HW_8_Answers - Miller Kierste – Homework 8 – Due 3:00...

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Unformatted text preview: 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 – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine if the improper integral I = Z ∞ 4 (3 x + 1) 2 dx is convergent, and if it is, find 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 infinite. To test for convergence we thus have to determine if lim t →∞ I t , I t = Z t 4 (3 x + 1) 2 dx, exists. To evaluate I t , set u = 3 x + 1, Then du = 3 dx, 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, infinite interval integration, rational function 002 (part 1 of 1) 10 points Determine if the improper integral I = Z-∞ 4 6 x- 1 dx converges, and if it does, find 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 infinite. To test for convergence, therefore, we have to check if the limit lim t →-∞ Z t 4 6 x- 1 dx exists. But Z t 4 6 x- 1 dx = h 2 3 ln | 6 x- 1 | i t = ‡ 2 3 ln 1- 2 3 ln | 6 t- 1 | · . 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 infinite. 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 . For 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 · ....
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HW_8_Answers - Miller Kierste – Homework 8 – Due 3:00...

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