS to PRACTICE FINAL 1y" 2...

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Unformatted text preview: SOLUTIONS to PRACTICE FINAL 1y" 2 5 (30395.12) ~l— sin :5 cos(c052 :5} 2‘ 11 [ea cos(e5)d.s 2 (let; u z 62") m sinks") + C b [3(8 — ezhdcc = [012.13 — me‘” = 6322/2 — we“ ~§~ 6:916 : 13/2 ~ 1 c‘ f WM 3 (let 11 =§nrc} =1/31n311;+C d j; 1W1: = {let u =1+12)= 1/3(1+t2)3/216=(5\/5 .1 11/3 e. I sin(3:1:)e2mdz = 1/13823{w3 cos(3m) —i— 2si11(3;1:)+ O f‘ [12 @dm = (let: :1: a sec 8) = f tam2 Brig = [sec2 —}.a'9 m $11110 w 0 m 1/11:2 — 1 ~— sec"1m|§ = x/g— 11/3 g. {51112.13 cos3 soda: $091.11 2: sin Mm 1/3 si113rc —1/551n5m + C 111 I ma 2 [W (let m+2=2ta120 or {11: (m+2)/2) :1/2ta11“{3%2-)+C f:*+5:1=-GW1/7IM—G_fi=I/7(lnl$+6lwlnlrull)+ ' 1 _ 1 __ _ Jlxi—ummdm m“ [m — tan 1(z -3)»+C 1:..[01 :1: In mdn: = CAREFULH! 1 lim m111z= lim 1/21;2 (1113 ~1/2)1§ 2 lim «~1/4— 1/2t2(lnt—- 1/2): —1/4 1—404- 1—411 1—0+ For lim —1/2t2§n 1mm» write Int lim —1/th t—wU‘l‘ and use L’Hospital’s Rule t—l M Ill—3351“ —1/2w2t'—31 = 0 °° : 1 1' I1 v'm7+3:3+7m7d'> — [.100 W: floor" WW Diverges 3‘ W = ””1600 w dem m 47560ft lbs 2 4;; yr ; J75? ‘1 1441‘“ 1+ (\A/EEW Dada: = Maw/“HG m 124/5 b SA =f11“ 27m\/1+ (x/x/E — mm m sn/gmgf‘ilifi = 40887119 .4. By trapezoidal rule: m 1/2 * 1/4(sin2(0) + 2 sin2{‘ 25) + 25in2(»5} + 2 sin2(\75) ~+ sin3(1))z.2774 By midpoint rule: m 1 /=1{sin2{ 125)+s§n2( 375)+sin2(.625)+sin2( .875) z 2703 e. «1342 — (4 m M)? m 24:21 — 3/2e23/315 = 24a“ — 3mm — 45/2 ...
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