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Homework 7 - miller(dkm645 w HW07 w hamrick w(57395 This...

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Unformatted text preview: miller (dkm645) w HW07 w hamrick w (57395) This print—out should have 22 questions. Multiple—choice questions may continue 01:1 /- @ 10.0 points the next column or page find all choices . w before answering. Evaluate the definite integral 001 10.0 points Determine the integral 004 10.0 points Determine the integral 1 M1 "f :tan 3(33—7) +0 ‘5‘ lfltarrlw 0 1 + 38 52112151 3(z' — 7) + C 7T ‘ 1. I z 1—6- U ‘"I"&“"{§J f L )4; 002 10.0 points 7T \ (I: ) Determine the integral 2 I 2 g lie“ “ (It *1 r 71.2 4.13%; . ., O 41‘ 5 a, I m / "——-——d$ . 3- I : "_' {fl \JHDr‘MEfl urge-1:. . 0 4 —- a: 16 j A g r;- gu‘ 4 I = m 1. I W —5- \\ 3 W2 , \ 5. I m- -——— z‘ ‘ 5 ‘ 4 / 2- I = ‘6‘ \M “““““ // ' I “2 5 6. w: w 5 005 10.0 points 4 I 2 Ti Determine the integral (W. » “WE; m3; 4 _ {L m m I m —-————— do: 6.12w7r 1.132tanmlxmgux/1m—m2—i-C miller (2112111545) — HWUTr’ hamick — (57395) . f(:r,-)F(:I;) 2. I = 4tan“1:cm\/1—sc2+0 >< 1:115:11 9(93) ’ 1 are indeterminate forms? 22" ~ 43111—1162!— —W+ C 2 1. A anti B only W” m. .BWW was" 451nmlw+v1~x2+0 “” "‘1 :1 x“, (:::2 B2222 5. I = 23in‘1a3—V1—332+C’ 6. I w 2tan“193+\/1——J;2+C 006 10.0 points 3. none of them 4. A and C only 5. all of them Determine the indefinite integral 6. B and C only 633; I m f1+66md$3 (m<0). '7. Conly “WV“...u.1,.1.72-2-2“""”"mmw—Imuhhwmm“‘~-2\§ ,Wl I z .1. 310mm 6321,; g) 8. A only ngwu’mww W 2. I = efixv’ 1 m 653: «+- O 008 10. 0 points Determine if // " z m 613 /’,‘.2""' “““ ’ 3 I V 1 8 “l" O /: - 5111m1(333) ,2 hm _1 1 arm->0 tan ‘ (533),/ 4. I “22 marcsin 6397 + C \ ,2 3 exists, and ifi-tdgg‘gghfipd its value. 1 I — e333 . . . 5. I = —ln 3 + C 1. hunt does not emst 3 +6 :2, 6% fl/ .1 «IOOPOifltS 2‘ 11mm 2 3 23-2 pkg/{WWW {ii 3. limit m 5 “M WW1. $2; When f, g, F and G are functions such that 5 4. limlt : 7" hmflmlm Bmm2)=m 3 W1 “*1 ‘ 5. limit 2 0 lim F(a*:) = 2, lim (Km) 2 00, :1; 222+ 1 m -> 1 _ _ 3 .' limit m —5- which if any, of \ 11119611009 100 points 12-41 f(cc ’ Find the value of . “9902) (:03 21; ~22 cos 3:1: ' B. 11m , ' —-————————-—- . ("V “fl .6133) $11599 7x2 ‘3‘... ...-., .. “ . 1.5"“; . $252 '3'" 25W” miller (dkm645) w HWO’T m hamick — (57395) 3 wfimg = ——oo 6. limit 7 0 .i'ir 5 WWW {312 10.0 points 3. limit = 3% Determine if 2 2 . . 5 1 (W ) 4. hn’llt = ~7- $1312 ln($—— 1) :L-~2 5 exists, and if it does, find its value. 5. limit 3 ~77— 1. limit = 2 limit —- —5— . I 14 11111; = 1 010 10-0 130111138 3. none of the other answers Determine . . 