Math HW6 - Perez (nap563) HW06 Zheng (56555) 1 This...

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Unformatted text preview: Perez (nap563) HW06 Zheng (56555) 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 + 25( x- 6) 2 dx . 1. I = 5 tan- 1 parenleftBig x- 6 5 parenrightBig + C 2. I = sin- 1 5( x- 6) + C 3. I = 5 sin- 1 parenleftBig x- 6 5 parenrightBig + C 4. I = tan- 1 5( x- 6) + C 5. I = 1 5 tan- 1 5( x- 6) + C correct 6. I = 1 5 sin- 1 5( x- 6) + C Explanation: Since d dx tan- 1 x = 1 1 + x 2 , the substitution u = 5( x- 6) is suggested. For then du = 5 dx , in which case I = 1 5 integraldisplay 1 1 + u 2 du = 1 5 tan- 1 u + C , with C an arbitrary constant. Consequently, I = 1 5 tan- 1 5( x- 6) + C . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay 1 4 1 1- 4 x 2 dx . Correct answer: 0 . 261799. 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 = 2 x . Then du = 2 dx , and x = 0 = u = 0 , while x = 1 4 = u = 1 2 . In this case I = 1 2 integraldisplay 1 2 1 1- u 2 du = bracketleftbigg 1 2 sin- 1 u bracketrightbigg 1 2 . Consequently, I = 1 2 arcsin parenleftbigg 1 2 parenrightbigg = 0 . 261799 . 003 10.0 points Determine the indefinite integral I = integraldisplay ( 1- x 2 )- 1 / 2 4- 3 sin- 1 x dx . 1. I = 1 3 ( 4- 3 sin- 1 x ) 2 + C 2. I =- 1 6 ( 4 + 3 sin- 1 x ) 2 + C 3. I =- 1 6 ln vextendsingle vextendsingle 4 + 3 sin- 1 x vextendsingle vextendsingle + C 4. I =- 1 3 ln vextendsingle vextendsingle 4- 3 sin- 1 x vextendsingle vextendsingle + C correct 5. I = 1 3 ln vextendsingle vextendsingle 4- 3 sin- 1 x vextendsingle vextendsingle + C 6. I = 1 6 ( 4 + 3 sin- 1 x ) 2 + C Perez (nap563) HW06 Zheng (56555) 2 Explanation: Set u = 4- 3 sin- 1 x . Then du =- 3 1- x 2 dx =- 3 ( 1- x 2 )- 1 / 2 dx , so I =- 1 3 integraldisplay 1 u du =- 1 3 ln vextendsingle vextendsingle 4- 3 sin- 1 x vextendsingle vextendsingle + C . Consequently, I =- 1 3 ln vextendsingle vextendsingle 4- 3 sin- 1 x vextendsingle vextendsingle + C . 004 10.0 points Evaluate the definite integral I = integraldisplay / 2 4 sin 1 + cos 2 d . 1. I = 5 4 2. I = correct 3. I = 3 4 4. I = 3 2 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 , = 2 = u = 0 , so that I =- 4 integraldisplay 1 1 1 + u 2 du = 4 integraldisplay 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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Math HW6 - Perez (nap563) HW06 Zheng (56555) 1 This...

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