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# hw6 - padilla(tp5647 HW06 Gilbert(56650 This print-out...

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padilla (tp5647) – HW06 – Gilbert – (56650) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the most general function f such that f ( x ) = 3 x + 8 4 - x 2 . 1. f ( x ) = 3 2 x - 4 tan 1 x 2 + C 2. f ( x ) = 3 x - 4 tan 1 x + C 3. f ( x ) = 3 x - 8 sin 1 x 2 + C 4. f ( x ) = 3 2 x 2 + 4 tan 1 x 2 + C 5. f ( x ) = 3 2 x 2 + 8 sin 1 x + C 6. f ( x ) = 3 2 x 2 + 8 sin 1 x 2 + C correct Explanation: Since d dx parenleftBig sin 1 x 2 parenrightBig = 1 4 - x 2 , we see that f ( x ) = 3 2 x 2 + 8 sin 1 x 2 + C with C an arbitrary constant. 002 10.0 points Determine the indefinite integral I = integraldisplay ( 1 - x 2 ) 1 / 2 1 - 2 sin 1 x dx . 1. I = - 1 4 ln vextendsingle vextendsingle 1 + 2 sin 1 x vextendsingle vextendsingle + C 2. I = 1 2 ( 1 - 2 sin 1 x ) 2 + C 3. I = 1 2 ln vextendsingle vextendsingle 1 - 2 sin 1 x vextendsingle vextendsingle + C 4. I = - 1 4 ( 1 + 2 sin 1 x ) 2 + C 5. I = 1 4 ( 1 + 2 sin 1 x ) 2 + C 6. I = - 1 2 ln vextendsingle vextendsingle 1 - 2 sin 1 x vextendsingle vextendsingle + C correct Explanation: Set u = 1 - 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 1 - 2 sin 1 x vextendsingle vextendsingle + C . Consequently, I = - 1 2 ln vextendsingle vextendsingle 1 - 2 sin 1 x vextendsingle vextendsingle + C . 003 10.0 points Determine the integral I = integraldisplay 1 0 8 + x 1 + x 2 dx . 1. I = 1 2 (4 π + ln 2) correct 2. I = 1 2 (4 π - ln 2) 3. I = 4 π - ln 4 4. I = 4 π - ln 2 5. I = 1 2 (4 π + ln 4) 6. I = 4 π + ln 4

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padilla (tp5647) – HW06 – Gilbert – (56650) 2 Explanation: We deal with the two integrals I 1 = integraldisplay 1 0 8 1 + x 2 dx, I 2 = integraldisplay 1 0 x 1 + x 2 dx separately. Now d dx tan 1 x = 1 1 + x 2 , so we see that I 1 = bracketleftBig 8 tan 1 x bracketrightBig 1 0 = 2 π . On the the other hand, to evaluate I 2 set u = 1 + x 2 . Then du = 2 x dx , and x = 0 = u = 1 , while x = 1 = u = 2 . In this case, I 2 = 1 2 integraldisplay 2 1 1 u du = 1 2 bracketleftBig ln u bracketrightBig 2 1 . Consequently, I = 1 2 (4 π + ln 2) . keywords: 004 10.0 points Determine the indefinite integral I = integraldisplay 3 t 3 1 - t 8 dt . 1. I = 3 4 sin 1 ( t 4 ) + C correct 2. I = 3 1 - t 8 + C 3. I = 3 4 radicalbig 1 - t 8 + C 4. I = 3 tan 1 ( t 4 ) + C 5. I = 3 sin 1 ( t 4 ) + C 6. I = 3 4 tan 1 ( t 4 ) + C Explanation: Since integraldisplay 1 1 - x 2 dx = sin 1 x + C , we need to reduce I to this form by changing t . Indeed, set x = t 4 . Then dx = 4 t 3 dt . In this case I = 1 4 integraldisplay 3 1 - x 2 dx = 3 4 sin 1 x + C with C an arbitrary constant. Consequently, I = 3 4 sin 1 ( t 4 ) + C . 005 10.0 points Determine the integral I = integraldisplay 1 1 / 2 4 x 2 - x dx . . 1. I = 2 3 π correct 2. I = 2 3 3. I = 4 3 4. I = 4 3 π 5. I = 1 3
padilla (tp5647) – HW06 – Gilbert – (56650) 3 6. I = 1 3 π Explanation: Set u 2 = x . Then 2 u du = dx , while x = 1 2 = u = 1 2 x = 1 = u = 1 .

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