HW11-solutions - husain(aih243 HW11 Gilbert(56215 This...

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husain (aih243) – HW11 – Gilbert – (56215) 1 This print-out should have 22 questions. Multiple-choice 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 0 (3 cos 2 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 0 (3 cos u + sin u ) du = 1 2 bracketleftBig 3 sin u - cos u bracketrightBig π/ 2 0 . 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 0 (sin 3 x - 5 cos 3 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 0 (sin u - 5 cos u ) du = 1 3 bracketleftBig - cos u - 5 sin u bracketrightBig 3 π/ 2 0 . 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,
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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 0 f ( x ) dx = 6 , determine the value of the integral I = integraldisplay 3 0 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 0 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 0 f ( u ) du = 14 . keywords: IntSubst, IntSubstExam, 005 10.0 points Evaluate the integral I = integraldisplay 1 0 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:
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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 0 1 u 3 du = bracketleftBig 1 4 u 4 bracketrightBig 0 1 . Consequently, I = - 1 4 . 006 10.0 points The graph of f has slope df dx = 2 x radicalbig 2 x 2 + 1 and passes through the point (2 , 3). Find the y -intercept of this graph. 1. y -intercept = - 6 2. y -intercept = - 5 3. y -intercept = - 17 3 correct 4. y -intercept = - 16 3 5. y -intercept = - 14 3 Explanation: The function f satisfies the equations f ( x ) = integraldisplay 2 x radicalbig 2 x 2 + 1 dx, f (2) = 3 . To evaluate the integral set u = 2 x 2 + 1. For then du = 4 x dx , in which case f ( x ) = 1 2 integraldisplay u 1 / 2 du = 1 3 u 3 / 2 + C = 1 3 (2 x 2 + 1) 3 / 2 + C , where C has to be chosen so that f (2) = 3 , i . e ., C + 9 = 3 .
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