HW14-solutions

# HW14-solutions - husain(aih243 – HW14 – Gilbert...

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Unformatted text preview: husain (aih243) – HW14 – Gilbert – (56215) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay 1 (2 e x − e − x ) 2 dx . 1. I = 2 e 2 + 7 + e − 2 2. I = 2 e 2 − 7 − e − 2 3. I = 2 e 2 + 4 + e − 2 4. I = 2 e 2 − 11 2 + 1 2 e − 2 5. I = 2 e 2 + 4 − 1 2 e − 2 6. I = 2 e 2 − 11 2 − 1 2 e − 2 correct Explanation: After expansion (2 e x − e − x ) 2 = 4 e 2 x − 4 + e − 2 x , in which case I = integraldisplay 1 ( 4 e 2 x − 4 + e − 2 x ) dx = bracketleftBig 2 e 2 x − 4 x − 1 2 e − 2 x bracketrightBig 1 . Consequently, I = 2 e 2 − 11 2 − 1 2 e − 2 . 002 10.0 points Determine the indefinite integral I = integraldisplay 4 x 2 − x − 2 x dx. 1. I = 2 x 2 − x − 2 ln | x | 2. I = 2 x 2 − x − 2 ln | x | + C correct 3. I = 4 x 2 + x + 2 ln | x | + C 4. I = 4 x 2 − x − 2 ln | x | + C 5. I = 4 x 2 + x + 2 ln | x | 6. I = 2 x 2 + x + 2 ln | x | Explanation: After division, 4 x 2 − x − 2 x = 4 x − 1 − 2 x , so we can now integrate term by term using the fact that integraldisplay x r dx = x r +1 r + 1 + C, ( r negationslash = − 1) and that integraldisplay 1 x dx = ln | x | + C . Thus integraldisplay 4 x 2 − x − 2 x dx = 4 integraldisplay x dx − integraldisplay dx − 2 integraldisplay 1 x dx . Consequently, I = 2 x 2 − x − 2 ln | x | + C with C an arbitrary constant. 003 10.0 points Evaluate the definite integral I = integraldisplay 3 π/ 4 π/ 4 7 cos x − 3 sin x 7 sin x + 3 cos x dx . 1. I = − ln parenleftbigg 5 2 parenrightbigg correct 2. I = ln(3) husain (aih243) – HW14 – Gilbert – (56215) 2 3. I = ln parenleftbigg 5 2 parenrightbigg 4. none of these 5. I = − ln (3) Explanation: Since the integrand is of the form f ′ ( x ) f ( x ) , f ( x ) = 7 sin x + 3 cos x this suggests the substitution u = 7 sin x + 3 cos x . For then du = (7 cos x − 3 sin x ) dx, while x = π 4 = ⇒ u = 10 √ 2 , x = 3 π 4 = ⇒ u = 4 √ 2 . Thus I = integraldisplay 4 / √ 2 10 / √ 2 1 u du = bracketleftBig ln | u | bracketrightBig 4 / √ 2 10 / √ 2 . Consequently, I = ln parenleftBig 2 5 parenrightBig = − ln parenleftbigg 5 2 parenrightbigg ....
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## This note was uploaded on 11/30/2010 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.

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HW14-solutions - husain(aih243 – HW14 – Gilbert...

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