On the other hand lim t 0 t 2 3 0 consequently i is

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On the other hand, lim t 0 + t 2 / 3 = 0 . Consequently, I is convergent and I = 3 . 017(part1of3)10.0points (i) Express the improper integral I = integraldisplay 2 xe - x/ 2 dx as lim t → ∞ I t with each I t a proper integral. 1. I = lim t → ∞ integraldisplay t - t xe - x/ 2 dx 2. I = lim t → ∞ integraldisplay 2+ t xe - x/ 2 dx 3. I = lim t → ∞ integraldisplay 2 xe - x/t dx 4. I = lim t → ∞ integraldisplay t xe - x/ 2 dx 5. I = lim t → ∞ integraldisplay t 2 xe - x/ 2 dx correct Explanation: The integral I = integraldisplay 2 xe - x/ 2 dx is improper because of the infinite range of integration. To overcome this we restrict to a finite interval of integration and consider the limit I = lim t → ∞ I t , I t = integraldisplay t 2 xe - x/ 2 dx . 018(part2of3)10.0points (ii) Compute the value of I t . 1. I t = 8 e - 1 - 2( t + 2) e - t/ 2 correct 2. I t = 2(2 - t ) { e - 1 - e - t/ 2 } 3. I t = - 8 e - 1 + 2( t + 2) e - t/ 2
white (jew3279) – Homework 03 – staron – (53545) 10 4. I t = 4 e - 1 - ( t + 2) e - t/ 2 5. I t = 8 e - 1 + 2( t + 2) e - t/ 2 020 10.0points Determine if the improper integral I = integraldisplay 021 10.0points
white (jew3279) – Homework 03 – staron – (53545) 11 Find the total area under the graph of y = 8 x 3 for x 1 . 1. Area = 2. Area = 4 correct 3. Area = 7 2 4. Area = 9 2 5. Area = 5 6. Area = 3 Explanation: The total area under the graph for x 1 is an improper integral whose value is the limit lim t → ∞ integraldisplay t 1 8 x 3 dx . Now integraldisplay t 1 8 x - 3 dx = - 4 bracketleftBig x - 2 bracketrightBig t 1 . Consequently, Area = lim t → ∞ 4 bracketleftBig 1 - t - 2 bracketrightBig = 4 . 022 10.0points Determine if the improper integral I = integraldisplay 0 2 (2 x + 3) 2 dx is convergent, and if it is, find its value.

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