Exam01-solutions-1 - Version 092 Exam01 gilbert (55485) 1...

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Unformatted text preview: Version 092 Exam01 gilbert (55485) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Let h be a continuous, positive, decreasing function on [3 , ). Compare the values of the series A = 13 summationdisplay n = 3 h ( n ) and the integral B = integraldisplay 14 3 h ( z ) dz . 1. A < B 2. A = B 3. A > B correct Explanation: In the figure 3 4 5 6 7 . . . a 3 a 4 a 5 a 6 the bold line is the graph of h on [3 , ) and the areas of the rectangles the terms in the series summationdisplay n =3 a n , a n = h ( n ) . Clearly from this figure we see that a 3 = h (3) > integraldisplay 4 3 h ( z ) dz, a 4 = h (4) > integraldisplay 5 4 h ( z ) dz , while a 5 = h (5) > integraldisplay 6 5 h ( z ) dz, a 6 = h (6) > integraldisplay 7 6 h ( z ) dz , and so on. Consequently, A > B . keywords: 002 10.0 points Determine if the improper integral I = integraldisplay 2 8 x (4 + x 2 ) 2 dx converges, and if it does, compute its value. 1. I = 2 2. I = 1 3. integral doesnt converge 4. I = 1 2 correct 5. I = 2 3 Explanation: The integral I = integraldisplay 2 8 x (4 + x 2 ) 2 dx is improper because of the infinite interval of integration. To overcome this, we truncate and consider the limit lim t I t , I t = integraldisplay t 2 8 x (4 + x 2 ) 2 dx . Version 092 Exam01 gilbert (55485) 2 To evaluate I t , set u = 4 + x 2 . Then du = 2 x dx , in which case integraldisplay 8 x (4 + x 2 ) 2 dx = 4 integraldisplay 1 u 2 du. Thus I t = integraldisplay t 2 8 x (4 + x 2 ) 2 dx = 4 bracketleftBig- 1 4 + x 2 bracketrightBig t 2 = 4 braceleftBig 1 8- 1 4 + t 2 bracerightBig . Consequently, since lim t 1 4 + t 2 = 0 , we see that I converges and that I = lim t integraldisplay t 2 8 x (4 + x 2 ) 2 dx = 1 2 . 003 10.0 points Determine whether the series summationdisplay n = 0 3 parenleftbigg 1 4 parenrightbigg n is convergent or divergent, and if convergent, find its sum. 1. divergent 2. convergent, sum = 12 5 3. convergent, sum = 4 correct 4. convergent, sum =- 13 3 5. convergent, sum = 13 3 Explanation: The given series is an infinite geometric series summationdisplay n =0 a r n with a = 3 and r = 1 4 . But the sum of such a series is (i) convergent with sum a 1- r when | r | < 1, (ii) divergent when | r | 1. Consequently, the given series is convergent, sum = 4 ....
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Exam01-solutions-1 - Version 092 Exam01 gilbert (55485) 1...

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