HW08-solutions - wei (jw35975) – HW08 – kalahurka –...

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Unformatted text preview: wei (jw35975) – HW08 – kalahurka – (55230) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points (i) Express the improper integral I = integraldisplay ∞ 3 2 xe- x/ 4 dx as lim t →∞ I t with each I t a proper integral. 1. I = lim t →∞ integraldisplay ∞ 3+ t 2 xe- x/ 4 dx 2. I = lim t →∞ integraldisplay ∞ t 2 xe- x/ 4 dx 3. I = lim t →∞ integraldisplay ∞ 3 2 xe- x/t dx 4. I = lim t →∞ integraldisplay t 3 2 xe- x/ 4 dx correct 5. I = lim t →∞ integraldisplay t- t 2 xe- x/ 4 dx Explanation: The integral I = integraldisplay ∞ 3 2 xe- x/ 4 dx is improper because of the infinite range of integration. To overcome this we restrict to a finite interval of integration and consider the limit I = lim t →∞ I t , I t = integraldisplay t 3 2 xe- x/ 4 dx . 002 (part 2 of 3) 10.0 points (ii) Compute the value of I t . 1. I t = 56 e- 3 4 − 8( t + 4) e- t/ 4 correct 2. I t = 8(3 − t ) { e- 3 4 − e- t/ 4 } 3. I t = 56 e- 3 4 + 8( t + 4) e- t/ 4 4. I t = 7 e- 3 4 − ( t + 4) e- t/ 4 5. I t = − 56 e- 3 4 + 8( t + 4) e- t/ 4 Explanation: To evaluate I t = integraldisplay t 3 2 xe- x/ 4 dx we integrate by parts. Then I t = bracketleftBig − 8 xe- x/ 4 bracketrightBig t 3 + 8 integraldisplay t 3 e- x/ 4 dx = − 8 bracketleftBig ( x + 4) e- x/ 4 bracketrightBig t 3 . Consequently, I t = 56 e- 3 4 − 8( t + 4) e- t/ 4 . 003 (part 3 of 3) 10.0 points (iii) Determine if lim t →∞ I t exists, and if it does, find its value. 1. I = 8 e 3 4 2. I = 56 e- 3 4 correct 3. I does not exist 4. I = 56 e 3 4 5. I = 8 e- 3 4 Explanation: By L’Hospital’s Rule, lim x →∞ x m e- x = 0 for any power m of x . It follows that lim t →∞ ( t + 4) e- t/ 4 = 0 . Consequently, lim t →∞ I t exists, and I = 56 e- 3 4 . wei (jw35975) – HW08 – kalahurka – (55230) 2 004 10.0 points Determine if the improper integral I = integraldisplay 1-∞ 1 √ 2 − z dz is convergent or divergent, and if convergent, find its value. 1. I = 2 2. I is divergent correct 3. I = 3 2 4. I = 1 2 5. I = 1 Explanation: The integral is improper because of the in- finite interval of integration. It will converge if lim t →∞ integraldisplay 1- t 1 √ 2 − z dz exists. Now integraldisplay 1- t 1 √ 2 − z dz = bracketleftBig − 2 √ 2 − z bracketrightBig 1- t = − 2 + 2 √ 2 + t . But we know that lim t →∞ √ 2 + t = ∞ . Consequently, I is divergent . 005 10.0 points Determine if the improper integral I = integraldisplay ∞ e 1 x (ln6 x ) 2 dx converges, and if it does, compute its value. 1. I = 6 2. I = ln 6 e 3. I = 1 ln 6 e correct 4. I = 1 6 e 5. I = 6 ln 6 e 6. I does not converge Explanation: The integral is improper because of the in- finite interval of integration, so we write I = lim t →∞ I t , I t = integraldisplay t e 1 x (ln6 x ) 2 dx, whenever the limit exists. To evaluate I t , first set u = ln6...
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This note was uploaded on 03/26/2012 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas.

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HW08-solutions - wei (jw35975) – HW08 – kalahurka –...

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