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Homework 5 - ejm753 Homework 5 Cepparo(58400 This print-out...

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ejm753 – Homework 5 – Cepparo – (58400) 1 This print-out should have 18 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 + 25( x - 8) 2 dx . 1. I = 5 sin 1 parenleftBig x - 8 5 parenrightBig + C 2. I = tan 1 5( x - 8) + C 3. I = 5 tan 1 parenleftBig x - 8 5 parenrightBig + C 4. I = 1 5 tan 1 5( x - 8) + C correct 5. I = 1 5 sin 1 5( x - 8) + C 6. I = sin 1 5( x - 8) + C Explanation: Since d dx tan 1 x = 1 1 + x 2 , the substitution u = 5( x - 8) is suggested. For then du = 5 dx , in which case I = 1 5 integraldisplay 1 1 + u 2 du = 1 5 tan 1 u + C , with C an arbitrary constant. Consequently, I = 1 5 tan 1 5( x - 8) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 2 0 6 4 + x 2 dx . 1. I = 9 8 π 2. I = 3 4 π correct 3. I = π 4. I = 5 4 π 5. I = 7 8 π Explanation: Since d dx tan 1 x = 1 1 + x 2 , the substitution x = 2 u is suggested. For then dx = 2 du , while x = 0 = u = 0 , x = 2 = u = 1 . Thus I = 3 integraldisplay 1 0 1 1 + u 2 du . Consequently. I = bracketleftBig 3 tan 1 u bracketrightBig 1 0 = 3 4 π . keywords: 003 10.0 points Determine the integral I = integraldisplay 1 0 2 4 - x 2 dx . 1. I = 1 2 2. I = 1 3
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ejm753 – Homework 5 – Cepparo – (58400) 2 3. I = 2 3 4. I = 1 2 π 5. I = 2 3 π 6. I = 1 3 π correct 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 = 2 u . Then dx = 2 du while x = 0 = u = 0 and x = 1 = u = 1 2 . In this case I = 4 integraldisplay 1 / 2 0 1 2 1 - u 2 du = 2 integraldisplay 1 / 2 0 1 1 - u 2 du . Consequently, I = bracketleftBig 2 sin 1 u bracketrightBig 1 / 2 0 = 1 3 π . keywords: 004 10.0 points Determine the integral I = integraldisplay (1 - x 2 ) 1 / 2 3 - 4 arcsin x dx . 1. I = - 1 8 (3 - 4 arcsin x ) 2 + C 2. I = - 1 3 ln | 3 - 4 arcsin x | + C correct 3. I = 1 8 ln | 3 - 4 arcsin x | + C 4. I = 1 3 (3 - 4 arcsin x ) 2 + C 5. I = 1 3 ln | 3 - 4 arcsin x | + C 6. I = - 1 8 ln | 3 - 4 arcsin x | + C Explanation: Set u = 3 - 4 arcsin x . Then du = - 4 1 - x 2 dx = - 4(1 - x 2 ) 1 / 2 dx, so I = - 1 3 integraldisplay 1 u du = - 1 3 ln | u | + C with C an arbitrary constant. Consequently, I = - 1 3 ln | 3 - 4 arcsin x | + C . keywords: 005 10.0 points Determine the integral I = integraldisplay π/ 2 0 4 cos θ 1 + sin 2 θ dθ . 1. I = 7 4 π 2. I = 5 4 π 3. I = 3 4 π
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ejm753 – Homework 5 – Cepparo – (58400) 3 4. I = π correct 5. I = 3 2 π Explanation: Since d sin θ = cos θ , the substitution u = sin θ is suggested. For then du = cos θ dθ , while θ = 0 = u = 0 , θ = π 2 = u = 1 , so that I = 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 .
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