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1b-2009-final_exam_sample_2_solutions

# 1b-2009-final_exam_sample_2_solutions - Math 1B Sample...

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Math 1B — Sample Final #2 Solutions Fall Program for Freshmen 2009 Fred Bourgoin 1. Evaluate each integral. (a) Z cos x 4 - sin 2 x dx Solution. Let u = sin x . Then du = cos x dx and Z cos x 4 - sin 2 x dx = Z 1 4 - u 2 du = 1 4 Z 1 2 - u - 1 2 + u du = 1 4 ln 2 - u 2 + u + C = 1 4 ln 2 - sin x 2 + sin x + C (b) Z π/ 2 0 cos 3 θ sin θ dθ Solution. Let u = cos θ . Then du = - sin θ dθ and Z π/ 2 0 cos 3 θ sin θ dθ = - Z 0 1 u 3 du = 1 4 u 4 1 0 = 1 4 . 2. Determine whether the series X n =2 1 n ln n converges or diverges. Solution. Let f ( x ) = 1 x ln x . Then f is positive, continuous, and decreasing on [2 , ), so we can use the integral test. Z 2 1 x ln x dx = lim t →∞ h 2 ln x i t 2 = lim t →∞ (2 ln t - 2 ln 2) = . The integral diverges, so the series diverges as well. 3. Evaluate Z 4 0 ln x x dx or show that it is divergent. Solution. For the integration, proceed by parts with u = ln x and dv = 1 x dx . Z ln x x dx = 2 x ln x - Z 2 x dx = 2 x ln x - 4 x + C = 2 x (ln x - 2) + C. Hence, Z 4 0 ln x x dx = lim t 0 + Z 4 t ln x x dx = lim t 0 + h 2 x (ln x - 2) i 4 t = lim t 0 + 8(ln 2 - 1) - 2 t (ln t - 2) = . (L’Hˆ opital’s rule was used there.) Therefore, the integral diverges. 1

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4. Consider the integral Z 4 0 f ( x ) dx , where f ( x ) = x 2 + 1. [Note that f 00 ( x ) = ( x 2 + 1) - 3 / 2 .] (a) If we use n = 4, which is more likely to give a better approximation: the midpoint rule or the trapezoidal rule? Why? Solution. The midpoint is in general more accurate. In this case, since the function is concave up, the trapezoidal rule will definitely give an overesti- mate.
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1b-2009-final_exam_sample_2_solutions - Math 1B Sample...

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