HW06-solutions - Van Ligten(hlv63 – HW06 – Gilbert...

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Unformatted text preview: Van Ligten (hlv63) – HW06 – Gilbert – (56650) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the most general function f such that f ′ ( x ) = 6 x- 7 √ 9- x 2 . 1. f ( x ) = 6 x + 7 sin − 1 x 3 + C 2. f ( x ) = 3 x 2- 7 sin − 1 x + C 3. f ( x ) = 6 x + 7 3 tan − 1 x + C 4. f ( x ) = 3 x + 7 3 tan − 1 x 3 + C 5. f ( x ) = 3 x 2- 7 3 tan − 1 x 3 + C 6. f ( x ) = 3 x 2- 7 sin − 1 x 3 + C correct Explanation: Since d dx parenleftBig sin − 1 x 3 parenrightBig = 1 √ 9- x 2 , we see that f ( x ) = 3 x 2- 7 sin − 1 x 3 + C with C an arbitrary constant. 002 10.0 points Determine the indefinite integral I = integraldisplay ( 1- x 2 ) − 1 / 2 4 + sin − 1 x dx . 1. I = ln vextendsingle vextendsingle 4 + sin − 1 x vextendsingle vextendsingle + C correct 2. I =- ln vextendsingle vextendsingle 4 + sin − 1 x vextendsingle vextendsingle + C 3. I = 1 2 ln vextendsingle vextendsingle 4 + sin − 1 x vextendsingle vextendsingle + C 4. I = 1 2 ( 4 + sin − 1 x ) 2 + C 5. I =- 1 2 ( 4 + sin − 1 x ) 2 + C 6. I =- ( 4 + sin − 1 x ) 2 + C Explanation: Set u = 4 + sin − 1 x . Then du = 1 √ 1- x 2 dx = ( 1- x 2 ) − 1 / 2 dx , so I = integraldisplay 1 u du = ln vextendsingle vextendsingle 4 + sin − 1 x vextendsingle vextendsingle + C . Consequently, I = ln vextendsingle vextendsingle 4 + sin − 1 x vextendsingle vextendsingle + C . 003 10.0 points Determine the integral I = integraldisplay 1 4- x 1 + x 2 dx . 1. I = 1 2 (2 π- ln4) 2. I = 1 2 (2 π + ln2) 3. I = 2 π- ln4 4. I = 2 π + ln4 5. I = 1 2 (2 π- ln2) correct 6. I = 2 π + ln2 Explanation: Van Ligten (hlv63) – HW06 – Gilbert – (56650) 2 We deal with the two integrals I 1 = integraldisplay 1 4 1 + x 2 dx, I 2 = integraldisplay 1 x 1 + x 2 dx separately. Now d dx tan − 1 x = 1 1 + x 2 , so we see that I 1 = bracketleftBig 4 tan − 1 x bracketrightBig 1 = π . On the the other hand, to evaluate I 2 set u = 1 + x 2 . Then du = 2 x dx , and x = 0 = ⇒ u = 1 , while x = 1 = ⇒ u = 2 . In this case, I 2 = 1 2 integraldisplay 2 1 1 u du = 1 2 bracketleftBig ln u bracketrightBig 2 1 . Consequently, I = 1 2 (2 π- ln2) . keywords: 004 10.0 points Determine the indefinite integral I = integraldisplay 4 t 3 √ 1- t 8 dt . 1. I = radicalbig 1- t 8 + C 2. I = 4 sin − 1 ( t 4 ) + C 3. I = tan − 1 ( t 4 ) + C 4. I = 4 √ 1- t 8 + C 5. I = sin − 1 ( t 4 ) + C correct 6. I = 4 tan − 1 ( t 4 ) + C Explanation: Since integraldisplay 1 √ 1- x 2 dx = sin − 1 x + C , we need to reduce I to this form by changing t . Indeed, set x = t 4 . Then dx = 4 t 3 dt ....
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This note was uploaded on 06/10/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.

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HW06-solutions - Van Ligten(hlv63 – HW06 – Gilbert...

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