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MAC 2312 Exam 2

# MAC 2312 Exam 2 - MAC 2312 Exam 2 Solutions July 7 2011...

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MAC 2312 Exam 2 Solutions July 7, 2011 Name: | {z } By writing my name, I swear by the honor code. Read all of the following information before starting the exam: Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct). Circle or otherwise indicate your final answers. This test has 5 problems and is worth 50 points, plus one bonus problem at the end. It is your responsibility to make sure that you have all of the pages! Since time is limited, it is crucial that you think before you compute. You must state any theorems or tests that you use. All hypotheses must be verified. Good luck!

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1. Evaluate the integral. Z 2 x 2 - x + 1 x 3 + x dx Solution: The integrand is a proper rational function, so we use partial fractions. 2 x 2 - x + 1 x 3 + x = 2 x 2 - x + 1 x ( x 2 + 1) = A x + Bx + C x 2 + 1 Multiplying both sides by x ( x 2 + 1) yields 2 x 2 - x + 1 = A ( x 2 + 1) + ( Bx + C ) x. Plugging in x = 0 gives that A = 1. Then, expanding and equating coefficients gives B = 1 and C = - 1. Thus, Z 2 x 2 - x + 1 x 3 + x dx = Z dx x + Z x - 1 x 2 + 1 dx = ln | x | + Z x x 2 + 1 dx - Z dx x 2 + 1 .
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