HW06 Solution - badruddin(ssb776 – HW06 – Radin...

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Unformatted text preview: badruddin (ssb776) – 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 ) + 3 } dx when f (0) = 1 and f (1) = 4. 1. I = 10 2. I = 11 3. I = 9 correct 4. I = 8 5. I = 12 Explanation: Set u = ln x . Then while x = 1 = ⇒ u = 0 , x = e = ⇒ u = 1 . In this case, I = integraldisplay 1 { 2 f ′ ( u ) + 3 } du = integraldisplay 1 { 2 f ′ ( u ) + 3 } du = bracketleftBig 2 f ( u ) + 3 u bracketrightBig 1 . Consequently, I = 2 { f (2)- f (1) } + 3 = 9 . 002 10.0 points Evaluate the definite integral I = integraldisplay 1 2 6- 5 x dx . 1. I = 2 ln 6 5 2. I = 2 5 ln 11 6 3. I = 2 ln 11 6 4. I = 2 5 ln6 correct 5. I = 2 ln6 6. I = 2 5 ln 6 5 Explanation: Set u = 6- 5 x ; then du =- 5 dx while x = 0 = ⇒ u = 6 x = 1 = ⇒ u = 1 . In this case, I =- 2 5 integraldisplay 1 6 1 du du = 2 5 integraldisplay 6 1 1 u du = 2 5 bracketleftBig ln | u | bracketrightBig 6 1 . Consequently, I = 2 5 (ln 6- ln 1) = 2 5 ln 6 . 003 10.0 points Determine the indefinite integral I = integraldisplay x 3 x 2 + 3 dx . 1. I = x 2 + x ln( x 2 + 3) + C 2. I = 1 2 x 2 + 3 2 ln( x 2 + 3) + C 3. I = 1 2 x 2- 3 2 ln( x 2 + 3) + C correct 4. I = 1 3 x 3 + 3 ln( x 2 + 3) + C badruddin (ssb776) – HW06 – Radin – (57410) 2 5. I = x 2- x ln( x 2 + 3) + C 6. I = 1 3 x 3- 3 ln( x 2 + 3) + C Explanation: After long division on the integrand, x 3 x 2 + 3 = x- 3 x x 2 + 3 . Thus I = integraldisplay x dx- integraldisplay 3 x x 2 + 3 dx . To evaluate the last integral we use the sub- stitution u = x 2 + 3. For then du = 2 x dx , so I = 1 2 x 2- 3 2 integraldisplay 1 u du ....
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This note was uploaded on 11/13/2011 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas.

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HW06 Solution - badruddin(ssb776 – HW06 – Radin...

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