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Unformatted text preview: momin (rrm497) Review2 cheng (58520) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay e 1 4 x 3 ln x dx. 1. I = (3 e 4 + 1) 2. I = 1 4 (3 e 4 + 1) correct 3. I = 3 4 e 4 4. I = (3 e 4 1) 5. I = 1 4 (3 e 4 1) Explanation: After integration by parts, I = bracketleftBig x 4 ln x bracketrightBig e 1 integraldisplay e 1 x 3 dx = e 4 integraldisplay e 1 x 3 dx , since ln e = 1 and ln1 = 0. But integraldisplay e 1 x 3 dx = 1 4 ( e 4 1) . Consequently, I = e 4 1 4 ( e 4 1) = 1 4 (3 e 4 + 1) . 002 10.0 points Find the volume, V , of the solid obtained by rotating the graph of y = sin x between x = 0 and x = / 2 about the xaxis. 1. V = 1 4 2. V = 1 2 3. V = 1 4 2 correct 4. V = 1 8 5. V = 1 2 2 6. V = 1 8 2 Explanation: The volume of the solid obtained by rotat ing the graph of y = sin x between x = 0 and x = / 2 about the axis is given by V = integraldisplay / 2 sin 2 x dx . But sin 2 x = 1 2 parenleftBig 1 cos 2 x parenrightBig , in which case V = 1 2 integraldisplay / 2 parenleftBig 1 cos 2 x parenrightBig dx = 1 2 bracketleftBig x 1 2 sin 2 x bracketrightBig / 2 . Consequently, V = 1 4 2 . 003 10.0 points Evaluate the definite integral I = integraldisplay 2 2 4 x 2 x 2 1 dx . 1. I = 4( 3 2 ) 2. I = 2( 3 2 ) correct momin (rrm497) Review2 cheng (58520) 2 3. I = 3 + 2 4. I = 4( 3 + 2 ) 5. I = 3 2 6. I = 2( 3 + 2 ) Explanation: Set x = sec u . Then dx = sec u tan u du , x 2 1 = tan 2 u , while x = 2 = u = 4 , x = 2 = u = 3 . In this case, I = 4 integraldisplay / 3 / 4 sec u tan u sec 2 u tan u du = integraldisplay / 3 / 4 4 cos u du = 4 bracketleftBig sin u bracketrightBig / 3 / 4 . Consequently, I = 2( 3 2 ) . 004 10.0 points Evaluate the definite integral I = integraldisplay 3 1 4 x 2 4 x + 5 dx . 1. I = 4 2. I = 3 4 3. I = 2 correct 4. I = 3 2 5. I = 5 2 Explanation: By completing the square we see that x 2 4 x + 5 = ( x 2 4 x + 4) + 5 4 = ( x 2) 2 + 1 . Thus I = integraldisplay 3 1 4 ( x 2) 2 + 1 . Since d dx tan 1 x = 1 1 + x 2 , this suggests the substitution x 2 = u . For then dx = du , while x = 1 = u = 1 , x = 3 = u = 1 ....
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 Fall '08
 RAdin
 Calculus

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