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M408L hwk7 solutions

M408L hwk7 solutions - Pham Quoc – Homework 7 – Due Mar...

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Unformatted text preview: Pham, Quoc – Homework 7 – Due: Mar 6 2007, 3:00 am – Inst: Eric Katerman 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Rewrite the expression f ( x ) = 3 x- 1 x 2 ( x- 2) using partial fractions. 1. f ( x ) = 1 2 x- 5 4 x 2- 5 4( x- 2) 2. f ( x ) =- 5 4 x + 1 2 x 2 + 5 4( x- 2) correct 3. f ( x ) = 5 x- 1 2 x 2 + 5 4( x- 2) 4. f ( x ) =- 1 x 2 + 5 x- 2 5. f ( x ) = 1 x 2- 5 x- 2 Explanation: We have to find A, B, and C so that 3 x- 1 x 2 ( x- 2) = A x + B x 2 + C x- 2 = Ax ( x- 2) + B ( x- 2) + Cx 2 x 2 ( x- 2) . Thus 3 x- 1 = Ax ( x- 2) + B ( x- 2) + Cx 2 . Now x = 0 = ⇒ B = 1 2 , while x = 2 = ⇒ C = 5 4 . But then on comparing coefficients of x 2 we see that A + C = 0 = ⇒ A =- 5 4 . Consequently, f ( x ) =- 5 4 x + 1 2 x 2 + 5 4( x- 2) . keywords: partial fractions 002 (part 1 of 1) 10 points Find the unique function y satisfying the equations dy dx = 2 ( x- 2)(7- x ) , y (3) = 0 . 1. y = 2 5 ‡ ln fl fl fl 7- x x- 2 fl fl fl- ln4 · 2. y = 2 ‡ ln fl fl fl x- 2 7- x fl fl fl + ln4 · 3. y = 2 ‡ ln fl fl fl 7- x x- 2 fl fl fl- ln4 · 4. y = 1 5 ‡ ln fl fl fl x- 2 7- x fl fl fl + ln4 · 5. y = 2 5 ‡ ln fl fl fl x- 2 7- x fl fl fl + ln4 · correct Explanation: We first find A, B so that 2 ( x- 2)(7- x ) = A x- 2 + B 7- x by bringing the right hand side to a common denominator. In this case, 2 ( x- 2)(7- x ) = A (7- x ) + B ( x- 2) ( x- 2)(7- x ) , and so A (7- x ) + B ( x- 2) = 2 . To find the values of A, B particular choices of x are made When x = 2, for instance, A = 2 5 , while when x = 7, B = 2 5 . Thus dy dx = 2 5 ‡ 1 x- 2 + 1 7- x · . Pham, Quoc – Homework 7 – Due: Mar 6 2007, 3:00 am – Inst: Eric Katerman 2 Hence after integration, y = 2 5 n ln | x- 2 |- ln | 7- x | o + C = 2 5 ln fl fl fl x- 2 7- x fl fl fl + C with C an arbitrary constant. But y (3) = 0 = ⇒ C =- 2 5 ln ‡ 1 4 · , i.e. , C = 2 5 ln4 . Consequently, y = 2 5 ‡ ln fl fl fl x- 2 7- x fl fl fl + ln4 · . keywords: partial fractions, log function 003 (part 1 of 1) 10 points Evaluate the integral I = Z 2 1 6 x 3 + 4 x dx. 1. I = 3 2 ln ‡ 8 5 · 2. I = 3 8 ln ‡ 5 2 · 3. I = 3 4 ln ‡ 8 5 · 4. I = 3 4 ln ‡ 5 2 · correct 5. I = 3 8 ln ‡ 8 5 · 6. I = 3 2 ln ‡ 5 2 · Explanation: By partial fractions, 6 x 3 + 4 x = A x + Bx + C x 2 + 4 . To determine A, B, and C multiply through by x 3 + 4 x : for then 6 = A ( x 2 + 4) + x ( Bx + C ) = ( A + B ) x 2 + Cx + 4 A, which after comparing coefficients gives A =- B , C = 0 , A = 3 2 . Thus I = 3 2 Z 2 1 ‡ 1 x- x x 2 + 4 · dx = 3 2 h ln x- 1 2 ln( x 2 + 4) i 2 1 = 3 4 h ln ‡ x 2 x 2 + 4 · i 2 1 ....
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M408L hwk7 solutions - Pham Quoc – Homework 7 – Due Mar...

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