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Exam 2 - Version 080 ~ EXAM 2 w hamrick ~(57395 This...

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Unformatted text preview: Version 080 ~ EXAM 2 w hamrick ~ (57395) This printwout should have 16 questions. Evaluate the integral Muitiple—choice questions may continue on the next column or page — find all choices before answering. [email protected] 10.0 points Determine 1 . (111202 elimoo 6m + 81110: exists, and if it doee, find its value. 1. limit 2 6 2. limit m 14 3. limit m 00 4. innit = Moo . @ iimit = 0 . 004 10.0 points 6. none of the other answers Determine the integral 002 10.0 points I = / 2(3—21nm)3dm. Evaluate the definite integral ”/3 secxtana: I m ————md:1:. 9 5+28eccc 1. I : ~1na:(3-m21n3:)2+0 2. I = 21nx(3—21m:)2+0 1 9 ’4 1 I m @an 4 \ 2 7 3. I Z —2(3—21m:) +0 » 1 7 w 2- I m *1 "‘ l ‘/ 2 115': —(3w-21nm)4+0 3 I — 21 9 ' “ 1‘16 5. I = ——21na:(3—21n;t)2~%-C 7 2 4 I M —21ng 1n:c(3—~21n$) +0 9 4 5 I m ~21nfi Cam—211103) +0 "\ <91 : gin»: 2(3~21nw)4+0 003 10.0 points 005 10.0 points Version 086 — EXAM 2 —~ hamrick w (57395) Find the value of the definite integral 1 2.1:41mm IW/wvl—m 251nm} 33d ( V3) 1% 3.I=2(\/§w1) m___2 1’1”“:127r 4.]m21n-1~ 2 12 2 1535’” 5 Im4ln2 1-..1 ii N E E0 008 10.0 points Determine the integral I. L1” I : 3W -3; <’ "If! I 1'\IZ : Evaiuate the integral W“‘3 9 / E m": :1 4 d "S 7 {a 4 2 I m —— 3 I — / £33 dt. 34?? it ui—t 1 fl 5 3 F f‘ W ’6‘; 3. I 3 E7? 6 1 1:8(in2_1) wet 23/ A, (in: : J~ {iv-i2" B *- 13 “— WW u 4:»: ‘- :1 (M 4 I m "—7? ’3’: 2 I :2 4(In4—i—1) 1 3‘ SM} 24 C K m , . lj {33.3“_ M U‘VU 5 I : Egg-[- 1? (4??) 3. I : 1:14 “1 M 24 v F f: 009 10.0 points 4 I 2 8(1n2+1) 2 } Mwém mm : «FEW: 01 cl 01"" Evaluate the definite integrai 4 — 1) E r V 3' M \w t 1 \\\10 I 3 —‘ 007 10.0 points ,9in i 2 ”a: , - Evaluate the integral 2.. K1” (L I \0 2 I 2 271* 1r/3 q”K/Id\flvi . O} f 3. 3: 7T — 1 I = / 4mtan(a:2)da: Kwnu‘ ' ”S" 0 {I 4. I 2 7r {1 1.1:4(\/§w1) \ ,,\' 53132 ,. 2 ‘W‘ U": U 71' I 2/ 2msinmdm. 0 Ii Version 086 w EXAM 2 w hamrick — (57395) 010 10.0 points Determine the integral 7r/2 f0 [2 1.126 M.