Mid 3 - 1) (20 points) Find the general form of the partial...

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Unformatted text preview: 1) (20 points) Find the general form of the partial fractions decomposition of (8 — 3:13)12 (2:2 + 555 + 4)3(CL“3 +102:2 + 285C)2 (DO NOT FIND THE NUMERICAL VALUES OF THE COEFFICIENTS.) The complete factorization of the denominator is (x2 + 51' + 4)3(m3 + 10:1:2 + 28$)2 2 (a: + 4)3(J: +1)33r2(:1:2 +101’ + 28)2 Also (degree of numerator) — (degree of denominator) = 12 — 12 = 0 Thus the general form of the decomposition is: Jx+K A} + + ' + l + Lw + M B C D E F GH£I $ (a:+4)3 (x+4)2 $+4+(x+1)3l($+1)2 x—l—l $2 ' (51:2 +109: + 28)2 + 3:2 + 103: + 28 n6 pf 2) (20 points) Determine if the following improper integral converges: /10 11r+5 ——————dr 0 5r2+4r— 1 If it converges, find its value. We first find the indefinite integral by partial fractions decomposition. The denominator factors as 57“2 +47“ #1:(5r— 1)(7” + 1) Thus the general form of the partial fractions decomposition is: 11r+5 A B 5r2+4r—1:r+1'57~—1 Clearing out fractions, we obtain 11r+5=A(5r*1)+B(7"+1) Plugging in 'r' = 1, we obtain ~11+5=A(~5—1):>A:1 Comparing degree one term on both sides, we obtain 11r:5Ar+Br:5r+Br'=>B:6 117+5 1 6 /5r2+4r—1 T /r—1r+/5r+1 I 6 :lnlr~1|+gln|5r—1l Thus From the factorization of the denominator, we see that the integrand has a vertical asymptote at ’r' : Hence 117' + 5 10 % 11 5 ‘10 11 r / —-————dr:/ T+ dri/ T+O d7" 0 5r2+4r—1 0 5r2+4T—1 r 5r2+4r—1 5 and both integrals on the right would have to converge. However 1 5 % 11r+5 6 —————d : l —1 —-l 5 —1 /05T2+4T~1r [nv |+5nlr 1L 1 g~11+g limlnlSrell—ln(1)#gln(1) r—>— 5 :ln and the limit tends to ~00. Thus the integral diverges. (3|... 3) (20 points) Determine the following integral /4xln (x2 ~ 10x + 24) doc We have $2—1Oac+24: (cc—6)(:t—4) Hence /4:L'ln (:02 ~ 103: + 24) dzr : /4:c1n [(33 — em — 4)] dx :/4xln(:r—6)da:+/4a:ln(x—4)d:13 Integrating each of these integrals by parts, we obtain 222 ln(x —— 6) —/ 2122 30—6 dx+2xgln(:v—4)—/l 2«3 2—16 1 ac 6+36d$ 2/x ' +66” :21n(x2—10x+24)—2/ $43 36 1 :2ln(:L‘2—103:+24)—2/ x+6+ dx—2/ x+4+ 6 dx 36—6 30—4 2 2 :21n(3:2 1036 l 24) 2(2 {fix I 36ln1:r—6l> —2(%—+4x+16lnlx—4l> =2ln(:v2 1090:24) 2x2 20:1: 721nlx—6l—321n|x—4l+0 :—2562——20$—7Olnl33~6|—301n|x—4[+C 4) (20 points) Find the length of the curve 1 y = —2—0(10x\/100$2 — 1 — ln [10$ + \/10O:c2 — 1D, 1 g :1: g 5 DEVIOUS HINT: You may find the following integration formula helpful 1 jfx/100x2——1dx== i6 (lex/lOOxQ-—1-—ln‘10x—%\/100m2—-1J)v+C7 (You don’t have to use this hint, but you Will save yourself from a lot of unnecessarily complicated algebra if you do.) d By the devious hint d—y 2 \/100m2 — 1. Hence a 5 2 5 L2/ 1+<—> dx=/\/1+100:c2—1d;v 1 dw 1 5 2 5 2/ 10xd35210$ 0 2 250 10 122 5) (20 points) Find the surface area generated by revolving the curve y:2:r2, 03x34 around the y—axis. 4 :/ 27r$\/1+ 16:1:2 dl‘ 0 d Let u : 1 + 16$? Then dx : Thus 25L 257 S : / 27T1'u1/2—dfl 1 32$ : 5% ((257)3/2 — 1) : 539.178905139473 ...
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Mid 3 - 1) (20 points) Find the general form of the partial...

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