102-2007&amp;2008-3-M20-July2008

# 102-2007&amp;2008-3-M20-July2008 - KU WAIT UNIVERSITY...

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Unformatted text preview: KU WAIT UNIVERSITY Department of Mathematics 85 Computer Science Math 102 (Calculus B) Second Examination July 19, 2008 Duration: 90 minutes Calculators, Cell phones and Pagers ARE NOT ALLOWED. Answer all of the following questions. 1‘ Evaluate the following integrals. (3 points each) (a) [cosﬂnﬂdm (b) f tani'seﬁ1 z + sin Edi.- sec3 9: “3) lid/12;“ 72:2 + x + 3 (d) [(x + 1) (23:2 +1)“ ‘8) fmd (13: (U [FT/"m (g)/(1+x)er homage. Ida: 2. Determine wether the integral 1 0 Ex — is convergent or divergent. Evaluate it, if convergent. {4 points) 99 Solution Key 1. (a) Using integration by parts twice [cosﬂnxjdm : f(a:)'cos(lna:)dm = xcosﬂnx) +/sin(ln ﬂab: :mcosﬂnx) +f(:c)’sin(lnz! II a: cosﬂn w) + xsinﬂn m) i fcosﬂn :r)d9: fcos(1n:1:)d5~3 : §(cos(ln x) + sin(1r1:t)) + C ('3) tan3 sec‘1 :5 + sin :c —.——d\$ : 3 (atan 1‘ sec :1: + sin :r: (1033 :13) d3: sec- :6 fgs ssc2 \$71 (“3603 )+f(sin\$—sin3 3:) d(sin3:) see 4 sin2\$ sin :5 sec :1: + 33 2 4 +0 ((3) Using the substitution u = :52 71 1 —1 1 1 I :sinu3 —i—C=::in:2 +0 ((1) Partial Fractions: ﬂzmﬁ1+iﬁzﬁ (14:3, B: 1, C: 0) 7112 +35+3 3 _— d i 1 2 /(\$+1 ““2532 1)ulas m—I—f\$+ld\$+/Qa:9+1 x— 3nix+1|+— :ln(2z +1)+C‘ (e ) Using the substitution u = tan? sinx— — 1i: 2, COSl': 11H“; dz "- 1+“ 22cm secs: 1 1 2_—2 oh: = fail: W 255“ gecf+tan\$+l +SIHI+COSLE 2+1+u2 }+—u21+u / 2 d~ 2 d 2(1+u2)+2u+1—u2“_ u2+2u+3u 2 __ 2 lad-1+ fmdu—stan 7* : ﬁtan’1ﬁ(tan§+1)+0 F! (f) Using the substitution 5: = u" dz 4113 11 6/3? —-—-: ﬁdu:2sec'1—+C=ZSE 1 +C fmx/ﬁ—Ll fu4x/u2—4 2 C 2 OO 99 (g) Using the substitution mex = \$1118. (meg) — (1 + 3:)?” /(1-i-:1:)eI 1H (meﬂzdm = f 1—( ((381)2(1 we I)— [(1052 6:19 Sin 229 = 2/(L1E=11+cos23)6 —(5+ )+C ' ‘1 I ' 28in e : s1n [me)+s1n( (9: ”2+0 2 4 2, The integral is divergent since 1 d2? , 1 d9: ,1 Ae\$_1=c13£cex_1=—Clirgir[ln(l—e )—1n(1*ec)]=00 OO 99 ...
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