This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
 Spring '08
 RAdin

Click to edit the document details