exam2 - Version 078 EXAM 2 BERG (56525) 1 This print-out...

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Unformatted text preview: Version 078 EXAM 2 BERG (56525) 1 This print-out should have 18 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 6 d 2. I = integraldisplay tan sec 5 d 3. I = integraldisplay tan 5 sec d 4. I = integraldisplay sin 5 sec 5 d 5. I = integraldisplay sin 5 d correct 6. I = integraldisplay sec 5 sin 6 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 integral I = integraldisplay 4 1 ln t 8 t dt . 1. I = ln 2 1 2. I = ln 2 + 1 3. I = 1 8 (ln2 + 1) 4. I = 1 8 (ln4 1) 5. I = 1 2 (ln4 + 1) 6. I = 1 2 (ln4 1) correct Explanation: After integration by parts, I = 1 4 bracketleftBig t ln t bracketrightBig 4 1 1 4 integraldisplay 4 1 t parenleftBig 1 t parenrightBig dt = 1 2 ln 4 1 4 integraldisplay 4 1 1 t dt . But integraldisplay 4 1 1 t dt = 2 bracketleftBig t bracketrightBig 4 1 . Consequently, I = 1 2 ln 4 1 2 = 1 2 (ln 4 1) . keywords: integration by parts, logarithmic functions 003 10.0 points Evaluate the integral I = integraldisplay / 2 sin 2 x cos 3 x dx . 1. I = 8 15 2. I = 1 15 Version 078 EXAM 2 BERG (56525) 2 3. I = 2 15 correct 4. I = 2 5 5. I = 4 15 Explanation: Since sin 2 x cos 3 x = (sin 2 x cos 2 x ) cos x = sin 2 x (1 sin 2 x )cos x = (sin 2 x sin 4 x )cos x , the integrand is of the form cos xf (sin x ), sug- gesting use of the substitution u = sin x . For then du = cos x dx , while x = 0 = u = 0 x = 2 = u = 1 . In this case I = integraldisplay 1 ( u 2 u 4 ) du . Consequently, I = bracketleftBig 1 3 u 3 1 5 u 5 bracketrightBig 1 = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 004 10.0 points Determine the indefinite integral I = integraldisplay 1 12 x x 2 dx . 1. I = 1 6 sin 1 ( x 6) + C 2. I = sin 1 parenleftBig x + 6 6 parenrightBig + C 3. I = 12 x x 2 6 + C 4. I = 2 radicalbig 12 x x 2 + C 5. I = sin 1 parenleftBig x 6 6 parenrightBig + C correct Explanation: After completing the square we see that x 2 12 x = ( x 2 12 x + 36) 36 = ( x 6) 2 36 . In this case I = integraldisplay 1 radicalbig 36 ( x 6) 2 dx . To evaluate this last integral, set x 6 = 6 sin ....
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This note was uploaded on 04/04/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas at Austin.

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exam2 - Version 078 EXAM 2 BERG (56525) 1 This print-out...

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