MATH 1120 Midterm: MATH 1120 Cornell e1sol3 4 18Exam1
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Math 1120, solution to prelim i, february 22, 2018. Problem 1 (20 pts): compute the de nite integrals or antiderivatives. (a)(5pts)z e. The expression inside the integral has 1 x dx and ln x. So substitute u = ln x, du = 1 u2 + 1 x((ln x)2 + 1) dx =z 1 du = arctan u(cid:12)(cid:12)(cid:12)(cid:12) Let u = x 2, then du = 1dx = dx, and x = u + 2. Noticing that x ranges from 2 to 4, so correspondingly u = x 2 ranges from 2 2 = 0 to 4 2 = 2. = arctan(1) arctan(0) = x x 2 dx. The denominator 2 2x + x2 is a quadratic polynomial in x, so we complete the square: 2 2x + x2 = (x 1)2 + 1. Change variables u = x 1. dx =z.
