q8-sol-1042-sp08

q8-sol-1042-sp08 - Calculus II-1042 3 Solution of Quiz 8...

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Calculus II—1042 Solution of Quiz # 8 Fri., 03/21/08 1. Compute the integral Z 3 0 10 ( y - 1)( y 2 + 9) dy Solution: Compute it using Partial Fraction. Since the degree in the numerator is smaller so NO long division. 10 ( y - 1)( y 2 + 9) = A y - 1 + By + C y 2 + 9 OR 10 ( y - 1)( y 2 + 9) = A y - 1 + By y 2 + 9 + C y 2 + 9 (clear the denominator from both sides by multiplying each term by ( y - 1)( y 2 + 9) ) 10 = A ( y 2 + 9) + By ( y - 1) + C ( y - 1) (Write the terms on the right in descending power of y ) 0 y 2 + 0 y + 10 = ( A + B ) y 2 + ( - B + C ) y + 9 A - C (The coefficients of y 2 , y and the constant terms on both sides must be equal) Which gives A + B = 0 , - B + C = 0 , and 9 A - C = 10 Solving these equations in A, B, and C together we get A = 1 , B = - 1 , C = - 1 Since 10 ( y - 1)( y 2 + 9) = 1 y - 1 + - y y 2 + 9 + - 1 y 2 + 9 So Z 3 0 10 ( y - 1)( y 2 + 9) = Z 1 y - 1 dy + Z - y y 2 + 9 dy + Z - 1 y 2 + 9 dy (Use the substitution u = y - 1 for the first term and the substitution u = y 2 + 9
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This homework help was uploaded on 04/18/2008 for the course MATH 1042 taught by Professor Dr.z during the Spring '08 term at Temple.

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