HW08-solutions

# HW08-solutions - wei(jw35975 HW08 kalahurka(55230 This...

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wei (jw35975) – HW08 – kalahurka – (55230) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 3) 10.0 points (i) Express the improper integral I = i 3 2 xe - x/ 4 dx as lim t →∞ I t with each I t a proper integral. 1. I = lim t i 3+ t 2 xe - x/ 4 dx 2. I = lim t i t 2 xe - x/ 4 dx 3. I = lim t i 3 2 xe - x/t dx 4. I = lim t i t 3 2 xe - x/ 4 dx correct 5. I = lim t i t - t 2 xe - x/ 4 dx Explanation: The integral I = i 3 2 xe - x/ 4 dx is improper because oF the infnite range oF integration. To overcome this we restrict to a fnite interval oF integration and consider the limit I = lim t I t , I t = i 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 = i t 3 2 xe - x/ 4 dx we integrate by parts. Then I t = b 8 xe - x/ 4 B t 3 + 8 i t 3 e - x/ 4 dx = 8 b ( x + 4) e - x/ 4 B 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, fnd 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 .

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wei (jw35975) – HW08 – kalahurka – (55230) 2 004 10.0 points Determine if the improper integral I = i 1 -∞ 1 2 z dz is convergent or divergent, and if convergent, Fnd 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- Fnite interval of integration. It will converge if lim t →∞ i 1 - t 1 2 z dz exists. Now i 1 - t 1 2 z dz = b 2 2 z B 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 = i 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- Fnite interval of integration, so we write I = lim t I t , I t = i t e 1 x x ) 2 dx, whenever the limit exists. To evaluate I t , Frst set u = ln6 x . Then i 1 x x ) 2 dx = i 1 u 2 du = 1 u + C , and so I t = b 1 ln 6 x B t e = p 1 ln6 e 1 ln 6 t P . On the other hand, lim t 1 t = 0 . Consequently, lim t I t exists, and I = lim t p 1 e 1 ln 6 t P = 1 ln 6 e . 006 10.0 points Determine if the improper integral I = i 1 6 tan - 1 x 1 + x 2 dx converges, and if it does, Fnd its value.
wei (jw35975) – HW08 – kalahurka – (55230) 3 1. I does not converge 2. I = 3 4 π 2 3. I = 9 16 4. I = 9 8 π 2 5. I = 3 4 6. I = 9 16 π 2 correct Explanation: The integral I is improper because the in- terval of integration is inFnite. Thus the integral will converge if lim t →∞ I t , I t = i t 1 6 tan - 1 x 1 + x 2 dx , exists, and its value will then be the value of the limit.

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HW08-solutions - wei(jw35975 HW08 kalahurka(55230 This...

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