# 1a - 3&amp;quot; A 241 Test 1 Form Spring 2 0 9. ....

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Unformatted text preview: 3" A 241 Test 1 Form Spring 2 0 9. . Integrate. x a f—dx c. fxzexsdx d. fxln(5x)aix x—2 . ——————dx e f(x2—4x+1)2 Continue on next page. Form D continued 10. Set up ONLY a, b, and c below. Insert all numerical values so that if you had a calculator, you could ﬁnish the arithmetic. a. Use Trapezoidal Rule with n=4 subintervals to approximate } x I 1 1+ (sin x)2 b. Find the area of the region R bounded by the line y=x, and the parabola 2 y = x — 12' c. Find the volume of the solid generated by revolving the region in part b (previous area problem) about the line y= - 15. 11. Decompose to partial fractions (this is not asking for you to integrate!) 2 x — x + 4 Jc(x2 + 4) 12. Determine if the improper integral converges or diverges. If it converges, evaluate. lldx 0x2+9 13. Use the appropriate formula, if needed, from the given ones list below from the Tables of Integrals to integrate 3 x ————— (176 f sz - 9 . Table of Integrals: a. fi— = lnlu+ [M2 _ a2 uzdu u (—2 2 a2 [—_‘2 2 u —a +—2—ln’u+ U —a +C V112 —a2 +C MA 241 Spring 2007 Test 1 solution key Yellow/White version 9(a) Integrate x /x+1dm Solution: This fraction is improper, hence we perform division. 1 x + 1) m —m—1 — 1 Therefore :0 1 = 1 _. x+1 m+1 And we now integrate :r 1 d 1— /x+1m w+1>dx /1dx—/ 1 dm x+1 x—ln|x+1|+C’ l l 9(b) Integrate Solution: Use u—substitution. Integrate: \ O o m A E + 83 % | l [\D \ O o m A \$1 + £13 2% 2x5 2/cos(\/§+ 3);dm 2%? 2 / cos(u)du 2 sin(u) + C 2 sink/E + 3) + C H I | 9(e) Integrate / 3326"”3 dx u = x3,du = 3m2d93 3 / x239” din Solution: Use u-substitution 1 3/ 3 33:2 ex doc 9(d) Integrate / £1n(5m)dac Solution: Use integration by parts (LIATE) -1-dac 5 u — ln(5m),du — Educ — a: 2 do=mdm,v=%— / x ln(5:c)dx _ 51:2 1n(52:) / £3 1 — 2 2 x x21n(5x) m = — —d 2 2 x 11:2 1n(5m) :02 _ 2 _ Z + C 9(e) Integrate / ——-————x - 2 dm (m2 — 49: + 1)2 Solution: Use u—substitution u=m2—4x+1,du=(2x—4)dm 93—2 d 1/ 2113—4 dx /‘(:c2—4191:+1)2 x 2 (\$2—4(B+1)2 II NH»—A \ Q .‘o R. if II II I | l | + Q 23 l | + Q 12 Determine of the improper integral converges or diverges. If it converges, evaluate. °° 1 (1 /0 (1324—9 33 00 1 ‘ t 1 [0 x2 + gdx £120 dx Solution: ll z=t t—>oo 3 1 95—0 lim — arctan t—>oo 3 g) -— g arctan(0) _ 1 t tlirglo g arctan —— O 0 502 + 9 , 1 a: 11m — arctan II II Form D 723.1 ’ 0V D \$ Oil/L Y 4‘“ ‘0 0~ got .10’Q L12. 2 .' 3 ' j" 1+(séhx)2 ' q 1 = 15' x +62? x + 52 x +0? x + x I: o - ‘_ " ‘ ___ : _L X: l ' ' =1 : x:3/g X:& :%x T: (\+2¥L%+a 2+a¥9+ a I I” + a +2 % + 1+,<s~m(s/z>)z --—-z~ I \ (9mm); \+(S‘m(9§)) “(330(3))2 b v a 5 MA Jim v" iVxILva/ﬁ - = x1- x2 _ o = 2- x- n9. “W. = 3 o- .1“ dx I _ l -— ,- I-I" ‘ CI .'.A ' ‘ V" V ' Vat/hon 6 (A: .54“ ~ 2‘ oL‘ ' 0.; ‘10 . 0 l0 — —“ x-c-w )2- xz-n — 46 . Fovm D) I X X2+ 4 . XCwa—i) I 2*X+ = x5e} + 5x46. X Ol-o+ = o+4+’\3(o+co ‘ J> - X N t X 4. H S (l FOVWD15>33 I ...
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## 1a - 3&amp;quot; A 241 Test 1 Form Spring 2 0 9. ....

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