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# XM2SOL - MAT 126 Calculus B Fall 2009 Midterm II Solutions...

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MAT 126: Calculus B, Fall 2009 Midterm II Solutions 1. Evaluate the integral: Z 1 0 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 0 1 - x 4 - x 2 dx = Z 1 0 1 4 - x 2 dx - Z 1 0 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 0 1 4 - x 2 dx = 1 4 Z 1 0 dx 2 - x - 1 4 Z 1 0 dx 2 + x = - 1 4 ln | 2 - x | 1 0 - 1 4 ln | 2 + x | 1 0 = - 1 4 ln | 2 - 1 | + 1 4 ln | 2 - 0 | - 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 0 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 . Combining everthing yields that Z 1 0 1 - x 4 - x 2 dx = - ln 2 - ln 4 3 + ln 2 - ln 3 .

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2. Find the derivative of the following function: s ( x ) = Z 1 2 1 2 sin x dt 1 - t 2 .
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