1b-2009-exam_1_solutions

# 1b-2009-exam_1_solutions - Math 1B Exam#1 Solutions Fall...

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Math 1B — Exam #1 Solutions Fall Program for Freshmen 2009 Fred Bourgoin 1. Evaluate the following integrals. (a) Z dx e x 1 - e - 2 x Solution. Let u = e - x . Then du = - e - x dx , so that dx = - du u and Z dx e x 1 - e - 2 x = - Z du 1 - u 2 = - arcsin u + C = - arcsin( e - x ) + C. (b) Z 1 y 2 - 4 y - 12 dy Solution. Proceed using partial fractions. Z 1 y 2 - 4 y - 12 dy = 1 8 Z ± 1 y - 6 - 1 y + 2 ² dy = 1 8 (ln | y - 6 | - ln | y + 2 | ) + C. 2. Determine whether the integral Z 2 dx x ln x converges. If it does, evaluate it. Solution. First rewrite the integral as a limit. The antiderivative is found using the substitution u = ln x . Z 2 dx x ln x = lim t →∞ Z t 2 dx x ln x = lim t →∞ h ln(ln x ) i t 2 = lim t →∞ h ln(ln t ) - ln(ln 2) i = . Therefore, the integral diverges. 3. Find the arc length of f ( x ) = x 4 16 + 1 2 x 2 for 1 x 2. Solution. The derivative is f 0 ( x ) = x 3 4 - 1 x 3 , so the arc length is s = Z 2 1 s 1 + ± x 3 4 - 1 x 3 ² 2 dx = Z 2 1 r 1 + x 6 16 - 1 2 + 1 x 6 dx = Z 2 1 r x 6 16 + 1 2 + 1 x 6 dx = Z 2 1 s ± x 3 4 + 1 x 3 ² 2 dx = Z 2 1 ± x 3 4 + 1 x 3 ² dx = ³ x 4 16 - 1 2 x 2 ´ 2 1 = ± 1 - 1 8 ² - ± 1 16 - 1 2 ² = 21 16 1 . 31 .

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## This note was uploaded on 02/19/2012 for the course MATH 1 taught by Professor Wilkening during the Spring '08 term at Berkeley.

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1b-2009-exam_1_solutions - Math 1B Exam#1 Solutions Fall...

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