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102-2006&2007-1-M20-Dec2006

102-2006&2007-1-M20-Dec2006 - Kuwait University...

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Unformatted text preview: Kuwait University Math 102 (Calculus II) December 14, 2006 Mathematics 8 Computer Science Dept. Second Examination Duration: 90 minutes m Calculators, Mobile telephones and Pagers ARE NOT ALLOWED. Answer all of the following questions. 1. Evaluate the following integrals. (each is worth 3 points) (210/358—331133. (b) /(seor)(sin2x)3 d3. l (”J “777 d?“ a: Cl --~———d jfxlm2+2$~3 I (e) /2:_____2_ —103:+15 (in: 4I2+5$ ' «is: (f) f—m—w—t . . /. anr+smsc «453 ”W“ 2, Determine whether the 1mpr0per integral / + d1: 15 convergent or divergent and find ES :5 . a value if convergent. (4 points) CO Solution Key 1. (a) fI58_I3 2 —%f3¢3(—3$28“I3) d3: 2 -§fm3(e—$a)’dz= —%m38'za +f322e_13d$ _ 3 _ 3 _ S = —%z% I — %e I + C {Nomenclature : u 2 3:3, ch; 2 $29 3” dz} (1)) j (sec as) (sin 23:)3 d2: 2 8 / cosgzsin‘a’ mix 2 8/ {(3052 3:) (1 — cos2 x) sinxdm _78(ms Leos ) +0 . 2 2 _ du mat—an 6' 83:2 age 0055 9 d9=1—sln 3 “:31“ 9 {1—1321sz (C) / :2 (144:2); H jfi— (tan2 6) sec 9 Sin 9 sin 6 C059 d9 _ u =-/'(;1§w2+1£2)du2—fi—2u:%u3+C:[email protected]§_ 2m +% 33 314-15 (2|.+:t2]v1+.1:2 + C d) f—(flmdxfil‘i—Esecg/23‘159—125366ten6d9—/(25ec26—sec6)d6 22tan921n|secfl+tan9|+02 \/$2+23—3+1n ”‘1 ”a 229373 +0 (e) :53 2 49:2 + 5m 2 a: (9:2 2 4m + 5), and 5:2 — 455 + 5 is irreducible. figjgg—jg = £1 +r§+fl5 => (A+B)x2+(24A+C)x+5A=2x2 —10:c+15 .1: —0 5_3 — A23,BI—1,C=2. SO —H’—m—_14:+é)——zzf4:+5. f(§_x_ —4:n+5) d3=31nbs|2-fflm—2Ldas231nlml—%1n(22—4x+5)+0 2—4m+5 [uh—- I 2tan- 1 _ct_r “,2 .. 2 _ 1—1? _1 an;_; 2a (f) jtanz+sinx — / 2n l+u2 21.; 1+u§dui / u dufi 211111; 2| 4tan 2 +0 1+u2 121i2 1+2»2 (g)/%I_ “T d3: mue==xf——“ "" Gust26 “3"‘la’u2 gm] (u4+4)—%tan_1”—2+C x( 3+4 f) 1:3 (115 +4112 u4+4 2 L, 1 I . . . . 2. z+fi 13 contmuous on (0,1], and $133+ 31+fi—OO7 so the lutegral is Improper at 0. 1 1 1 ft I+Ifid$=ft fifidx=21nfl+fiflt22[ln2—1n(1+\/E)] l '111/1 main:— 2 2 1112. The integral converges and f0 “Ti/Eda: = 2111 2. tliD+ “W ...
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