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MATH 151 TEST 2 SPRING, 2010 4/15/2010Remember to keep your work neat and orderly. Show all of your work. NO WORK = NO CREDIT! Read each question carefully and be sure to answer the question that was asked. Good luck!Name:____________________________________________________________ pts1. Evaluate each of the following or show why they diverge:a.∫1∞(16 x4 /3)dxb. ∫03(17 x)dx6,6= ½ div2. Given that a+ar+ar2+...+arn−1=a(1−rn)1−r, evaluate each of the following:a.∑n=2∞(−1)n+153nb. 1. 2+(1 .2 )2+(1 . 2)3+. . .+(1.2 )307,6= -5/12≅ 1418.263. Write the repeating decimal 0 .7777 . .. as a ratio of integers (fraction in lowest terms). You must use an infinite series to solve! 7= 7/94. Set up an integral that will give the area (A=∫αβ12r2dθ) of one leaf of the 12-leaf roser=3 sin 6 θ . Be sure to have the correct limits on your integral. Do not evaluate! 6∫0π /692sin26 θdθ5. Use the integration formula∫√x2−a2xdx=√x2−a2−a sec−1|xa|+C to evaluate the integral∫√9 x2−72 xdx .7¿32[√x2−79−√79sec−1|x√97|]+C6. For the two points with polar coordinates, P1(−5 , π ) and P2(2 , π /4 ), a. Plot and label the two points on the same set of axes. 4b. Find two other equivalent polar coordinates for P2, at least one of which has r<0 .4(2, 2 π +π4),(−2,3 π4)c. Find the rectangular coordinates for each point. 6P1(5, 0), P2(2√2 ,2√2)7. Given that limn→∞(1+xn)n=ex, find limn→∞(n−3n)n4¿e−38. Write the terms a1 thru a5 for each sequence below:a.a0=4a1=−2ai=(−1)i3 ai−2b. an= (−1 )n+1sin(nπ2)5,4-2, 12, 6, 36, -18 1, 0, -1, 0, 19. State whether the given sequence {an} or series converges or diverges. Give a reason for each answer!a.∑n=6∞(−73)nb. ∑n=8∞n3−n6+7 n25 n6−7 n+16Div – geom with r > 1 div – limit of nth term not 0c. ∑n=1∞3√nnd. ∑n=1∞156div – p-series, p = 2/3 < 1 div - limit of nth term not 0e. an=n4−3 n+72 n5−n3+4f. an=16 n6conv to 0 conv to 010.Give an example of a convergent p-series and a divergent geometric series. 4p-series: geometric series:∑n=1∞1n2∑n=1∞(53)n11.Use the integration formula∫dx√2ax−x2=sin−1(x−aa)+C to evaluate the integral∫7√x −x2dx . Simplify your answer. 6 ¿7

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