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Unformatted text preview: Calculus II V1102 Section 007, Fall 2007 Exam 2 Name: November 8, 2007 Do all problems, in any order. Show your work. An answer alone may not receive full credit. No notes, texts, or calculators may be used on this exam. Problem Possible Points Points Earned 1 10 2 9 3 5 4 10 5 10 6 10 7 10 8 10 TOTAL 74 1 1. (10 points) Test for convergence Z 2 e cos x p x ( x 2) dx INTUITION: Integral is improper at 2 and . e cos x shouldnt affect convergenceits bounded. near 2, the integrand looks like 1 x 2 , which has convergent integral. near , the integrand looks like 1 x 2 = 1 x which has divergent integral. SOLUTION: Z 2 e cos x p x ( x 2) dx Z 2 e 1 x dx which diverges. Hence, by comparison, the given integral diverges. 2 2. (9 points) True or false. If true, briefly explain why. If false, provide a specific counterex ample. (a) If x 2 n converges then so does x n . (b) Suppose a n > 0 for all n . If a n diverges then 1 a n converges. (c) Suppose a n 0 for all n . Suppose that s N = N n =0 a n 100 for all N . Then n =0 a n converges. SOLUTION: (a) False, for example x n = ( 1) n diverges while x 2 n = ( 1) 2 n = 1 converges. COMMENT: We cannot say x 2 n > x n . This is only true if x n < 0 or x n > 1. Further, even if it were the case that x 2 n > x n we cannot conclude by comparison that x n converges. The reason is that x n could be negative (cf the statement of the comparison theorem if this doesnt make sense). (b) False. Some possible answers are a n = 1, a n = n , a n = 1 /n ......
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 Fall '08
 Mosina
 Calculus

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