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Unformatted text preview: Version 061 – Exam01 – Gilbert – (56380) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x → parenleftBig 5 x 5 e x 1 parenrightBig exists, and if it does, find its value. 1. limit = 5 3 2. limit does not exist 3. limit = 0 4. limit = 5 4 5. limit = 5 2 correct 6. limit = 5 Explanation: Now 5 x 5 e x 1 = 5 parenleftBig e x 1 x x ( e x 1) parenrightBig = f ( x ) g ( x ) where f, g are everywhere differentiable func tions such that lim x → f ( x ) = 0 , lim x → g ( x ) = 0 . Thus L’Hospital’s rule can be applied. But f ′ ( x ) = 5 e x 5 , while g ′ ( x ) = ( e x 1) + xe x , so lim x → f ( x ) g ( x ) = lim x → f ′ ( x ) g ′ ( x ) = lim x → parenleftBig 5( e x 1) e x + xe x 1 parenrightBig . As f ′ and g ′ are differentiable functions such that lim x → f ′ ( x ) = 0 , lim x → g ′ ( x ) = 0 we have to apply L’Hospital’s rule again. But f ′′ ( x ) = 5 e x , g ′′ ( x ) = 2 e x + xe x , from which it follows that lim x → f ′′ ( x ) = 5 , lim x → g ′′ ( x ) = 2 . Consequently, the limit exists and limit = 5 2 . 002 10.0 points The region R is bounded by the xaxis and the graphs of y = 4 x , x = 1 . A part of R is shown as the shaded region in x 1 y Compute the volume of the solid of revolution obtained by rotating R around the xaxis. 1. volume = 19 π 2. volume = 20 π 3. volume infinite 4. volume = 16 π correct Version 061 – Exam01 – Gilbert – (56380) 2 5. volume = 17 π 6. volume = 18 π Explanation: Since R extends all the way to x = ∞ , the volume of the solid of revolution obtained by rotating R around the xaxis is given by the improper integral V = π integraldisplay ∞ 1 y 2 dx = π integraldisplay ∞ 1 16 x 2 dx . Now integraldisplay t 1 16 x 2 dx = bracketleftBig 16 x bracketrightBig t 1 = 16 parenleftBig 1 1 t parenrightBig . But lim t →∞ 1 t = 0 . Consequently, V = lim t →∞ π integraldisplay t 1 16 x 2 dx = 16 π . 003 10.0 points Find the n th term, a n , of an infinite series ∑ ∞ n =1 a n when the n th partial sum, S n , of the series is given by S n = 4 n n + 1 . 1. a n = 2 n 2. a n = 1 2 n 3. a n = 2 n 2 4. a n = 1 2 n 2 5. a n = 1 n ( n + 1) 6. a n = 4 n ( n + 1) correct Explanation: Since S n = a 1 + a 2 + ··· + a n , we see that a 1 = S 1 , a n = S n S n − 1 ( n > 1) ....
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This note was uploaded on 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas.
 Spring '07
 TextbookAnswers
 Multivariable Calculus

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