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# solution_2pdf - armington(kma786 – Chapter 7 – Stepp...

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Unformatted text preview: armington (kma786) – Chapter 7 – Stepp – (55860) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of f (- 1) when f ′ ( x ) = 8 xe − 7 x 2 , f (0) = 8 . 1. f (- 1) =- 4 7 e 7 2. f (- 1) = 60 7- 4 7 e 7 3. f (- 1) = 4 7 e 7- 60 7 4. f (- 1) = 60 7 + 4 7 e − 7 5. f (- 1) =- 4 7 e − 7 6. f (- 1) = 60 7- 4 7 e − 7 correct Explanation: By integration f ( x ) = integraldisplay 8 xe − 7 x 2 dx. After substitution u = x 2 the integral be- comes 4 integraldisplay e − 7 u du =- 4 7 e − 7 u + C with C an arbitrary constant. Thus f ( x ) =- 4 7 e − 7 x 2 + C. The condition f (0) = 8 determines C since f (0) = 8 = ⇒ C = 60 7 , i.e. , f ( x ) = 60 7- 4 7 e − 7 x 2 . At x =- 1, therefore, f (- 1) = 60 7- 4 7 e − 7 . 002 10.0 points Find the value of the integral I = integraldisplay π 4 5 e tan x + 2 cos 2 x dx ....
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## This note was uploaded on 02/21/2011 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas.

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solution_2pdf - armington(kma786 – Chapter 7 – Stepp...

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