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Unformatted text preview: Pham, Quoc – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Eric Katerman 1 This printout should have 23 questions. Multiplechoice 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 nonzero for all x > 3. This means that x 5 √ 5 x 2 exists and is nonzero 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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 Spring '08
 RAdin
 Derivative, lim, dx, Pham, Eric Katerman

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