“ \ 16 3.Im——— 3 19 .Im— 4 3 5.125 Determine the indefinite integral i! Ii (3 cos 6 + 4(2033 9) (£69. (-655: (:3): 4- W tfisléx (“5' (342 at ( §~S\mjg may (7 - “admit (A F \\ j 011 10.0 points 012 10.0 points Evaluate the definite integral 1 6 I = d . f“ (1+m2)3/2 m 1. I m 6 3 l 2. I z _2_ p) M K} twig”) 3‘ I z 6% \J § 4 I : Fifi (q m o 2 3; V 3’" a“; (52:, -_':.: 2 r? “17}; \u/ 3f ‘L f»; 6. I I: 3 a“ 013 103 POintg Determine if the improper integral 1 f 2111456033: 0 converges, and if it does, find its value. I 1. I m 401124) 2.1 : 41n2+2 3.1: 21n4~w4 4. I does not convergefij // “My" 2(154 + 1) 4 _Q. 100 points Version 086 ~- EXAM 2 M hamrick w (57395) Determine fyx when flat, :4) m 9281mm- 1. fm 2 ~m2(3 sin my + my cos my) 2. fyw : ~2y2(3 cos my +- my sin‘my) El 2m2(3 sin my —- my cos my) my2 (3 sin my + my cos my) 332(3 cos my w my sin my) ll y2(3 cos my _. my sin my) enclosed by the graphs of m ~2m2(3 cos my 4» my sin my) 8. fym : 2y2(3 sin my — my cos my) 015 10.0 points Find the volume of the solid lying under the plane 2: 7» 9~m~2y 1. I = 2111}: and above the rectangle A={(m,y):1§m52, 09,132}. 13 2. I = 111—” 9 1. volume m 12 cu. units _ 17 2. volume = 15011.11mts 3. I = 1113“ dolume m 11 cu. units ‘ . 13 4. volume r 13 cu. units 2 2111—33» 5. volume x 14011. units . 14 016 10.0p01nts 5. I x Zln—g» Evaluate the double integral 8m 14 I = ————d d. . m w— //:4 36+y2 y m 6 I ln 9 when A is the shaded region in “MW—”‘1‘ O '1‘; 472. M '32 3§Cvsfi6 + U fit)? 16 (9‘6 0 c7 '4 COS-QCCGS 6) “ C,G5€Cf“5:m1€> T "’ 1" 3f 73’ iii c0513 Casfa‘Sw 19 o (as '6) (056 gm .336) “it: 0 “IE: 0 M —: Siwfi 3 Jam-6 t qsmaef 02A,“; saw 0&6 O l G' .J L“, f 2L u M 3(1’0) L'EC§”G) o z a. "1’ i «*2 3L3“ 0’ J) _ W %’ Lg w— i ”(g->0 «c5 ,, _ 3 :23 a: J .11 GI) j? J M» «333‘ 3 3 " ‘3 _ hm J71. _ 7/ "m 5 w a W” M m K i=6 J: «x 1“}, K: > 22"“ J 3: $3,; WM. _, M'H’ m .5 J 6'; QW- Q “14’5“ JrL *~ N ,cw L'(§»,h\ C: fifi/ . ‘ , y: mtm «SUN 1 A E) Q \L JJJfl-M'S. M 7‘ «rm + W 0“}. if} I [‘Ji-‘v-PL ”kg-.1 KL: 3 AJ » + em! if» Mr:- ACsAvW‘V BOW") 3 f4“) "fink-ma ,4— ;- F ‘1 PC ‘1’ {2' ”on v ’” Q53 J)“ L {4 \ : pm??? $3.?— & z W \ 01313: (73" f UV? \ V J x {aw/~71" J, WW WM 6 .a~ _ i C ”7.. . «+4 -——~“—"" ”‘1’ J\ ‘N (A \m a” 1 ~ \“ l3: Luge (\2 “a Lg-m‘kiy ‘71 A — i pf; 7:" 3 \I[ w‘} '15 If"? :3 a E ? 3: ”3‘ {m {.3 " “‘1' ‘- :: , _..— ‘ I i , t“ ‘3 4 JJ 5.. 7- \ 55A h C". ’ p} l (90') m; ”k ‘3 3 2 WW ‘ “'*" 1 \y- 1<_. 5sm2£31fi>4~ Li (0335‘? 1% 1 1% (3033(9) .J km; .6 1' 'SJ 4 X g 1 W L‘ Wm 7. < 5:» 5 g L] $.92“ 55W; '3b#7 C3 [AIS ‘11“‘(1‘Mmy 9? -F\ ’2, m M 31%? gm 1 Mflxu : am PW ‘ t") {a ./ ...
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