Gilbert_Exam01sol

# Gilbert_Exam01sol - Version 017 – Exam01 – Gilbert –...

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Unformatted text preview: Version 017 – Exam01 – Gilbert – (56380) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x → parenleftBig 4 x- 12 e 3 x- 1 parenrightBig exists, and if it does, find its value. 1. limit = 4 2. limit = 0 3. limit does not exist 4. limit = 3 5. limit = 12 6. limit = 6 correct Explanation: Now 4 x- 12 e 3 x- 1 = 4 parenleftBig e 3 x- 1- 3 x x ( e 3 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 ) = 12 e 3 x- 12 , while g ′ ( x ) = ( e 3 x- 1) + 3 xe 3 x , so lim x → f ( x ) g ( x ) = lim x → f ′ ( x ) g ′ ( x ) = lim x → parenleftBig 12( e 3 x- 1) e 3 x + 3 xe 3 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 ) = 36 e 3 x , g ′′ ( x ) = 6 e 3 x + 9 xe 3 x , from which it follows that lim x → f ′′ ( x ) = 36 , lim x → g ′′ ( x ) = 6 . Consequently, the limit exists and limit = 6 . 002 10.0 points The region R is bounded by the x-axis and the graphs of y = 4 x , x = 2 . A part of R is shown as the shaded region in x 2 y Compute the volume of the solid of revolution obtained by rotating R around the x-axis. 1. volume = 12 π 2. volume = 11 π 3. volume = 10 π 4. volume = 9 π 5. volume infinite Version 017 – Exam01 – Gilbert – (56380) 2 6. volume = 8 π correct Explanation: Since R extends all the way to x = ∞ , the volume of the solid of revolution obtained by rotating R around the x-axis is given by the improper integral V = π integraldisplay ∞ 2 y 2 dx = π integraldisplay ∞ 2 16 x 2 dx . Now integraldisplay t 2 16 x 2 dx = bracketleftBig- 16 x bracketrightBig t 2 = 16 parenleftBig 1 2- 1 t parenrightBig . But lim t →∞ 1 t = 0 . Consequently, V = lim t →∞ π integraldisplay t 2 16 x 2 dx = 8 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 = n n + 1 . 1. a n = 1 2 n 2 2. a n = 1 n ( n + 1) correct 3. a n = 5 n ( n + 1) 4. a n = 5 2 n 5. a n = 1 2 n 6. a n = 5 2 n 2 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 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.

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Gilbert_Exam01sol - Version 017 – Exam01 – Gilbert –...

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