ReviewExam3SolnsSp07

# ReviewExam3SolnsSp07 - Create assignment, 93205, Homework...

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Unformatted text preview: Create assignment, 93205, Homework 3, Apr 26 at 12:49 am 1 This print-out should have 40 questions. Multiple-choice 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, wording-variable. 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, wording-variable. 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, wording-variable. 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, wording-variable....
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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.

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ReviewExam3SolnsSp07 - Create assignment, 93205, Homework...

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