cal13 - choi (kc24547) HW13 BERG (56525) 1 This print-out...

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Unformatted text preview: choi (kc24547) HW13 BERG (56525) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Compare the values of the series A = summationdisplay n = 2 6 n 1 . 7 and the improper integral B = integraldisplay 1 6 x 1 . 7 dx . 1. A = B 2. A > B 3. A < B correct Explanation: In the figure 1 2 3 4 5 . . . a 2 a 3 a 4 a 5 the bold line is the graph of the function f ( x ) = 6 x 1 . 7 on [1 , ) and the area of each of the rectan- gles is one of the values of a n = 6 n 1 . 7 . Clearly from this figure we see that a 2 = f (2) < integraldisplay 2 1 f ( x ) dx, a 3 = f (3) < integraldisplay 3 2 f ( x ) dx , while a 4 = f (4) < integraldisplay 4 3 f ( x ) dx, a 5 = f (5) < integraldisplay 5 4 f ( x ) dx , and so on for all n . Consequently, A = summationdisplay n = 2 6 n 1 . 7 < integraldisplay 1 6 x 1 . 7 dx = B . 002 10.0 points Which of the following series are convergent: A . summationdisplay n = 1 2 n 2 / 3 B . summationdisplay n = 1 3 n + 1 C . 1 + 1 4 + 1 9 + 1 16 + . . . 1. B and C only 2. A and B only 3. all of them 4. B only 5. A and C only 6. C only correct 7. A only choi (kc24547) HW13 BERG (56525) 2 8. none of them Explanation: By the Integral test, if f ( x ) is a positive, decreasing function, then the infinite series summationdisplay n = 1 f ( n ) converges if and only if the improper integral integraldisplay 1 f ( x ) dx converges. Thus for the three given series we have to use an appropriate choice of f . A. Use f ( x ) = 2 x 2 / 3 . Then integraldisplay 1 f ( x ) dx is divergent. B. Use f ( x ) = 3 x + 1 . Then integraldisplay 1 f ( x ) dx is divergent (log integral). C. Use f ( x ) = 1 x 2 . Then integraldisplay 1 f ( x ) dx is convergent. keywords: convergent, Integral test, 003 10.0 points Determine whether the series summationdisplay n = 8 2 n 4 converges or diverges. 1. series is divergent correct 2. series is convergent Explanation: We apply the integral test with f ( x ) = 2 x 4 . Now f is continuous, positive and decreasing on [8 , ). Thus the series summationdisplay n = 8 2 n 4 converges if and only if the improper integral integraldisplay 8 2 x 4 dx converges. But integraldisplay 8 2 x 4 dx = lim t integraldisplay t 8 2 x 4 dx = lim t bracketleftBig 2 ln | x 4 | bracketrightBig t 8 = lim t 2 ln vextendsingle vextendsingle vextendsingle vextendsingle t 4 8 4 vextendsingle vextendsingle vextendsingle vextendsingle = . Consequently, the series summationdisplay n = 8 2 n 4 is divergent ....
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This note was uploaded on 04/04/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas at Austin.

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cal13 - choi (kc24547) HW13 BERG (56525) 1 This print-out...

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