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int ca l6

# int ca l6 - pokharel(yp624 HW06 Radin(57410 This print-out...

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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 0 { 2 f ( u ) - 1 } du = integraldisplay 1 0 { 2 f ( u ) - 1 } du = bracketleftBig 2 f ( u ) - u bracketrightBig 1 0 . Consequently, I = 2 { f (2) - f (1) } - 1 = 3 . 002 10.0 points Evaluate the definite integral I = integraldisplay 1 0 1 3 - 2 x dx . 1. I = ln 3 2 2. I = 1 2 ln 3 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

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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 . Consequently, I = 1 2 x 2 - 11 2 ln( x 2 + 11) + C . 004 10.0 points Evaluate the definite integral I = integraldisplay 4 1 2 x ( x + 5) dx . 1. I = 4 ln 7 6 correct 2. I = 2( 7 - 6) 3. I = 2 ln 7 6 4. I = 4 ln 4 3 5. I = 4(2 2 - 6) 6. I = 2(2 2 - 6) 7. I = 4( 7 - 6) 8. I = 2 ln 4 3 Explanation: Set u 2 = x . Then 2 u du = dx , while x = 1 = u = 1 x = 4 = u = 2 .
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