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PostClass 7.4-solutions

# PostClass 7.4-solutions - eakin(ace669 PostClass 7.4...

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eakin (ace669) – PostClass 7.4 – morales – (56430) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Evaluate the integral I = i 4 2 2 ( x - 1)(5 - x ) dx . 1. I = 2 ln(9) 2. I = 1 2 ln p 15 7 P 3. I = 1 3 ln (9) 4. I = 1 3 ln p 15 7 P 5. I = 1 2 ln (9) correct 6. I = 2 ln p 15 7 P Explanation: To fnd A, B so that 2 ( x - 1)(5 - x ) = A x - 1 + B 5 - x , we frst bring the right hand side to a common denominator. In this case, 2 ( x - 1)(5 - x ) = A (5 - x ) + B ( x - 1) ( x - 1)(5 - x ) , and so A (5 - x ) + B ( x - 1) = 2 . To fnd the values oF A, and B , we can make particular choices oF x : x = 1 = A = 1 2 , and x = 5 = B = 1 2 . Thus I = i 4 2 1 2 ± 1 x - 1 + 1 5 - x ² dx. Hence aFter integration, I = b 1 2 (ln( x - 1) - ln(5 - x )) B 4 2 = 1 2 b ln p x - 1 5 - x PB 4 2 . Consequently, I = 1 2 ln (9) . 002 10.0 points Evaluate the defnite integral I = i ln 2 0 3 e x + 4 dx . 1. I = 4 3 ln 2 2. I = 3 4 ln 5 3 correct 3. I = 3 ln 2 4. I = 3 ln 5 3 5. I = 3 4 ln 2 6. I = 4 3 ln 5 3 Explanation: Set u = e x + 4 . Then du = e x dx = ( u - 4) dx, while x = 0 = u = 5 , x = ln 2 = u = 6 .

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eakin (ace669) – PostClass 7.4 – morales – (56430) 2 In this case, I = i 6 5 3 u ( u - 4) du . To evaluate this last integral we need to use partial fractions: for then 1 u ( u - 4) = 1 4 p 1 u - 4 - 1 u P .
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PostClass 7.4-solutions - eakin(ace669 PostClass 7.4...

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