Homework 8

# Homework 8 - miller(dkmﬁtiﬁ M HWOS w hamrick e(57395...

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Unformatted text preview: miller (dkmﬁtiﬁ) M HWOS w hamrick e (57395) This printmout should have 23 questions. wwewé-mﬁv Muitipiewehoice questions may continue on . I :2 3: m w D the next column or page ~ ﬁnd ail choices wwﬁwwﬁw before answering. W 2. I m ~— 001 10.0 points 3 Evaluate the integral . 3. I m 35+ 87?- 1T/2 I : / 51113130082330156. 7T 2 0 ‘ 4. I = m + w , .’ 1 2... > 4 3 5‘T%§i‘m ‘A CAMS i”: 1- - t ‘ e. gﬁ 3,3 14z;_of9;~f'}£e5 3“ ' V“ I 2 .7: W W S 5 I‘m-melt:— “5"” 59°“ '2 7T 3 6. I = ~——-— 3 2 10.0 points t. 4 p 5' I 2 {5 Evaluate the deﬁniteintegrai 802 10.0 paints w I 2/ m(5 c0523: + sin2:5)drrz Evaluate the deﬁnite integral . 0 “M 2 com: ~ 4 sinz [3/0 00532: dm' LIﬁ’n’Z 1 "M3 1 I m — _ m" ‘‘‘‘ d .x' 2 (,M-v- 2 I \W m, , n «M2 2. I m —1 3 3 I m 2 _ 1 -:2 24> be H 1-H H H H I w: E i [Us % 5 003 10.0 points Find the value of Z t I. a _, {X I = tan4\$dzc. w “5 Feb ‘3‘ * . . 0 3M” U ‘ The shaded reglon m J a. M K s 45"" '5’ ‘ 16 (’L Y‘ ‘ San 2 2 «UM-u \ miller (dkmﬁ45) m HWOS hamrick - (57395) 2 (not drawn to scale) is the one below the graph of ﬁx) = 481nm- H on the interval [0, 7r/2]. Find the volume of the SID—solid obtained by rotating this region about the dashed line 3; x ml. 1. Volume 2 8?? m 271* cu.units 8W2 + 871' cu.units 2. Volume 3. Volume 8% m 2 summits ll 4. Volume = 477 + 8 cu.units // /m .--”"g ,..--'/ gig/Volume amp/4W2 + 877 cu.units ./ a “k Emmm .7 6. Volume = 4r ~ 2 cunnits 006 10.0 points Determine the indeﬁnite integral I m feec92artan2mdm. 1 MA KM" Elwin/M > 2.x.wLHm—fN-ig‘eecg 23; y n 1 8 ‘3 ‘ ' I w 3. I w EGan 2r—ln]secZ.rf)-n—C 4. I = 8se69293+C 5. I m 9(sec8 2184—1331118 2:13) +0 007 10.0 points Evaluate the deﬁnite integral arr/3 seextanzr I m / wake- 0 5mseca: 7 1. I zl w n6 2 1"me 4 3. I —— —-En§ 4 I m ~ln2 “M . 6 I — wln: ' " 6 “W /3 E :11) points 5 To which one of the following does the inte- gral \$2 I = / —_— Ce V as? + 1 reduce after an appropriete trig substitution? )(i'ltxmvx (w a , aux Essex: ‘4 Gk??? 1. 1' m f ta1126sec39d6 2. 3 9 sec2 9 d9 17 '1: K: S—ﬂﬁ/ Shell 9S8039d9 4. I m fseegéldﬂ 5. I z: ten36d9 6.1: miller (dkm645) e HWOS — hamrick w (57395) f 31113 9019 009 10.0 points Evaluate the integral 1/3 4 I m WWW—elm. 0 Vilmgazg 4 1. I w 571’ 1 2. I = g M “"7 3. I = 67? ,/ 2 . I 2 — 5 9 4 6. I m — 9 010 10.0 points Evaluate the integral 1/2 -——1 I = / 8m x div. . 9 V1 m :62 / “I_/»‘ -1 = 3‘3 “M’ZQI 2 7r 2. I m wig 2 7r 3. I = ——— 8 2 1r 4. = w I 4 2 7T 5. m —« I 9 011 10.0 points Determine the integral 3 ‘ m (Quit/2 dw' 1. I m “3;? +Sin_1(%3) +0 2.1 m ﬁwmmlgl +0 ( +0 95:2 \$2 + Emil + C 5. I = 935\$ “Mlél +0 6.1 m \$Wsmd<gl +0 012 10,0 points Evaluate the deﬁnite integral 2 2 I m/ {E +2023. .0 4+3? 1 1 1. I m A24?) 