int ca l6 - pokharel (yp624) HW06 Radin (57410) 1 This...

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Unformatted text preview: pokharel (yp624) HW06 Radin (57410) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay e 1 1 x { 2 f (ln x )- 1 } dx when f (0) = 2 and f (1) = 4. 1. I = 2 2. I = 4 3. I = 5 4. I = 3 correct 5. I = 6 Explanation: Set u = ln x . Then while x = 1 = u = 0 , x = e = u = 1 . In this case, I = integraldisplay 1 { 2 f ( u )- 1 } du = integraldisplay 1 { 2 f ( u )- 1 } du = bracketleftBig 2 f ( u )- u bracketrightBig 1 . Consequently, I = 2 { f (2)- f (1) }- 1 = 3 . 002 10.0 points Evaluate the definite integral I = integraldisplay 1 1 3- 2 x dx . 1. I = ln 3 2 2. I = 1 2 ln3 correct 3. I = 1 2 ln 5 3 4. I = 1 2 ln 3 2 5. I = ln 3 6. I = ln 5 3 Explanation: Set u = 3- 2 x ; then du =- 2 dx while x = 0 = u = 3 x = 1 = u = 1 . In this case, I =- 1 2 integraldisplay 1 3 1 du du = 1 2 integraldisplay 3 1 1 u du = 1 2 bracketleftBig ln | u | bracketrightBig 3 1 . Consequently, I = 1 2 (ln 3- ln 1) = 1 2 ln 3 . 003 10.0 points Determine the indefinite integral I = integraldisplay x 3 x 2 + 11 dx . 1. I = 1 2 x 2 + 11 2 ln( x 2 + 11) + C 2. I = x 2- x ln( x 2 + 11) + C 3. I = 1 2 x 2- 11 2 ln( x 2 + 11) + C correct 4. I = 1 3 x 3- 11 ln( x 2 + 11) + C pokharel (yp624) HW06 Radin (57410) 2 5. I = 1 3 x 3 + 11 ln( x 2 + 11) + C 6. I = x 2 + x ln( x 2 + 11) + C Explanation: After long division on the integrand, x 3 x 2 + 11 = x- 11 x x 2 + 11 . Thus I = integraldisplay x dx- integraldisplay 11 x x 2 + 11 dx . To evaluate the last integral we use the sub- stitution u = x 2 + 11. For then du = 2 x dx , so I = 1 2 x 2- 11 2 integraldisplay 1 u du ....
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This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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int ca l6 - pokharel (yp624) HW06 Radin (57410) 1 This...

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