4.1111(11’5 = 0 65:13—- 65 9311331 ”may; 5'2") 2: 5.11mit m ——oo 6.111111%; 3 +00 013 10.0 points Determine if the limit W i 9 7L. 1 ~ 2 " m $113+ (1 33:) A a} / ,3 . . . . [17% w» / masts, and if it does, find its value. v/ , A El 3 a: limit $4“! .limit does not exist .0 points 3. limit Z6 3:5; ‘ «x/ W Determine L 0‘5 4. limit m e2 MM M e w} ll ('0 i C» \F lim :1: (4 711111;) We? ”Lg _ 7L ‘ 5. limit = (3% g/gfl wwmx; M 2’“ ‘ . . I \WW 1. none of the other answefsx ' 6. hunt = 1 \ 2. limit = ,3 7. 1mm, = 5% 3. limit 2 oo 8. limit m 00 4. limit m —7 014 10.0 points miller (dkm645) w HWO'? —— hamrick w (57395) 4 Evaluate the definite integral (ii) Find the volume of the solid generated 1 by revolving this region about the (weeds. I = f (5932 + 2):? dm. 3 2 9 ‘ l.4valume m 36(1 + (1n 3) ) + 2 1. I m 76 -1- 12 2. volume : m3(ln3)2 — 211' 2. I m 56+8 @ume : 31re(1 + (1113)?) W 211' 3 I = 5e+ 12 4. volume = 6(1 +(lr13)2) +2 4. I = 56 W 12 5. volume = 37re(1 + (1:13P) + 2% (E I = 76 f ::3 6. volume = 8(1 + (11:13)2) - 2 6. I m 76 —- 8 017 10.0 points 015 (part 1 of 2) 10.0 points Determine the indefinite integral The shaded region in I = /(4m+ 5) cosQazdzv. V W “\Tk f} . 1;?“ - :2 K ’ afi‘w ‘V _ . . . .._. more" ............ . , ~ 77 , ‘___.__.._N, ,W'M “we“ 5 c 2 Imeos2mm2m sin2m_ —2— sirl2m+0 gmnflu-n-h--—-'- ' '. 5 3. I m 2x0032mw0032m+§ sian+O 5 4. I M 2:13 (3032a: - 60823: w» :- sin2x+ C‘ is bounded by the graphs of 3 ymlnm, 9:0? $m35' 5.I=c032m+2msin2m~::—sin2$+C (i) Find the area of this region. _.,,If.‘“.“-- my.:'.:.;'§.j.‘;'j;......m.«::::3_3 M 018 100 DOthS 1. area m 36 1133 m g Evaluate the definite integral 2aree = e 1113 -i- 1 JI/ 5x 2 1 I :2 / 681n(5\/E)dm. 3. area 2 8—1 '0 4. area = 361fl3*1 1. I = §(%sin5—cos‘,5) emB—l 5. area 2 2. I m }Bw(cos5 ~— 35—31115) t 6. area = 361113 5 . 1 016 (part 2 of 2) 10.0 points 3' I m ”5' (31115 ”l" “5"" COS 5) miller 611011645) ~ l “(E sin5 +3035) 019 10.0 points Determine the indefinite integral LI: 9" h.‘ I! I 4 ~——e 5 4 5 2 g8 a) 2 _. 4 5 2 mt!) “We 5 f 6% sin 23: da: . 5"; (cos 23: m— % sin 2:: (cos 2:1: m % sin 2:1: 1 (cos 2:5 + 5 sin 23; 1 (cos 2% + is 851123;) + C ' mem (cos 2:13 W 2 sin 2x) + C "$ (cos 23: + 2 sin 23:) —i— C W )+0 )+0 HW07 ~ hamrick ~— (57395) 5 Evaluate the integral “fr/4 I m/ (3wm2)sec2a:d:c. 0 3 ' 3 / 2 21-71“ + 2 __ "— 2113 W2/ fig; , M = §n~ + 2 w 3 2 / A 2 3/0 \9 = ~37? —§~ 2 +23 \VU /’f 1.5L.Lma__§1rw 2 + 31H? 6.I 2m 0 2 war — 2 m '- 022 10.0 points Evaluate the definite integral 1 I 2 f0 a:f(:13)d:r when 020 10.0 points Evaluate the definite integral 021 4 :1: t —1 [a an 4dr; 10.0 points 1 1 1 : ‘51—‘54 wgf”($)d l 1 1 3 w 2 f2" ‘” 76 0 33 f (@111: l 1 1 2 r “W” ”3“ '2' 4) a: f (:3) dm mfi'i'g/O $f($)dmmg ...
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