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solution_pdf5 - Horstman(mdh995 – HW05 – Radin...

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Unformatted text preview: Horstman (mdh995) – HW05 – Radin – (58505) 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 Determine the integral I = integraldisplay 2 x (3 + 2 ln x ) 3 dx. 1. I =- 1 4 (3 + 2 ln x ) 4 + C 2. I =- 1 4 ln x (3 + 2 ln x ) 2 + C 3. I =- 1 2 ln x (3 + 2 ln x ) 2 + C 4. I = 1 2 ln x (3 + 2 ln x ) 2 + C 5. I = 1 2 (3 + 2 ln x ) 4 + C 6. I = 1 4 ln x (3 + 2 ln x ) 2 + C 7. I = 1 4 (3 + 2 ln x ) 4 + C correct 8. I =- 1 2 (3 + 2 ln x ) 4 + C Explanation: Set u = 3 + 2 ln x . Then du = 2 x dx, so I = integraldisplay u 3 du = 1 4 u 4 + C . Consequently, I = 1 4 (3 + 2 ln x ) 4 + C with C an arbitrary constant. 002 10.0 points Determine the indefinite integral I = integraldisplay 2 x ( x- 3) 2 dx . 1. I = ln( x- 3) 2- 6 x- 3 + C correct 2. I =- 2 x- 3 + C 3. I = 3 ln( x- 3) 2 + C 4. I = 6 ( x- 3) 2 + C 5. I = ln( x- 3) 2 + 6 ( x- 3) 2 + C Explanation: Set u = x- 3 ; then du = dx , so I = 2 integraldisplay x ( x- 3) − 2 dx = 2 integraldisplay ( u + 3) u − 2 du = 2 integraldisplay du u + 6 integraldisplay u − 2 du . But 2 integraldisplay du u = 2 ln | u | + C = ln u 2 + C, while 6 integraldisplay u − 2 du =- 6 u − 1 + C. Consequently, I = ln( x- 3) 2- 6 x- 3 + C . 003 10.0 points Evaluate the definite integral I = integraldisplay 4 1 2 √ x ( √ x + 6) dx . Horstman (mdh995) – HW05 – Radin – (58505) 2 1. I = 2 ln 9 7 2. I = 4(2 √ 2- √ 7) 3. I = 2(3- √ 7) 4. I = 2 ln 8 7 5. I = 4 ln 9 7 6. I = 4 ln 8 7 correct 7. I = 4(3- √ 7) 8. I = 2(2 √ 2- √ 7) Explanation: Set u 2 = x . Then 2 u du = dx , while x = 1 = ⇒ u = 1 x = 4 = ⇒ u = 2 . In this case, I = 4 integraldisplay 2 1 1 u + 6 du = 4 bracketleftBig ln | u + 6 | bracketrightBig 2 1 ....
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This note was uploaded on 11/30/2008 for the course M 58510 taught by Professor Radin during the Fall '08 term at University of Texas.

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solution_pdf5 - Horstman(mdh995 – HW05 – Radin...

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