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Unformatted text preview: Calculus II, Review Problems for Exam 2 October 27, 2007 1. Test for convergence Z 2 e cos x p x 3 (2 x ) dx . SOLUTION: Since 2 x 2 for all x [0 , 2] we have Z 2 e cos x p x 3 (2 x ) dx Z 1 e 1 2 x 3 / 2 which diverges. Hence the given integral diverges. 2. Give a complete argument that the harmonic series diverges (saying pseries, p = 1 is not enough!). If s n denotes the nth partial sum, show that s n 1 + ln n . Conclude that the sum of the first billion terms is less than 22. Is the sequence of partial sums unbounded? SOLUTION: Notice that f ( x ) = 1 x is continuous, positive, and decreasing on [1 , ). Further f ( n ) = 1 n and R 1 1 x dx diverges. Hence by the integral test, the harmonic series diverges. s n 1 + R n 1 dx x = 1 + ln n (draw the picture!!). 3. For which and 0 does Z x 1 + x dx converge? INTUITION: Near 0 the integrand looks like x which has convergent integral if and only if > 1. Near the integrand looks like x  which has convergent integral if and only if  < 1 SOLUTION: Since 0, for x 1 we have x 1 + x x + x = 2 x hence Z...
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This note was uploaded on 10/18/2011 for the course MATH V1102 taught by Professor Mosina during the Fall '08 term at Columbia.
 Fall '08
 Mosina
 Calculus

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