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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 doesn’t 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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This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas.
 Spring '11
 Gilbert

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