ma241exam1-c-key

ma241exam1-c-key - Spring 2007 Therkelsen MA 241 - Exam 1...

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Spring 2007 Therkelsen MA 241 - Exam 1 (C) KEY 1. Evaluate 1 Z 0 x 4 x 2 dx Solution: Let u =4 x 2 , then du = 2 xdx . If we also transform the limits of integration, 4 (1) 2 = 3 and 4 (0) 2 ,wehave 1 Z 0 x 4 x 2 dx = 1 2 3 Z 4 1 u du = 1 2 (2 u ) 3 4 = u 3 4 = 3 ( 4) =2 3 2. Integrate Z x sin 4 xdx Solution: Let u = x and dv = sin 4 . Then du = dx and v = 1 4 cos 4 x . Integration by parts (and appropriate substitutions) yields the result Z x sin 4 = 1 4 x cos 4 x + Z 1 4 cos 4 = 1 4 x cos 4 x + 1 16 sin 4 x + C 1
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3. Integrate Z x 2 2 x 1 ( x 1)( x 2 +1) dx Solution: We fnd the coefficients in the Following partial Fractions decomposition x 2 2 x 1 ( x 1)( x 2 = A x 1 + Bx + C x 2 +1 Multiplying both sides by ( x 1)( x 2 + 1) yields x 2 2 x 1= A ( x 2 +1)+( + C )( x 1) = Ax 2 + A + 2 + Cx C =( A + B ) x 2 +( B + C ) x A C ) We thereFore have the Following system oF linear equations A + B =1 B + C = 2 A C = 1
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ma241exam1-c-key - Spring 2007 Therkelsen MA 241 - Exam 1...

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