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int cal7 - pokharel(yp624 HW07 Radin(57410 This print-out...

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pokharel (yp624) – HW07 – Radin – (57410) 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 Determine the integral I = integraldisplay 1 1 + 4( x - 7) 2 dx . 1. I = 1 2 sin 1 2( x - 7) + C 2. I = 2 sin 1 parenleftBig x - 7 2 parenrightBig + C 3. I = 2 tan 1 parenleftBig x - 7 2 parenrightBig + C 4. I = sin 1 2( x - 7) + C 5. I = 1 2 tan 1 2( x - 7) + C correct 6. I = tan 1 2( x - 7) + C Explanation: Since d dx tan 1 x = 1 1 + x 2 , the substitution u = 2( x - 7) is suggested. For then du = 2 dx , in which case I = 1 2 integraldisplay 1 1 + u 2 du = 1 2 tan 1 u + C , with C an arbitrary constant. Consequently, I = 1 2 tan 1 2( x - 7) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 3 2 0 5 9 - x 2 dx . 1. I = 5 4 2. I = 5 6 3. I = 5 6 π correct 4. I = 5 4 π 5. I = 5 3 π 6. I = 5 3 Explanation: Since integraldisplay 1 1 - x 2 dx = sin 1 x + C , we need to reduce I to an integal of this form by changing the x variable. Indeed, set x = 3 u . Then dx = 3 du while x = 0 = u = 0 and x = 3 2 = u = 1 2 . In this case I = 15 integraldisplay 1 / 2 0 1 3 1 - u 2 du = 5 integraldisplay 1 / 2 0 1 1 - u 2 du . Consequently, I = bracketleftBig 5 sin 1 u bracketrightBig 1 / 2 0 = 5 6 π . keywords: 003 10.0 points
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pokharel (yp624) – HW07 – Radin – (57410) 2 Evaluate the definite integral I = integraldisplay π/ 2 0 4 sin θ 1 + cos 2 θ dθ . 1. I = π correct 2. I = 1 4 π 3. I = 3 4 π 4. I = 1 2 π 5. I = 5 4 π Explanation: Since d cos θ = - sin θ , the substitution u = cos θ is suggested. For then du = - sin θ dθ , while θ = 0 = u = 1 , θ = π 2 = u = 0 , so that I = - 4 integraldisplay 0 1 1 1 + u 2 du = 4 integraldisplay 1 0 1 1 + u 2 du , which can now be integrated using the fact that d du tan 1 u = 1 1 + u 2 . Consequently, I = 4 bracketleftBig tan 1 u bracketrightBig 1 0 = π since tan 1 1 = π 4 . 004 10.0 points Determine the integral I = integraldisplay 1 0 2 tan 1 x 1 + x 2 dx . 1. I = π 8 2. I = π 2 8 3. I = π 2 16 correct 4. I = π 16 5. I = π 4 6. I = π 2 4 Explanation: Set u = tan 1 x . Then du = 1 1 + x 2 dx , while x = 0 = u = 0 , x = 1 = u = π 4 . In this case, I = integraldisplay π/ 4 0 2 u du = bracketleftBig u 2 bracketrightBig π/ 4 0 . Consequently, I = 1 16 π 2 . keywords: 005 10.0 points Determine the integral I = integraldisplay 4 - x 1 - x 2 dx .
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pokharel (yp624) – HW07 – Radin – (57410) 3 1. I = 2 tan 1 x - 1 2 radicalbig 1 - x 2 + C 2. I = 4 sin 1 x + radicalbig 1 - x 2 + C correct 3. I = 2 sin 1 x - radicalbig 1 - x 2 + C 4. I = 2 tan 1 x + radicalbig 1 - x 2 + C 5. I = 4 tan 1 x - radicalbig 1 - x 2 + C 6. I = 4 sin 1 x + 1 2 radicalbig 1 - x 2 + C Explanation: We deal with the two integrals I 1 = integraldisplay 4 1 - x 2 dx, I 2 = x 1 - x 2 dx separately. Now d dx sin 1 x = 1 1 - x 2 , so we see that I 1 = 4 sin 1 x + C . On the the other hand, to evaluate I 2 set u = 1 - x 2 . Then du = - 2 x dx , and so I 2 = - 1 2 integraldisplay u 1 / 2 du = - u + C .
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