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midterm-solution2

midterm-solution2 - 1 Sketch the region bounded by the...

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Unformatted text preview: 1, Sketch the region bounded by the curves y = 2:1: + 1, y = 1 and a: = 2 clearly marking the coordinates of the corners and find the volume of the solid obtained by revolving the region around the w—axis. Express your answer as a fraction. (10 points) a 7621wa -031] olx ,fl 7:- . ., ~71 SO @XZ+Q)(>CLK fi . 2. E l t . va ua e / 1 + fidx (10 pomts) 3. Use integration by parts to evaluate /(1n 2:)2dx. (10 points) a 2' 9 ob< ‘ , ' u (M w x, 3:3 mm- xmi ngmx 3) ML: lUZ/«Mbg \r I: . .3...“ >‘ 2 , ; XQ’V‘X)” 251/“ A?“ [La QMX / AVIS-(AX L & h . A »_’_<, 59> Aw=gbg / “3% j a. xQZ/Mc) QKmeJrQSX X “X x r meJ’; ZNQMX‘JVZX '**C- 4. Evaluate /e+1 1222—1152—5514-ldx‘ (10 points) 27‘“1 1: .14; 4 E) @«D1 7‘”\ CK“; ‘ e-H QH :5 ZKWK :1, MK-O '+* B ‘ X—ZL '3) I: - V 64H x20 2:) «\2 ~—A+\ : WNW-”MAS 5-17 A :1 Z k x: 2/ H , rrrr "1 _I_, a - x - 2&35 Q \ 6 +l r 2 E Jr ‘ . 5: 3 fi’L 3 €\ 5. Evaluate/e 3°° OS%()sin(~a: 2)d:2:. (10 points) LL 2 Sam (3:: g :3 3163:10ng 9%ng 1 M 3 h 3: e; + C 3’ ea) A ZCOS L _, r}, {a '\' C 3 p 6. Evgluate the limit $13131 W. (10 points) . l1 1 m x——> .4, W .1 E” + 24—03300 ~M4) 4! ; \—27(-3+£«—Lc = 0 W a Q03 1443"“ “(—0 4:: 0. $3 LWAM 36" ,’ Mm Axgwyéxwéx'fii __’-—- x—e \ 3% Jer- \ MHW} 4/ WM) 4903 +603) -é(‘> 3 a, W » «fix—«>4 '2 O 3wLWigw W2 QM“ 4\‘lxlx llx-‘é m f1(~01+\2(~t) H6 ,/ rm \ 65w; 6&0 +2, ,4 h 3 .,__) ~zt 3, N "x du 7. Use one of the two inte ration formulas = 1n u + Vu2 :l: a2 + C g \/ U2 :l: a2 I l (10 points) / dcc to evaluate ————~— . V m2 + 4x + 2 (ix . 7. '1. l ' —:. Xi?” "‘1'fi 5W X khxkl C 3 W1 leu+lttlwaltl+ C, OV"; : QM x l— 2 t \W‘x. A +1 c (fié— 2 . 8. Evaluate the definite integral / —$d$ in two ways7 first by using properties -2 (.752 — 1)3/2 of odd or even functions (2 points), and second, by actually pelforrning the integral (6 points). Do both answers match? (2 points) $09 5 yr ()0 '7' 7C :3 ‘H' x)? can; to» >311 e 6’2 :7 Oil lmncl’mv‘ :fi SEPUQCXK BIO 4 9. Evaluate the definite integral /0 $2da‘: in four ways. (20 points) (a) Exactly 3:10“ : [3&5]? :4. : 2\-\§. (b) By the \midpoint rule with n = 4 A X ‘3' 4'0 7:. l ‘ 3L lxDe 3— : A X Q¥ L‘ " 4(l/ +l¥g+fl> X), ;:3sz?zl fire , Le a q a >93: 5'2)“. '3) $093); 2; '3: 8.5L: 2'\ xgc 37;?» w“) r it; ‘C :/ (c)By the trapezmdtl rule withn== 4 szw 3, \. x0 O :23 “*6 C" “ >92 C 5”" Wt): C l S N \ o+2€l+ci+<i>+ie> M2 A guy Lg "’ "2? x323 :2) W33 ‘16} ,_, 21 KLLQC: «23 £00”? a / (d) By Simpson 8 rule with n- ~— 4 mm SN \ qu.\t2_.q+q~nl«té) ...
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