HW14-solutions - husain (aih243) HW14 Gilbert (56215) 1...

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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 / 4 / 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 = 4 = u = 10 2 , x = 3 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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HW14-solutions - husain (aih243) HW14 Gilbert (56215) 1...

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