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 for the second term) = ln | y - 1 | - 1 2 ln | y 2 + 9 | - 1 3 tan - 1 y 3 | 3 0 = ln | 3 - 1 | - 1 2 ln | 9 + 9 | - 1 3 tan - 1 3 3 - ln | - 1 | - 1 2 ln | 9 | - 1 3 tan - 1 0 3 (Use the fact that ln | 1 | = 0 , tan - 1 0 = 0 , and tan - 1 1 = π 4 ) = ln | 2 | - ln(18) 2 - π 12 - 0 + ln(9) 2 + 0 = ln ˆ 2 9 18 ! - π 12 = ln( 2) - π 12 2. Compute the integral Z 2 x 4 + x 3 - x 2 - 1 x ( x + 1) 2 dx Solution: We want to compute it using Partial Fraction. Since the degree in the numerator is largerer we can’t go for the Partial Fration right away. First we compute the long division.
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