M408L hwk8 solutions

# M408L hwk8 solutions - Pham Quoc – Homework 8 – Due...

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Unformatted text preview: Pham, Quoc – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Eric Katerman 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 ∞ 3 ( x + 4) 2 dx is convergent, and if it is, find its value. 1. I = 1 2. I is not convergent 3. I = 1 2 4. I = 5 4 5. I = 1 4 6. I = 3 4 correct 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 3 ( x + 4) 2 dx, exists. To evaluate I t , set u = x + 4, Then du = dx, while x = 0 = ⇒ u = 4 x = t = ⇒ u = t + 4 . In this case, I t = Z t +4 4 3 u 2 du = h- 3 u i t +4 4 , so that lim t →∞ I t = lim t →∞ ‡ 3 4- 3 t + 4 · = 3 4 . Consequently, the integral is convergent and I = 3 4 . . keywords: improper integral, infinite interval integration, rational function 002 (part 1 of 1) 10 points Determine if the improper integral I = Z-∞ 3 5 x- 1 dx converges, and if it does, find its value. 1. I = 1 3 2. I = 5 3 3. I = 3 5 4. I is not convergent correct 5. I = 3 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 3 5 x- 1 dx exists. But Z t 3 5 x- 1 dx = h 3 5 ln | 5 x- 1 | i t = ‡ 3 5 ln 1- 3 5 ln | 5 t- 1 | · . Pham, Quoc – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Eric Katerman 2 On the other hand, lim t →-∞ ln | 5 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 √ 5- x 2 dx converges, and if it does, compute its value. 1. I does not converge correct 2. I =- 4 4 / 5 3. I = (- 4) 4 / 5 4. I = 5 8 (- 4) 4 / 5 5. I =- 5 · 4 4 / 5 8 Explanation: As a 1 / 5 exists for all a , both positive and negative, we see that 5 p 5- x 2 exists and is non-zero for all x > 3. This means that x 5 √ 5- 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 √ 5- x 2 dx, exists. To evaluate I t we use substitution, setting u = 5- x 2 . For then du =- 2 xdx, while x = 3 = ⇒ u =- 4 , x = t = ⇒ u = 5- t 2 . In this case Z t 3 x 5 √ 5- x 2 dx =- 1 2 Z 5- t 2- 4 1 u 1 / 5 du =- 5 8 h u 4 / 5 i 5- t 2- 4 = 5 8 ‡ (- 4) 4 / 5- (5- t 2 ) 4 / 5 · . But (- 4) 4 / 5 = 4 4 / 5 , (5- t 2 ) 4 / 5 = ( t 2- 5) 4 / 5 , so Z t 3 x 5 √ 5- x 2 dx = 5 8 ‡ 4 4 / 5- ( t 2- 5) 4 / 5 · ....
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M408L hwk8 solutions - Pham Quoc – Homework 8 – Due...

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