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Unformatted text preview: Rehman (aar638) – HW09 – sachse – (56620) 1 This printout should have 20 questions. Multiplechoice 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 , x ≤ 1 , x 2 , 1 ≤ x ≤ 2 , and find its value if convergent. 1. I = 4 2. I = 3 3. I not convergent correct 4. I = 7 2 5. I = 1 2 6. I = 5 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 dx + integraldisplay 2 1 x 2 dx = bracketleftBig ln x bracketrightBig 1 t + bracketleftBig 1 3 x 3 bracketrightBig 2 1 = lnt + 5 3 . On the other hand, lim t → + ln t =∞ . Consequently, I is not convergent . 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 · 5 4 / 5 8 2. I = 5 4 / 5 8 3. I = 5 4 / 5 4. I = 5 · 5 4 / 5 8 5. I does not converge correct 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 . Rehman (aar638) – HW09 – sachse – (56620) 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 8 x (9 + x 2 ) 2 dx converges, and if it does, compute its value. 1. I = 4 9 2. I = 2 9 correct 3. I = 8 9 4. integral doesn’t converge 5. I = 8 27 Explanation: The integral I = integraldisplay ∞ 3 8 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 8 x (9 + x 2 ) 2 dx . To evaluate I t , set u = 9 + x 2 . Then du = 2 x dx , in which case integraldisplay 8 x (9 + x 2 ) 2 dx = 4 integraldisplay 1 u 2 du. Thus I t = integraldisplay t 3 8 x (9 + x 2 ) 2 dx = 4 bracketleftBig 1 9 + x 2 bracketrightBig t 3 = 4 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 8 x (9 + x 2 ) 2 dx = 2 9 . 004 10.0 points Determine if the improper integral I = integraldisplay ∞ −∞ 5 x e − 2 x 2 dx is convergent or divergent, and if it is conver gent, find its value....
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This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.
 Fall '09
 GOGOLEV
 Calculus

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