# HW_8_Answers - Miller Kierste Homework 8 Due 3:00 am Inst...

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Miller, Kierste – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Determine iF the improper integral I = Z 0 4 (3 x + 1) 2 dx is convergent, and iF it is, fnd its value. 1. I = 2 2. I = 4 3 correct 3. I = 7 3 4. I = 1 5. I = 5 3 6. I is not convergent Explanation: The integral is improper because the in- terval oF integration is infnite. To test For convergence we thus have to determine iF lim t →∞ I t , I t = Z t 0 4 (3 x + 1) 2 dx, exists. To evaluate I t , set u = 3 x + 1, Then du = 3 while x = 0 = u = 1 x = t = u = 3 t + 1 . In this case, I t = 1 3 Z 3 t +1 1 4 u 2 du = 1 3 h - 4 u i 3 t +1 1 , so that lim t I t = lim t 4 3 - 4 3(3 t + 1) · = 4 3 . Consequently, the integral is convergent and I = 4 3 . . keywords: improper integral, infnite interval integration, rational Function 002 (part 1 oF 1) 10 points Determine iF the improper integral I = Z 0 -∞ 4 6 x - 1 dx converges, and iF it does, fnd its value. 1. I = 3 2 2. I is not convergent correct 3. I = 1 4 4. I = 2 3 5. I = 4 Explanation: The integral is improper because the in- terval oF integration is infnite. To test For convergence, thereFore, we have to check iF the limit lim t →-∞ Z 0 t 4 6 x - 1 dx exists. But Z 0 t 4 6 x - 1 dx = h 2 3 ln | 6 x - 1 | i 0 t = 2 3 ln 1 - 2 3 ln | 6 t - 1 | · .

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Miller, Kierste – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Gary Berg 2 On the other hand, lim t →-∞ ln | 6 t - 1 | = . Consequently, the integral I is not convergent . keywords: improper integral, limit 003 (part 1 of 1) 10 points Determine if I = Z 3 x 5 6 - x 2 dx converges, and if it does, compute its value. 1. I = - 5 · 3 4 / 5 8 2. I = - 3 4 / 5 3. I = ( - 3) 4 / 5 4. I does not converge correct 5. I = 5 8 ( - 3) 4 / 5 Explanation: As a 1 / 5 exists for all a , both positive and negative, we see that 5 p 6 - x 2 exists and is non-zero for all x > 3. This means that x 5 6 - x 2 exists and is non-zero on [3 , ), and so the only reason I is improper is because the in- terval of integration is inFnite. Thus I will converge if lim t →∞ I t , I t = Z t 3 x 5 6 - x 2 dx, exists. To evaluate I t we use substitution, setting u = 6 - x 2 . ±or then du = - 2 xdx, while x = 3 = u = - 3 , x = t = u = 6 - t 2 . In this case Z t 3 x 5 6 - x 2 dx = - 1 2 Z 6 - t 2 - 3 1 u 1 / 5 du = - 5 8 h u 4 / 5 i 6 - t 2 - 3 = 5 8 ( - 3) 4 / 5 - (6 - t 2 ) 4 / 5 · . But ( - 3) 4 / 5 = 3 4 / 5 , (6 - t 2 ) 4 / 5 = ( t 2 - 6) 4 / 5 , so Z t 3 x 5 6 - x 2 dx = 5 8 3 4 / 5 - ( t 2 - 6) 4 / 5 · . However, lim t ( t 2 - 6) 4 / 5 = . Consequently, I does not converge . keywords: improper integral, divergent, odd rational power 004 (part 1 of 1) 10 points Determine if the improper integral I = Z -∞ 3 xe - 2 x 2 dx is convergent or divergent, and if it is conver- gent, Fnd its value.
Miller, Kierste – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Gary Berg 3 1. I = 3 8 2. I = 3 4 3. I is divergent 4. I = 3 2 5. I = 0 correct Explanation: The integral is improper because the range if integration is inFnite in both directions.

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HW_8_Answers - Miller Kierste Homework 8 Due 3:00 am Inst...

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