MATH 132 Midterm: MATH 132 UMass Amherst math132_spring06_html solution-exam2

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31 Jan 2019
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I :=z 1 9x2dx =z r1 . 3z 1 + cos(2 ) cos( ) = 1 9x2. d = [arcsin(3x) + 3x 1 9x2] + c = Proceed using the identities sin(2 ) = 2 sin( ) cos( ), sin( ) = 3x, and. Exam 2 solution: (16 points) for each of the following improper integrals, determine if it converges. Otherwise, explain why it or diverges. diverges: the integral is divergent. Substitute u = 1 + x2 to get: x. 0: using integration by parts, with u = ln(x) and dv = x 2dx, we get that the integral is convergent. lim t . [ln(t) ln(1)] = . ln(x) x2 dx = lim. + ln(1)(cid:21) (cid:20) 1 t 1(cid:21)(cid:27) l hopital. = [0 + 0] [0 1] = 1. 2. (a) (12 points) use the trigonometric substitution x = 1. + c (b) (3 points) the area enclosed by the ellipse 9x2 + y2 = 1.

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