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# e2 - Calculus II V1102 Section 007 Fall 2007 Exam 2 Name...

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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

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1. (10 points) Test for convergence 2 e cos x x ( x - 2) dx INTUITION: Integral is improper at 2 and . e cos x shouldn’t affect convergence–it’s 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: 2 e cos x x ( x - 2) dx 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 doesn’t make sense). (b) False. Some possible answers are a n = 1, a n = n , a n = 1 /n ... COMMENT: If a n diverges, we cannot conclude that a n → ∞ (for instance (1), 1 n diverge). Even if it did, this is not enough to ensure that 1 a n 0 “fast enough” for 1 a n to

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