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Unformatted text preview: moore (jwm2685) – 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 1 1 √ 1 + x dx , I 2 = integraldisplay 2 x − 2 x − 1 dx , I 3 = integraldisplay ∞ 1 1 1 + x 2 dx . 1. I 1 and I 2 only 2. all of them 3. I 1 and I 3 only 4. I 3 only 5. I 1 only 6. none of them 7. I 2 only 8. I 2 and I 3 only correct 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 1 1 √ 1 + x dx is not improper; integraldisplay 2 x − 2 x − 1 dx is improper; integraldisplay ∞ 1 1 1 + x 2 dx is improper. 002 10.0 points Determine if the improper integral I = integraldisplay ∞ 6 9 ( x + 6) 2 dx converges, and if it does, compute its value. 1. I = 10 7 2. I = 3 4 correct 3. I = 9 13 4. I = − 3 4 5. I doesn’t converge Explanation: The integral is improper because of the infi nite interval of integration. To overcome this we set I = lim t →∞ integraldisplay t 6 9 ( x + 6) 2 dx moore (jwm2685) – HW02 – gilbert – (55485) 2 whenever the limit on the right hand side exists. Now integraldisplay t 6 9 ( x + 6) 2 dx = bracketleftBig − 9 x + 6 bracketrightBig t 6 = − 9 t + 6 + 3 4 . Since lim t →∞ 9 t + 6 = 0 , it follows that the improper integral converges and that I = integraldisplay ∞ 6 9 ( x + 6) 2 dx = 3 4 . 003 10.0 points Find the total area under the graph of y = 9 x 4 for x ≥ 1 . 1. Area = 3 correct 2. Area = 10 3 3. Area = 4 4. Area = 11 3 5. Area = 8 3 6. Area = ∞ Explanation: The total area under the graph for x ≥ 1 is an improper integral whose value is the limit lim t →∞ integraldisplay t 1 9 x 4 dx . Now integraldisplay t 1 9 x 4 dx = − 3 bracketleftBig x 3 bracketrightBig t 1 . Consequently, Area = lim t →∞ 3 bracketleftBig 1 − t 3 bracketrightBig = 3 . 004 10.0 points Determine if I = integraldisplay ∞ 3 x 3 √ x 2 − 2 dx converges, and if it does, compute its value. 1. I does not converge correct 2. I = 7 2 / 3 3. I = 7 2 / 3 4 4. I = − 3 · 7 2 / 3 4 5. I = − 3 · 7 2 / 3 2 6. I = 3 · 7 2 / 3 4 Explanation: The integral I is improper because of the infinite interval of integration. Thus I will converge if lim t →∞ integraldisplay t 3 x 3 √ x 2 − 2 dx exists. To evaluate this last integral, we use substitution, setting u = x 2 − 2. For then du = 2 x dx , while x = 3 = ⇒ u = 7 , x = t = ⇒ u = t 2 − 2 . moore (jwm2685) – HW02 – gilbert – (55485) 3 In this case integraldisplay t 3 x 3 √ x 2 − 2 dx = 1 2 integraldisplay t 2 2 7 1 u 1 / 3 du = 3 4 bracketleftBig u 2 / 3 bracketrightBig...
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This note was uploaded on 05/02/2011 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas at Austin.
 Spring '07
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