HW 06 solution - gutierrez (ig3472) HW06 Neitzke (56585) 1...

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Unformatted text preview: gutierrez (ig3472) HW06 Neitzke (56585) 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 Determine the integral I = integraldisplay 1 1 + 9( x- 1) 2 dx . 1. I = tan- 1 3( x- 1) + C 2. I = 3 sin- 1 parenleftBig x- 1 3 parenrightBig + C 3. I = sin- 1 3( x- 1) + C 4. I = 1 3 tan- 1 3( x- 1) + C correct 5. I = 1 3 sin- 1 3( x- 1) + C 6. I = 3 tan- 1 parenleftBig x- 1 3 parenrightBig + C Explanation: Since d dx tan- 1 x = 1 1 + x 2 , the substitution u = 3( x- 1) is suggested. For then du = 3 dx , in which case I = 1 3 integraldisplay 1 1 + u 2 du = 1 3 tan- 1 u + C , with C an arbitrary constant. Consequently, I = 1 3 tan- 1 3( x- 1) + C . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay 1 8 1 1- 16 x 2 dx . Correct answer: 0 . 1309. Explanation: Since integraldisplay 1 1- x 2 dx = sin- 1 x + C , a change of variable x is needed to reduce I to this form. Set u = 4 x . Then du = 4 dx , and x = 0 = u = 0 , while x = 1 8 = u = 1 2 . In this case I = 1 4 integraldisplay 1 2 1 1- u 2 du = bracketleftbigg 1 4 sin- 1 u bracketrightbigg 1 2 . Consequently, I = 1 4 arcsin parenleftbigg 1 2 parenrightbigg = 0 . 1309 . 003 10.0 points Determine the indefinite integral I = integraldisplay ( 1- x 2 )- 1 / 2 1 + 2 sin- 1 x dx . 1. I =- 1 2 ( 1 + 2 sin- 1 x ) 2 + C 2. I = 1 4 ( 1 + 2 sin- 1 x ) 2 + C 3. I = 1 4 ln vextendsingle vextendsingle 1 + 2 sin- 1 x vextendsingle vextendsingle + C 4. I = 1 2 ln vextendsingle vextendsingle 1 + 2 sin- 1 x vextendsingle vextendsingle + C correct 5. I =- 1 2 ln vextendsingle vextendsingle 1 + 2 sin- 1 x vextendsingle vextendsingle + C 6. I =- 1 4 ( 1 + 2 sin- 1 x ) 2 + C gutierrez (ig3472) HW06 Neitzke (56585) 2 Explanation: Set u = 1 + 2 sin- 1 x . Then du = 2 1- x 2 dx = 2 ( 1- x 2 )- 1 / 2 dx , so I = 1 2 integraldisplay 1 u du = 1 2 ln vextendsingle vextendsingle 1 + 2 sin- 1 x vextendsingle vextendsingle + C . Consequently, I = 1 2 ln vextendsingle vextendsingle 1 + 2 sin- 1 x vextendsingle vextendsingle + C . 004 10.0 points Evaluate the definite integral I = integraldisplay 2 sin 1 + cos 2 d . 1. I = correct 2. I = 5 4 3. I = 3 2 4. I = 3 4 5. I = 7 4 Explanation: Since d d cos =- sin , the substitution u = cos is suggested. For then du =- sin d , while = 0 = u = 1 , = = u =- 1 , so that I =- 2 integraldisplay- 1 1 1 1 + u 2 du = 2 integraldisplay 1- 1 1 1 + u 2 du , which can now be integrated using the fact that d du tan- 1 u = 1 1 + u 2 ....
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This note was uploaded on 04/13/2010 for the course MATH 408L taught by Professor Gogolev during the Spring '09 term at University of Texas at Austin.

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HW 06 solution - gutierrez (ig3472) HW06 Neitzke (56585) 1...

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