hw9 - padilla (tp5647) HW09 Gilbert (56650) 1 This...

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Unformatted text preview: padilla (tp5647) HW09 Gilbert (56650) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine if I = integraldisplay 2 f ( x ) dx is convergent or divergent when f ( x ) = braceleftBigg x 1 / 2 , x 1 , x , 1 x 2 , and find its value if convergent. 1. I not convergent 2. I = 4 3. I = 7 2 correct 4. I = 5 2 5. I = 3 6. I = 1 2 Explanation: The integral is improper because f ( x ) as x + . Thus I = lim t + I t , I t = integraldisplay 2 t f ( x ) dx . But for t < 1, I t = integraldisplay 1 t 1 x 1 / 2 dx + integraldisplay 2 1 x dx = bracketleftBig 2 x 1 / 2 bracketrightBig 1 t + bracketleftBig 1 2 x 2 bracketrightBig 2 1 = 2- 2 t 1 / 2 + 3 2 . On the other hand, lim t + t 1 / 2 = 0 . Consequently, I is convergent and I = 7 2 . 002 10.0 points Determine if I = integraldisplay 3 x 5 x 2- 4 dx converges, and if it does, compute its value. 1. I = 5 4 / 5 8 2. I does not converge correct 3. I = 5 5 4 / 5 8 4. I =- 5 5 4 / 5 8 5. I = 5 4 / 5 6. I =- 5 5 4 / 5 4 Explanation: The integral I is improper because of the infinite interval of integration. Thus I will converge if lim t integraldisplay t 3 x 5 x 2- 4 dx exists. To evaluate this last integral, we use substitution, setting u = x 2- 4. For then du = 2 x dx , while x = 3 = u = 5 , x = t = u = t 2- 4 . padilla (tp5647) HW09 Gilbert (56650) 2 In this case integraldisplay t 3 x 5 x 2- 4 dx = 1 2 integraldisplay t 2 4 5 1 u 1 / 5 du = 5 8 bracketleftBig u 4 / 5 bracketrightBig t 2 4 5 = 5 8 parenleftBig ( t 2- 4) 4 / 5- 5 4 / 5 parenrightBig . However, lim t ( t 2- 4) 4 / 5 = . Consequently, I does not converge . 003 10.0 points Determine if the improper integral I = integraldisplay 3 4 x (9 + x 2 ) 2 dx converges, and if it does, compute its value. 1. I = 4 9 2. I = 2 9 3. integral doesnt converge 4. I = 1 9 correct 5. I = 4 27 Explanation: The integral I = integraldisplay 3 4 x (9 + x 2 ) 2 dx is improper because of the infinite interval of integration. To overcome this, we truncate and consider the limit lim t I t , I t = integraldisplay t 3 4 x (9 + x 2 ) 2 dx . To evaluate I t , set u = 9 + x 2 . Then du = 2 x dx , in which case integraldisplay 4 x (9 + x 2 ) 2 dx = 2 integraldisplay 1 u 2 du. Thus I t = integraldisplay t 3 4 x (9 + x 2 ) 2 dx = 2 bracketleftBig- 1 9 + x 2 bracketrightBig t 3 = 2 braceleftBig 1 18- 1 9 + t 2 bracerightBig . Consequently, since lim t 1 9 + t 2 = 0 , we see that I converges and that I = lim t integraldisplay t 3 4 x (9 + x 2 ) 2 dx = 1 9 ....
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hw9 - padilla (tp5647) HW09 Gilbert (56650) 1 This...

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