XM2SOL - MAT 126: Calculus B, Fall 2009 Midterm II...

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Unformatted text preview: MAT 126: Calculus B, Fall 2009 Midterm II Solutions 1. Evaluate the integral: Z 1 1- x 4- x 2 dx. Sol. The key to this solution is to recall that x + y z for z 6 = 0 separates as x z + y z . Hence, we may rewrite the given integral as Z 1 1- x 4- x 2 dx = Z 1 1 4- x 2 dx- Z 1 x 4- x 2 dx. In the first integral, we proceed by partial fractions. A computation follows: = A + B 4- x 2 = A + B (2- x )(2 + x ) = A 2- x + B 2 + x = A (2 + x ) + B (2- x ) = 2 A + Ax + 2 B- Bx = (2 A + 2 B ) + ( A- B ) x From this we form a system of equations 2 A + 2 B = 1 A- B = 0 Solving this system yields A = 1 4 and B =- 1 4 . Hence, the integration yields Z 1 1 4- x 2 dx = 1 4 Z 1 dx 2- x- 1 4 Z 1 dx 2 + x =- 1 4 ln | 2- x | 1- 1 4 ln | 2 + x | 1 =- 1 4 ln | 2- 1 | + 1 4 ln | 2- | - 1 4 ln | 2 + 1 | + 1 4 ln | 2 + 0 | =- 1 4 ln | 1 | + 1 4 ln | 2 | - 1 4 ln | 3 | + 1 4 ln | 2 | = 1 2 ln 2- 1 4 ln 3 = ln 2- ln 4 3 For the second integral, we proceed by u-substitution with u = 4- x 2 . Then du =- 2 x 2 dx, so dx =- 1 2 du. Moreover, we find u (0) = 4 and u (1) = 3 . Hence, we have Z 1 x 4- x 2 dx =- 1 2 Z 3 4 du u =- 1 2 ln | u | 3 4 =- 1 2 ln 3 + 1 2 ln 4 =- 1 2 ln 3 + ln 2 = ln 2- ln 3 ....
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XM2SOL - MAT 126: Calculus B, Fall 2009 Midterm II...

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