Solutions_Assignment3

Solutions_Assignment3 - Assignment 3 Solutions MATH 138...

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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 . Therefore, A = 2 and C = - 2 . So the integral can be written as, Z 1 s 2 ( s - 1) 2 ds = Z 2 s + 1 s 2 - 2 s - 1 + 1 ( s - 1) 2 ds, = 2 ln | s | - 1 s - 2 ln | s - 1 | - 1 ( s - 1) + C. 1
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24) Evaluate the integral, Z x 2 - x + 6 x 3 + 3 x dx.
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