Quiz12-8.40-soln. - MATH 101-B Fall 2004 Quiz 12(8:40-10:30 Group Time 20 minutes Name Student No Follow the directions No work No credit Problem

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Unformatted text preview: MATH 101-B Fall 2004 Quiz 12 (8:40-10:30 Group) December 30, 2004 Time: 20 minutes Name: Student No: Follow the directions. No work No credit!! Problem 1 (5 pts.) x x x2 6 Solution. We use partial fractions. x 2 x x x6 x 3x 2 x x 3x 2 x x 3x dx 2 ? A B x3 x2 Ax 2 Bx 3 x 3x 2 A B x 2A 3B x 3x 2 A B 1 and 2A 3B 0 3, B 2. A 5 5 So, the integral can be rewritten and solved as follows: x 3 1 dx 2 1 dx dx 5 x3 5 x2 x2 x 6 3 ln|x 3| 2 ln|x 2| C 5 5 Problem 2 (5 pts.) x 1 x Solution. Let u x 1 2x 1. Then, du transformed integral will be 2 u du 3 2 u32 2 4 3 x 1 3 ? dx and hence 2du 4 u3 3 C dx 1 x dx. Therefore the C C (back to the original variable) ...
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This note was uploaded on 03/24/2012 for the course MATHEMATIC 101 taught by Professor Many during the Spring '10 term at Sabancı University.

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