# hw06 - padilla (tp5647) – HW06 – cheng – (57455) 1...

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Unformatted text preview: padilla (tp5647) – HW06 – cheng – (57455) 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) = 1 and f (1) = 5. 1. I = 6 2. I = 9 3. I = 7 correct 4. I = 5 5. I = 8 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 = 7 . 002 10.0 points Evaluate the definite integral I = integraldisplay 1 3 5- 2 x dx . 1. I = 3 2 ln 5 2 2. I = 3 ln 5 3 3. I = 3 ln 5 2 4. I = 3 2 ln 7 5 5. I = 3 ln 7 5 6. I = 3 2 ln 5 3 correct Explanation: Set u = 5- 2 x ; then du =- 2 dx while x = 0 = ⇒ u = 5 x = 1 = ⇒ u = 3 . In this case, I =- 3 2 integraldisplay 3 5 1 du du = 3 2 integraldisplay 5 3 1 u du = 3 2 bracketleftBig ln | u | bracketrightBig 5 3 . Consequently, I = 3 2 (ln 5- ln 3) = 3 2 ln 5 3 . 003 10.0 points Determine the indefinite integral I = integraldisplay x 3 x 2 + 5 dx . 1. I = 1 3 x 3- 5 ln( x 2 + 5) + C 2. I = x 2- x ln( x 2 + 5) + C 3. I = 1 3 x 3 + 5 ln( x 2 + 5) + C 4. I = 1 2 x 2- 5 2 ln( x 2 + 5) + C correct padilla (tp5647) – HW06 – cheng – (57455) 2 5. I = x 2 + x ln( x 2 + 5) + C 6. I = 1 2 x 2 + 5 2 ln( x 2 + 5) + C Explanation: After long division on the integrand, x 3 x 2 + 5 = x- 5 x x 2 + 5 . Thus I = integraldisplay x dx- integraldisplay 5 x x 2 + 5 dx . To evaluate the last integral we use the sub- stitution u = x 2 + 5. For then du = 2 x dx , so I = 1 2 x 2- 5 2 integraldisplay 1 u du ....
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## This note was uploaded on 03/31/2010 for the course M 408 K m 408 k taught by Professor G during the Spring '09 term at University of Texas-Tyler.

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hw06 - padilla (tp5647) – HW06 – cheng – (57455) 1...

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