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30 Pages

### Problem set ch 15 16

Course: ECON 1110, Fall 2007
School: Cornell
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Word Count: 4389

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ALL Government BLACKBOARD SITES &gt; ECON 101-2 INTRODUCTORY MICROECONOMIC (FALL 2007 - BURKHAUSER) &gt; ASSIGNMENTS &gt; REVIEW ASSESSMENT: PROBLEM SET #10 Review Assessment: Problem Set #10 Name: Status : Score: Problem Set #10 Completed 65 out of 100 points Instructions: Further instructions on how to complete online problem sets can be found in then Problem Set Guide in the Course Information...

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Cornell - MATH - 1910
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Math 191 2x3 dx 1 5.3: 6. 0 4 - x2 dx 5.3: 2. 5.3: 14. (a)3 1 0 -1Problem Set 2 Solutionsh(r)dr =1 33 -1h(r)dr -3-11 h(r)dr = 6 - 0 = 63h(u)du = -(- 1 h(u)du) = 1 h(u)du = 6 5.3: 22. y = 1 + 1 - x2 y - 1 = 1 - x2 (y - 1)2 = 1 -
Cornell - MATH - 1910
Math 191 5.5: 4 Let u = 1-cos 1 du = 2 2 t 3 3 (1 - cos 2 ) + C1 2HW 3 Solutions sin 1 dt 2du = sin 1 dt. Then (1-cos 2 2t2 2 t sin 2 )dt = 2 2u2 du = 3 u3 +C =5.5: 8 Let u = y 4 +4y 2 +1 du = (4y 3 +8y)dy 3du = 12(y 3 +2y)dy. Then 3u2 du =
Cornell - MATH - 1910
Math 191 6.2: 4 For the sketch given, c = 0, d = 2 3 0 0 1 0HW 4 Solutions 3; V =d c 2(shell radius)(shell height)dy =302y (y 2 )dy =y 3 dy = 2y4 4 b a3= 9/2. 2x[(2 - x2 ) - x2 ]dx = 21 06.2: 10 a = 0, b = 1; V = 41 (x 02
Cornell - MATH - 1910
Math 191 6.5: 6 a.dx dyHW 5 Solutions = 1 + y -1/2 dx dy 2= 1 + y -1/22 S = 22 (y 1 + 2 y) 1 + (1 + y -1/2 )2 dx.2y+2 sqrt(y)=x1.91.81.71.61.5y1.41.31.21.1133.23.43.63.8 x44.24.44.64.8b. c
Cornell - MATH - 1910
Math 191 6.7: 17 When the water reaches the top of the tank, the force on the movable side is0 0 0 (62.4)(2 -2HW 6 Solutions 4 - y 2 )(-y)dy =2 (62.4) -2 (4 - y 2 )1/2 (-2y)dy = (62.4) 3 (4 - y 2 )3/2 -2 = 332.8ft lb. The force compressing the
Cornell - MATH - 1910
Math 191 7.4: 6 a. b. c.log9 x log3 xHW 7 Solutions =ln x ln 9ln x ln 3=ln x 2 ln 3ln 3 ln x1 2= 1 21 2log10 x log2 x loga b logb a=ln x ln 10ln x ln 2= ln x ln 10ln 2 ln x=ln 2 ln 10=ln b ln aln a
Cornell - MATH - 1910
Math 191 7.6: 10 a. true; b. true;1 x+3 1 xHW 8 Solutions =x x+3&lt; 1 if x &gt; 1 (or sufficiently large)1 x1 1 x + x2 1 x=1+&lt; 2 if x &gt; 1 (or sufficiently large)1 = limx (1 - x ) = 1c. false; limxx1 1 x - x2 1 xd. true; 2 + cos x 3
Cornell - MATH - 1910
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Cornell - MATH - 1910
Math 191 sin x x 8.2: 42 (+)HW 10 Solutions- - -(+) (-)- cos x - sin x cos x sin-1 ydy = y sin-1 y - ydy = y sin-1 y + 1 - y2 + C t2 cos tdt = t2 sin t + 2t cos t - 2 sin t + C2x 2 08.2: 8 u = sin-1 y, du = dy 8.2: 241-y 2; dv = dy,
Cornell - MATH - 1910
Math 191 8.7: 12HW 11 Solutions3 a. n = 8 x = 3 x = 16 ; mf(i ) = 1(0) + 2(0.09334) + 2(0.18429) + 2(0.27075) + 2(0.3512) + 8 2 3 2(0.42443) + 2(0.49026) + 2(0.58466) + 1(0.6) = 5.3977 T = 16 (5.3977) = 1.01207b. n = 8 x = 3 x = 1 ; mf(i )
Cornell - MATH - 1910
Math 191 11.1: 16 an =n (-1) n2n+1HW 12 Solutions , n = 1, , 2, . 1 ) does not exist diverges. n = limn11.1: 32 lim (-1)n (1 - 11.1: 50 lim 1- 1 nnn1+(-1) nn= e-1 converges. (Theorem 5, #5) 1 1 n2 - 1 + n2 + n n2 - 1 + n2 + n
Cornell - MATH - 1910
Math 191 11.3: 2 converges; a geometric series with r = 1 eHW 13 Solutions &lt;111.3: 6 converges;1 -2 = -2 , which is a convergent p-series (p = 3 ). 2 3/2 n n n n=1 n=1n11.3: 18 diverges; lim an = limn1+1 nn=e=011.3: 28 diverge
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Math 191 11.6: 6 converges by the Alternating Series Test since f (x) =ln x xHW 14 Solutions f (x) = 1-ln x &lt; 0 when x &gt; e f (x) is x2 1 ln n decreasing un un+1 ; also un 0 for n 1 and lim un = lim = lim n = 0 n n n n 1 1 3 3 1+ n n+1 = l
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