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Unformatted text preview: husain (aih243) – HW11 – Gilbert – (56215) 1 This printout should have 22 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 π/ 4 (3 cos2 x + sin 2 x ) dx . 1. I = 4 2. I = 3 3. I = 5 4. I = 6 5. I = 2 correct Explanation: To reduce the integral to one involving just cos u and sin u set u = 2 x . Then du = 2 dx , while x = 0 = ⇒ u = 0 , and x = π 4 = ⇒ u = π 2 . In this case, I = 1 2 integraldisplay π/ 2 (3 cos u + sin u ) du = 1 2 bracketleftBig 3 sin u cos u bracketrightBig π/ 2 . But sin π 2 = 1 , sin 0 = 0 , while cos π 2 = 0 , cos 0 = 1 . Consequently, I = 1 2 (3 ( 1)) = 2 . keywords: IntSubst, IntSubstExam, 002 10.0 points Evaluate the integral I = integraldisplay π/ 2 (sin 3 x 5 cos3 x ) dx . 1. I = 2 correct 2. I = 0 3. I = 1 4. I = 3 5. I = 1 Explanation: To reduce the integral to one involving just sin u and cos u set u = 3 x . Then du = 3 dx , while x = 0 = ⇒ u = 0 , and x = π 2 = ⇒ u = 3 π 2 . In this case, I = 1 3 integraldisplay 3 π/ 2 (sin u 5 cos u ) du = 1 3 bracketleftBig cos u 5 sin u bracketrightBig 3 π/ 2 . But cos 3 π 2 = 0 , cos 0 = 1 , while sin 3 π 2 = 1 , sin 0 = 0 . Consequently, I = 1 3 parenleftBig ( 1) + (5) parenrightBig = 2 . keywords: IntSubst, IntSubstExam, husain (aih243) – HW11 – Gilbert – (56215) 2 003 10.0 points Evaluate the integral I = integraldisplay 4 1 4 + x 1 / 2 x 1 / 2 dx . 1. I = 14 2. I = 15 3. I = 13 4. I = 12 5. I = 11 correct Explanation: Set u = x 1 / 2 . Then du = 1 2 x 1 / 2 dx , while x = 1 = ⇒ u = 1 , x = 4 = ⇒ u = 2 . In this case, I = 2 integraldisplay 2 1 (4 + u ) du = bracketleftBig 8 u + u 2 bracketrightBig 2 1 . Consequently, I = 11 . keywords: IntSubst, IntSubstExam, 004 10.0 points If f is a continuous function such that integraldisplay 9 f ( x ) dx = 6 , determine the value of the integral I = integraldisplay 3 7 f (3 x ) dx . 1. I = 10 2. I = 13 3. I = 12 4. I = 11 5. I = 14 correct Explanation: To reduce I to an integral of the form integraldisplay 9 f ( x ) dx = 6 we change variable. Set u = 3 x . Then du = 3 dx , while x = 0 = ⇒ u = 0 , x = 3 = ⇒ u = 9 . In this case I = 7 3 integraldisplay 9 f ( u ) du = 14 . keywords: IntSubst, IntSubstExam, 005 10.0 points Evaluate the integral I = integraldisplay 1 2 x ( x 2 1) 3 dx . 1. I = 1 4 correct 2. I = 2 5 3. I = 1 4 4. I = 1 2 5. I = 1 5 6. I = 1 2 Explanation: husain (aih243) – HW11 – Gilbert – (56215) 3 Set u = x 2 1. Then du = 2 x dx while x = 0 = ⇒ u = 1 , x = 1 = ⇒ u = 0 . In this case, I = integraldisplay − 1 u 3 du = bracketleftBig 1 4 u 4 bracketrightBig − 1 ....
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This note was uploaded on 11/30/2010 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.
 Spring '06
 McAdam

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