MTH 161 Midterm: MTH 161 University of Rochester 161 Midterm 2 Fall 2006

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31 Jan 2019
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We use u-substitution, with u = x2 + 3, so du = 2xdx. 2 (cid:18)2 u3/2(cid:19) + c (x2 + 3)5/2 (x2 + 3)3/2 + c. u5/2 3. We use the trigonometric substitution x = (3/2) sin u, so dx = (3/2) cos u du. 18 z sec2 u du cos3 u du. Drawing a triangle, we nd that tan u = Z x2 + 3x x2 + 3x + 2 dx and so the answer is: (8 points) Using long division of fractions, we nd that x2 + 3x x2 + 3x + 2. 2 x + 2 and so: (10 points) Z x2 + 3x x2 + 3x + 2 dx = z dx z. = x 2 ln(x + 1) + 2 ln(x + 2) + c. The density of water is 1000 kg/m3 and the gravitational constant is 9. 8 m/s2.

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