Exam 2 - Version 018 EXAM 2 sachse (56620) 1 This print-out...

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Unformatted text preview: 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 sec 5 d 2. I = integraldisplay sec 5 sin 6 d 3. I = integraldisplay sin 5 d correct 4. I = integraldisplay sin 5 sec 6 d 5. I = integraldisplay tan 5 sec d 6. I = integraldisplay tan sec 5 d Explanation: Set x = sin . Then dx = cos d, radicalbig 1- sin 2 = cos . In this case I = integraldisplay sin 5 cos cos d . Consequently, I = integraldisplay sin 5 d . 002 10.0 points Evaluate the definite integral I = integraldisplay 1 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 2 x- 3 dx- integraldisplay 1 4 2 x + 3 dx . But integraldisplay 1 2 x- 3 dx = bracketleftBig 2 ln | x- 3 | bracketrightBig 1 , while integraldisplay 1 4 2 x + 3 dx = bracketleftBig 2 ln | 2 x + 3 | bracketrightBig 1 . Consequently, I = bracketleftbigg 2 ln vextendsingle vextendsingle vextendsingle vextendsingle x- 3 2 x + 3 vextendsingle vextendsingle vextendsingle vextendsingle bracketrightbigg 1 = 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 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 ln1 = 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 / 2 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 / 2 (1- sin 2 x )cos x dx ....
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Exam 2 - Version 018 EXAM 2 sachse (56620) 1 This print-out...

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