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# HW6 - momin(rrm497 Homework 06 cheng(58520 This print-out...

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momin (rrm497) – Homework 06 – cheng – (58520) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find f ( x ) when f ( x ) = 7 6 1 + x 2 and f (1) = 0 . 1. f ( x ) = 7( x 1) + 6 parenleftBig tan 1 x π 4 parenrightBig 2. f ( x ) = 6 tan 1 x 7 x + 1 3. f ( x ) = 6 parenleftBig tan 1 x π 4 parenrightBig 7( x 1) 4. f ( x ) = 7 x + 6 tan 1 x 7 5. f ( x ) = 7( x 1) 6 parenleftBig tan 1 x π 4 parenrightBig correct 6. f ( x ) = 7 x 6 tan 1 x 7 Explanation: Since d dx tan 1 x = 1 1 + x 2 , we see that f ( x ) = 7 x 6 tan 1 x + C . Now f (1) = 0 = 7 3 2 π + C = 0 . Consequently, f ( x ) = 7( x 1) 6 parenleftBig tan 1 x π 4 parenrightBig . 002 10.0 points Determine the integral I = integraldisplay 1 1 + 16( x 5) 2 dx . 1. I = 4 sin 1 parenleftBig x 5 4 parenrightBig + C 2. I = 4 tan 1 parenleftBig x 5 4 parenrightBig + C 3. I = sin 1 4( x 5) + C 4. I = 1 4 sin 1 4( x 5) + C 5. I = tan 1 4( x 5) + C 6. I = 1 4 tan 1 4( x 5) + C correct Explanation: Since d dx tan 1 x = 1 1 + x 2 , the substitution u = 4( x 5) is suggested. For then du = 4 dx , in which case I = 1 4 integraldisplay 1 1 + u 2 du = 1 4 tan 1 u + C , with C an arbitrary constant. Consequently, I = 1 4 tan 1 4( x 5) + C . keywords: 003 10.0 points Evaluate the definite integral I = integraldisplay 1 8 0 1 1 25 x 2 dx . Correct answer: 0 . 135026. Explanation: Since integraldisplay 1 1 x 2 dx = sin 1 x + C ,

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momin (rrm497) – Homework 06 – cheng – (58520) 2 a change of variable x is needed to reduce I to this form. Set u = 5 x . Then du = 5 dx , and x = 0 = u = 0 , while x = 1 8 = u = 5 8 . In this case I = 1 5 integraldisplay 5 8 0 1 1 u 2 du = bracketleftbigg 1 5 sin 1 u bracketrightbigg 5 8 0 . Consequently, I = 1 5 arcsin parenleftbigg 5 8 parenrightbigg = 0 . 135026 . 004 10.0 points Determine the indefinite integral I = integraldisplay ( 1 x 2 ) 1 / 2 3 + 2 sin 1 x dx . 1. I = 1 2 ( 3 + 2 sin 1 x ) 2 + C 2. I = 1 4 ( 3 + 2 sin 1 x ) 2 + C 3. I = 1 2 ln vextendsingle vextendsingle 3 + 2 sin 1 x vextendsingle vextendsingle + C 4. I = 1 4 ln vextendsingle vextendsingle 3 + 2 sin 1 x vextendsingle vextendsingle + C 5. I = 1 2 ln vextendsingle vextendsingle 3 + 2 sin 1 x vextendsingle vextendsingle + C correct 6. I = 1 4 ( 3 + 2 sin 1 x ) 2 + C Explanation: Set u = 3 + 2 sin 1 x . Then du = 2 1 x 2 dx = 2 ( 1 x 2 ) 1 / 2 dx , so I = 1 2 integraldisplay 1 u du = 1 2 ln vextendsingle vextendsingle 3 + 2 sin 1 x vextendsingle vextendsingle + C . Consequently, I = 1 2 ln vextendsingle vextendsingle 3 + 2 sin 1 x vextendsingle vextendsingle + C . 005 10.0 points Determine the integral I = integraldisplay π/ 2 0 7 cos θ 1 + sin 2 θ dθ . 1. I = 5 4 π 2. I = π 3. I = 3 2 π 4. I = 3 4 π 5. I = 7 4 π correct 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 = 7 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 .
momin (rrm497) – Homework 06 – cheng – (58520) 3

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