2 I 4+1 . : "7r 8 3 I - 2+17r ‘ “ 8 013 10.0 points Determine the integral 1 Im . (mg-F4)? miller (d1un645) ~ HWOS w harmick m (57395) 4 "(vi/f lllll v-r-W‘” m “'mm—LMMmH-“X 1 2.I:\W-——~——+C> _ :m —1 ~ ~ ( 4 x2+4/// 3 I 23111 (a, 2)+C ‘WM _. V\$2+4 4.1m2x/4a7~.L2+C’ .3 I - a: +0 2 m Ml \$+2 w 4:.I: czar/4+0 51 5111(2) C 016 10.0 points Evaiuate the deﬁnite integral 4 8 Im Wail. m2—6x+10 m 1. I m Zﬂ' Evaluate the deﬁnite integra} 4 I 3 1 d 2 I m 871: m “Mm—«WWW»- m_ WWW“ \$2V\$2 — 1 /”/M __ C" I w m? 1. I m x/Ei + x/i M..»...«w’-'”“" 4:. I = 27? E" 1-..4 H 1 1(\/§+x/§) 7 5. I w ‘2—71' 9" A; H 1 §(\/§+\/§) 017 10.0 points 4. I m» ﬁ~¢§ 5.1 m ENE—Mi) 1 4 m" I m f m”- ,“ Lﬂwrmmﬁ m m :,»r-’”"” \ K6. 1° Imln!1/m2+6\$+13_\$+3i / M Determine the incieﬁnite integral 015 10'0 Paints 2. I x In] V132 + 6:1: + 13 —~ 39 + 3 Determine the indeﬁnite integral +0 __ . m1 :13 + 3 I / 1 d 3. I g 8m e). 2 WW 3; ' “vﬂwuwmw— k "may 1/ m -2 \\ I _____ ,7 \$ 4'.” I = lrﬂ :32 + 63: + ---- ~- w — 2 N. M‘w___h__vw,,_‘_,_r 1. I 2 sin”1 + c > . K'N—a ( 2 ) “HM-J 5- I m :ESiIl—‘J-(‘IL—lhg) "'Hn-«r "NJ... 1—“ . u. “HM—“W” ' miller (dkm645) — HWOS m hamrick M (57395) 5 Find the unique function y satisﬁring the equations 2. I = dy 2 \NWMW..W.H.MN,.,.,“,..M_w———-”"'" 1. y a 2(1n‘:”_:_|_m1n4) 4- I m 1*51112 Jaw/“2 m 5. I = 2+51n2 2.” y :2 ~(1nl Jrﬁl—n'z} \M-~§~’M’_ saw—“ax \$.41: 6.I=2+61n2 3. y m 2(111‘ ‘4“1114) 9 —— a: 2 g m 021 10.0 points 4. y = 5011" 4 «1114) “1‘”! \ 1 ‘5 ~ 4 Evaluate the deﬁnite integral (w: c m w n f. m "’ 1 1 ‘4 ._ \«L'L 0 y 5(nt9—1E‘ n [2/ ﬂ dw' -B ,1 m—16 gut-mt 019 10.0 points I {VLF or" “:3 Evaluate the integral 1. I = 41:19— W 1 h 5‘” me W 5 g- (3% J x .. 9' C * IL ‘a? ‘ 2. I w 41:1L — 2 g a; 5 a: ‘,_';§'I.";T( : ,I'IZT; f" 3. I m 1 ~ 41113 2n; We - N3," i-in’u-f “J m»,ﬂ...,,..ue._..mwm._w‘_ M.— M» f/“Zi. I "Wm-m 2 _ L‘WWWWWMMWMWW:V . MM 0% 1 2! h ‘ /a/1 = lint?) 9 5' I = 41113“ 1 K“: 4% “CM-M ./ 4 2 (“W iii“ W<;{"‘givlw 3) I 1} 5 3m 6.1m2~4ln3 l g 3 4' = 5 11%) Wet-W “ he ‘ 1 022 10.0 points “a .i. ' .71 5. 1 m 3111(2) Ma “3, Evaluate the deﬁnite integral 6. I m 1116?) 21:2 3 32, 5’“ 5 I 2 dm. w“ M e . 0 e”: +~ 4 V ! f/«ﬁgx 020 10.0 points W W ‘ 3 - :f; 1., Evaluate the deﬁnite integral - 1‘ I 2 21m 2 5:" a z, E" 1 2__ , 4 5 Lyme I m/ Wm. 2. I m gin— 7\»,\W‘\o ,0 a: wit—2 3 ,ha‘w“ ‘x" t WM 4 w \ 7 «he. 4» '2, _ [-' 1' I m meﬁI—jJZ 3. I -—— (/ Mqlm; ’ “15316 A is; Km ; («~— : .— 13:” -,.r-- a“ k i '22: EU: F1! ‘_ ‘1 \ 1; \‘IV‘: I — (WW/W's: r75“? { \l miller (dkm645) w HW08 w hamrick W (57395) 4 I = 31n2 5 5 I z: 3} — x ”’ 3 5 . 3 f _, / . \ ‘ “MM 023 10.0 points \ié: Find the bounded area enclosed by the :9. "2, graphs of f and 9 when ‘ {Lg w— I ' “fa” _ 25m\$21 g(\$) "m 1 1,..6-w‘5 1“ Eu “‘ 2 g “V 01500 1112) sq.units “i ,, c, 3 + — 3114) sq.units :r 35' - ( ( 3. Area = (3 —— E; 1114) sq.units R\/ (15. ( ...
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## This note was uploaded on 03/24/2010 for the course M 57395 taught by Professor Hamrick during the Fall '09 term at University of Texas.

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Homework 8 - miller(dkmﬁtiﬁ M HWOS w hamrick e(57395...

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