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Unformatted text preview: Miller, Kierste Homework 8 Due: Mar 20 2007, 3:00 am Inst: Gary Berg 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 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 nonzero for all x > 3. This means that x 5 6 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 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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 Spring '08
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