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Exam 2 - Version 018 EXAM 2 sachse(56620 This print-out...

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Version 018 – EXAM 2 – sachse – (56620) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points To which of the following does the integral I = integraldisplay x 5 1 - x 2 dx reduce after an appropriate trig substitution? 1. I = integraldisplay sin 5 θ sec 5 θ dθ 2. I = integraldisplay sec 5 θ sin 6 θ dθ 3. I = integraldisplay sin 5 θ dθ correct 4. I = integraldisplay sin 5 θ sec 6 θ dθ 5. I = integraldisplay tan 5 θ sec θ dθ 6. I = integraldisplay tan θ sec 5 θ dθ Explanation: Set x = sin θ . Then dx = cos θ dθ, radicalbig 1 - sin 2 θ = cos θ . In this case I = integraldisplay sin 5 θ cos θ cos θ dθ . Consequently, I = integraldisplay sin 5 θ dθ . 002 10.0 points Evaluate the definite integral I = integraldisplay 1 0 18 2 x 2 - 3 x - 9 dx . 1. I = 4 ln 2 5 2. I = 2 ln 4 5 3. I = 2 ln 3 5 4. I = 2 ln 2 5 correct 5. I = 4 ln 4 5 6. I = 4 ln 3 5 Explanation: By partial fractions, 18 2 x 2 - 3 x - 9 = 2 x - 3 - 4 2 x + 3 . Thus I = integraldisplay 1 0 2 x - 3 dx - integraldisplay 1 0 4 2 x + 3 dx . But integraldisplay 1 0 2 x - 3 dx = bracketleftBig 2 ln | x - 3 | bracketrightBig 1 0 , while integraldisplay 1 0 4 2 x + 3 dx = bracketleftBig 2 ln | 2 x + 3 | bracketrightBig 1 0 . Consequently, I = bracketleftbigg 2 ln vextendsingle vextendsingle vextendsingle vextendsingle x - 3 2 x + 3 vextendsingle vextendsingle vextendsingle vextendsingle bracketrightbigg 1 0 = 2 ln 2 5 . 003 10.0 points Evaluate the definite integral I = integraldisplay e 1 18 ln x x 4 dx. 1. I = - 4 e 3

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Version 018 – EXAM 2 – sachse – (56620) 2 2. I = 2 parenleftBig 1 + 4 e 3 parenrightBig 3. I = 6 5 parenleftBig 1 - 4 e 5 parenrightBig 4. I = 6 5 parenleftBig 1 + 4 e 5 parenrightBig 5. I = 2 parenleftBig 1 - 4 e 3 parenrightBig correct Explanation: After integration by parts integraldisplay e 1 3 ln x x 4 dx = bracketleftBig - ln x x 3 bracketrightBig e 1 + integraldisplay e 1 1 x 4 dx. But bracketleftBig - ln x x 3 bracketrightBig e 1 = - 1 e 3 , since ln e = 1 and ln 1 = 0, while integraldisplay e 1 1 x 4 dx = 1 3 parenleftbigg 1 - 1 e 3 parenrightbigg . Thus I = 2 parenleftbigg 1 - 4 e 3 parenrightbigg . 004 10.0 points Evaluate the integral I = integraldisplay π/ 2 0 cos 3 x dx . 1. I = 2 3 correct 2. I = 1 3. I = 1 6 4. I = 5 6 5. I = 1 3 Explanation: Since cos 2 x = 1 - sin 2 x , we see that I = integraldisplay π/ 2 0 (1 - sin 2 x ) cos x dx . This suggests using the substitution u = sin x . For then du = cos x dx , while x = 0 = u = 0 , x = π 2 = u = 1 .
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