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Unformatted text preview: Create assignment, 93205, Homework 3, Apr 26 at 12:49 am 1 This printout should have 40 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. version 728 CalC12c04a 55:03, calculus3, multiple choice, > 1 min, wordingvariable. 001 Determine which of the following series are convergent: A . 1 + 1 4 + 1 9 + 1 16 + ... B . X n = 1 3 n 2 + 1 C . X n = 1 2 n 3 / 2 1. C only 2. A and C only 3. all of them correct 4. none of them 5. B and C only 6. A only 7. B only 8. A and B only Explanation: A. Series is X n =1 1 n 2 . Use f ( x ) = 1 x 2 . Then Z 1 f ( x ) dx convergent. B. Use f ( x ) = 3 x 2 + 1 . Then Z 1 f ( x ) dx convergent (tan 1 integral). C. Use f ( x ) = 2 x 3 / 2 . Then Z 1 f ( x ) dx convergent. keywords: CalC12c15s 55:03, calculus3, multiple choice, < 1 min, wordingvariable. 002 Determine whether the series X n = 1 1 n 2 + 4 converges or diverges. 1. series diverges 2. series converges correct Explanation: We apply the integral test with f ( x ) = 1 x 2 + 4 . Now f is continuous, positive and decreasing on [0 , ). Thus the series X n = 1 1 n 2 + 4 Create assignment, 93205, Homework 3, Apr 26 at 12:49 am 2 converges if and only if the improper integral Z 1 x 2 + 4 dx converges. But Z 1 x 2 + 4 dx = lim t Z t 1 x 2 + 4 dx. To evaluate this last integral, set x = 2 tan u . Then dx = 2 sec 2 udu while x = 0 = u = 0 , x = t = u = tan 1 t 2 . In this case Z t 1 x 2 + 4 dx = 1 2 Z tan 1 ( t/ 2) sec 2 u sec 2 u du = 1 2 h u i tan 1 ( t/ 2) = 1 2 tan 1 t 2 . But by properties of tan 1 t we know that lim t tan 1 t 2 = 2 . Consequently, by the Integral Test, the series X n = 1 1 n 2 + 4 converges . keywords: CalC12c20a 55:03, calculus3, multiple choice, < 1 min, wordingvariable. 003 Determine whether the series X m = 1 3 ln(4 m ) m 2 is convergent or divergent. 1. series converges correct 2. series diverges Explanation: The function f ( x ) = 3 ln(4 x ) x 2 is continous and positive on [ 1 2 , ); in addi tion, since f ( x ) = 3 1 2 ln 4 x x 3 < on [ 1 2 , ), f is also decreasing on this interval. This suggests applying the Integral Test, for then the series X m = 1 3 ln(4 m ) m 2 is convergent if and only if the improper inte gral Z 1 f ( x ) dx = Z 1 3 ln(4 x ) x 2 dx converges. Now, after Integration by Parts, we see that Z t 1 f ( x ) dx = 3 h ln(4 x ) x 1 x i t 1 = 3 n ln(4 t ) t 1 t + ln 4 + 1 o . Consequently, Z 1 f ( x ) dx = lim t Z 1 f ( x ) dx = lim t 3 n ln(4 t ) t 1 t + ln 4 + 1 o = 3(1 + ln 4) . The Integral Test thus ensures that the given series converges . Create assignment, 93205, Homework 3, Apr 26 at 12:49 am 3 keywords: CalC12d15s 55:04, calculus3, multiple choice, > 1 min, wordingvariable....
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This note was uploaded on 09/23/2009 for the course M 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.
 Fall '09
 RAdin
 Calculus

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