Solutions_Assignment3

Solutions_Assignment3 - Assignment 3 Solutions MATH 138...

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Unformatted text preview: Assignment 3: Solutions MATH 138 2012 Section 7.4: 14) Evaluate the integral, Z 1 ( x + a )( x + b ) dx. Solution: If a 6 = b then we have the following decomposition, 1 ( x + a )( x + b ) = 1 b- a 1 x + a- 1 x + b With this it is easy to integrate the problem above, Z 1 ( x + a )( x + b ) dx = 1 b- a (ln | x + a | - ln | x + b | ) + C = 1 b- a ln x + a x + b + C. In the other case where a = b we can integrate the problem directly, Z 1 ( x + a ) 2 dx =- 1 x + a + C. 22) Evaluate the integral, Z ds s 2 ( s- 1) 2 . Solution We must begin by rewriting the integrand, 1 s 2 ( s- 1) 2 = A s + B s 2 + C s- 1 + D ( s- 1) 2 We multiply the equation by the denominator on the left and obtain, 1 = As ( s- 1) 2 + B ( s- 1) 2 + Cs 2 ( s- 1) + Ds 2 If we substitute s = 0 then B = 1 . If we set s = 1 then D = 1 . The coefficient of s 3 yields A + C = 0 or C =- A . Finally, if we set s = 2 then 2 A + B + 4 C + 4 D = 1 , 2 A + 1- 4 A + 4 = 1 ,- 2 A =- 4 ....
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This note was uploaded on 02/22/2012 for the course CS cs136 taught by Professor Cormack during the Winter '10 term at Waterloo.

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Solutions_Assignment3 - Assignment 3 Solutions MATH 138...

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