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111_quiz6_spr08_key

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APMA 111 Quiz 6 March 11, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. [5 pts] Compute the following integral. Your answer may be , -∞ , or DNE (does not exist). Z e 1 dx x ln x Solution: Use u -substitution: set u = ln x , then du = dx/x , and so Z e 1 dx x ln x = Z 1 0 du u = lim t 0 + Z 1 t du u = lim t 0 + ln u 1 t = lim t 0 + (ln1 - ln t ) = . 2. [5 pts] Use the Comparison Test to determine whether the following improper integral converges or diverges. If it converges, ﬁnd an upper bound for the integral. Z 1 dx x 4 + x + 1 Solution: Since x 1, x 4 + x + 1 x 4 0, and so 0 1 / ( x 4 + x + 1) 1 /x 4 . This implies that 0 Z 1 dx x 4 + x + 1 Z 1 dx x 4 = 1 4 - 1 = 1 3 ( p integral with p = 4) . Thus we conclude that the given improper integral converges to a value between 0 and the upper bound 1 / 3.

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Quiz 6, Page 2 of 2 March 11, 2008 3. [5 pts] Find the area of the surface obtained by rotating the curve about the
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Unformatted text preview: x-axis. Your answer should include a sketch of the surface. x = 1 + 2 y 2 , 1 ≤ y ≤ 2 Solution: The curve is rotated about the x-axis, so the equation for surface area is S = Z 2 πy ds . We are given x = 1 + 2 y 2 , so it will be easiest to use ds = p 1 + ( dx/dy ) 2 dy = p 1 + (4 y ) 2 dy = p 1 + 16 y 2 dy . This implies that S = Z 2 πy ds = Z 2 1 2 πy p 1 + 16 y 2 dy . Now use u-substitution: set u = 1 + 16 y 2 , then du = 32 y dy , and so S = Z 2 1 2 πy p 1 + 16 y 2 dy = π 16 Z 65 17 u 1 / 2 du = π 16 · 2 3 u 3 / 2 ﬂ ﬂ ﬂ ﬂ 65 17 = π 24 ‡ 65 √ 65-17 √ 17 · . 2 1-2-1 3 1 4 2 5-1 6 7 8-2 9...
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