MATH 141 Midterm: MATH 141 TAMU Fall 17 Exam 3RevPrbl

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31 Jan 2019
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Solutions to: math 152, fall 2008, examination i, version a: evaluate r e. We have du = 1 u = 1. = z 1 (ln x)1/2 dx x u3/2 1. 2 (u + 1) and dx = 1: evaluate r 1. We have x = 1 and when x = 1, u = 1. 18: compute the area under the curve y = xe x, between x = 0 and x = 1. The area under the curve is the integral. We compute the integral using integration by parts with the choice u = x, dv = e xdx. We have du = dx and v = e x. Applying the integration by parts formula we have. 0 = e 1 e 1 + 1 = 1 2e 1. 1: compute the area enclosed by the parabola y = x2 and the straight line y = 2x.

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