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Unformatted text preview: zakaria (mmz255) HW02 gilbert (55485) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which of the following integrals are im proper? I 1 = integraldisplay 2 x + 2 x + 1 dx , I 2 = integraldisplay 100 1 1 1 + x 2 dx , I 3 = integraldisplay 1 1 1 + x dx . 1. none of them correct 2. I 1 and I 2 only 3. I 2 only 4. all of them 5. I 1 only 6. I 1 and I 3 only 7. I 3 only 8. I 2 and I 3 only Explanation: An integral I = integraldisplay b a f ( x ) dx is improper when one or more of the following conditions are satisfied: (i) the interval of integration is infinite, i.e. , when a = or b = , or both; (ii) f has a vertical asymptote at one or more of x = a , x = b or x = c for some a < c < b . Consequently, integraldisplay 2 x + 2 x + 1 dx is not improper; integraldisplay 100 1 1 1 + x 2 dx is not improper; integraldisplay 1 1 1 + x dx is not improper. 002 10.0 points Determine if the improper integral I = integraldisplay 5 9 ( x + 7) 2 dx converges, and if it does, compute its value. 1. I = 5 3 2. I doesnt converge 3. I = 3 4 4. I = 9 13 5. I = 3 4 correct Explanation: The integral is improper because of the infi nite interval of integration. To overcome this we set I = lim t integraldisplay t 5 9 ( x + 7) 2 dx zakaria (mmz255) HW02 gilbert (55485) 2 whenever the limit on the right hand side exists. Now integraldisplay t 5 9 ( x + 7) 2 dx = bracketleftBig 9 x + 7 bracketrightBig t 5 = 9 t + 7 + 3 4 . Since lim t 9 t + 7 = 0 , it follows that the improper integral converges and that I = integraldisplay 5 9 ( x + 7) 2 dx = 3 4 . 003 10.0 points Find the total area under the graph of y = 5 x 3 for x 1 . 1. Area = 2. Area = 5 2 correct 3. Area = 3 4. Area = 7 2 5. Area = 2 6. Area = 3 2 Explanation: The total area under the graph for x 1 is an improper integral whose value is the limit lim t integraldisplay t 1 5 x 3 dx . Now integraldisplay t 1 5 x 3 dx = 5 2 bracketleftBig x 2 bracketrightBig t 1 . Consequently, Area = lim t 5 2 bracketleftBig 1 t 2 bracketrightBig = 5 2 . 004 10.0 points Determine if I = integraldisplay 3 x 5 x 2 4 dx converges, and if it does, compute its value. 1. I = 5 4 / 5 8 2. I = 5 4 / 5 3. I = 5 5 4 / 5 8 4. I does not converge correct 5. I = 5 5 4 / 5 4 6. I = 5 5 4 / 5 8 Explanation: The integral I is improper because of the infinite interval of integration. Thus I will converge if lim t integraldisplay t 3 x 5 x 2 4 dx exists. To evaluate this last integral, we use substitution, setting u = x 2 4. For then du = 2 x dx , while x = 3 = u = 5 , x = t = u = t 2 4 . zakaria (mmz255) HW02 gilbert (55485) 3 In this case integraldisplay t 3 x 5 x 2 4 dx = 1 2 integraldisplay t 2 4 5 1 u 1 / 5 du = 5 8 bracketleftBig u 4 / 5 bracketrightBig...
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 Spring '11
 Gilbert

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