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Math 101 Apr 00 Solutions

Math 101 Apr 00 Solutions - MATH 101 Final Examination...

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Unformatted text preview: MATH 101 Final Examination, April 2000, SOLUTIONS l—i (2—i) 2—3i—1 1 3. 1. = - =—=___ (an 2+i (2—i) 4+1 5 51 The real part ofzis 11;. (b):z:2 =1+/ f(t)dt 2z=f(x), by the FTC, Part1. $+1 1 3+1 °° a:+1 (c)/0°° 2:1/3(_—$——2+:c+1)dz_11+12_/(; $1/3(z2+x+1)d$+/1 x1/3(x2+z+1) da: Notice that ———$—+1—-— < —1——— which converges for 0 < a: < 1. :131/3(:1:2 + :1: + 1) 31/3 By the comparison theorem, I1 converges. . . x+l 2a: 2a: _ 2 A150 notice that 1f :6 > 1, then m < m < W — W Wthh converges when 1 < a; < 00. Again by the comparison theorem, 12 converges. Since our original integral is the sum of two convergent integrals, it will also converge. d (d) —y+ay+y=0 r2 2+r+1=0=>r=—_1—ifl=-l+fl 2 2 2 The solution is y = ole—’3’ cos 39:; + C2612" sin 3.92:. (e) y = ole“? cos 3251: + aze—éz Sin 1231; + 2 1'2 (f)hm n—roo n—2 1 + F = also; (9 W G) Letxo=n0,a:n=1, thenscj=i,A:z:= =nlgr<§°ZzH/1 + zfiAm 2 =/‘;1m:/1+$2d$=-/2 uzdu=[%u %]1 =‘/§—1 1 Y 1 2. (a)/(;ey—‘/1——ydy 3. (c) /01 27r(:c + 2)(1 — 2:3) d9: (d) f(z) = §<1 +m2)% 1 HI) = 3 (g) (1 +x2)§<22) (rm)? = 42:20 + m2) 2 L:/ \/1+4:z:2(1+a:2)d:1: 0 (a) Let u = sin 21:, du : 2 cos 2a: (11: § 1 / sin5(2x) cos(2z) dz = l / u5 du o o (b) Let a; = 2sin0, d2: = 2cosOd6 /\/4-a:2d:r=2/V4—4sin29cosfld6 =4/cos20d0 1 =4/§(c0529+1)d6 =2<s1n26+0)+c 2 _ 0 =sm29+§+C =2si119cos9+g+0 = <;>( )< (c) Letu=ln(1+z2) and dvzdx, thenduzl‘i—Zgandv=a: 2:2:2 1 + x2 2—2 2 =xln(1+$2)—/2L——+—d$ da: /ln(l+z2)dx = xln(1 +252) —/ 1+:I:2 2(1+a:2) _ 2 —$ln(1+x)—/1—+;:2—dz—/ =xln(1+z2)—2x—2arctanx+C )+0 2 1+3:2 dz @) z+1 A B C EGTB=E+E+z—1 2+1=Aa:(:z:—1)+B(.11:—1)+C':c2 letx=021=-—B=>B=—1 letz=12 2=C lam-=2: 3=2A+B+4C ©322A—1+8:>A=—2 z+1 2 1 2 /22(z—1)d$_/_;_-z3+x——ldx 1 =—2ln|$|+;+2ln|m—1I+C dz 4. E =k(3—m)(2—z),z(0) = 1 (a 1 A B atzmiza=3_z+2_x 1=A(2—2:)+B(3—:c) 1etz=2=>B=1 letz=3=>A=——1 Therefore, hat—$704k” -1 1 /3_x+2_$_m+c 1n|3—x|—1n|2—2:|=kt+C' Subina:(0)=1 => ln2—1n1=C=>C=1n2 1n =kt+ln2 2—11: 3—2: —k mka-m t 3—:1: _ kt 2(2—z) 6 3—15 :43,“ —2:ce"‘ _4e"‘—3 :1: 28kt—1 kt_ kt 4e 3=lim4ke =2 (b) 3310101)“) = 131—1330 23’“ — 1 t—roo 2,66“ 5. (a) I z T4 = -A2—0 (8(90) + 28(01) + 28(02) + 25(03) + 8(94» = 2(100664 + 2(100543) + 2(100435) + 2(100331) + 1.00233) = 8.03515 I z 54 = %(3(90) + 48(01) + 23(92) + 43(63) + 5(04» 2 = 3(100664 + 4(1.00543) + 2(1.00435) + 4(1.00331) + 1.00233) w 8.035087 (b) Given that Is“) (9)[ g 1—0'3—0 therefore |s"(0)| 5 0.002 = K2 and [5(4>(0)| g 0.004 = K4. 1’ 3 0.002(8 — 0) _3 = — < — z _ Eu [1 f(z)dz T4 _ 12(4), 5 33 x 10 b 5 0.004(8 — 0) _3 = — < —— z ' Est /a f(x)dx S4 _ 180(4),, 284x10 6. NOTE: There are inconsistencies in this question. We will solve the initial value problem at the end of the question and ignore the values given in the text of the question. y"(t) + 91/0) = 4cost, y(0) = 3, y’(0) = -6 The auxiliary equation is r2 + 9 = 0 => 1‘ 2 i312 Therefore, yh = 01 sin(3t) + C2 cos(3t) Let yp = Acost+Bsint y; = ——Asint+ Bcost y: = —Acost— Bsint Substitute these values into our differential equation: —Acost — Bsint+9Acost+QBsint = 4cost 9A—A=4=>A=1/2 9B — B = 0 => B = 0 Therefore, 1 y(t) = 01 sin(3t) + 62 cos(3t) + 5 cost 1 y’ (t) = 361 cos(3t) — 3c2 sin(3t) — 5 sint 1 5 Subiny(0) =3=>3=0+C2(1)+ 5(1) =>62 = a Subiny’(0) = ~6=> —6=3cl(1)—0-—0=>c1 = —2 1 Therefore, y(t) = —2 sin 3t + gees 3t + 5 cost _ k(1—z3) if 05x31 7'f($)_{0 if $<00rz>1 (a) 1 / k(1 — $3) d1: = 1 o 4 1 k [2: — i] = 1 4 o k(§)=1 (filth (b) 1 §x(l—z3)dx [$2 15]] 2 5 0 ...
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