# Homework 4 - pacheco(jnp926 Homework 4 staron(52840 This...

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pacheco (jnp926) – Homework 4 – staron – (52840) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points (i) Express the improper integral I = integraldisplay 3 2 xe x/ 2 dx as lim t → ∞ I t with each I t a proper integral. 1. I = lim t → ∞ integraldisplay t 3 2 xe x/ 2 dx correct 2. I = lim t → ∞ integraldisplay t t 2 xe x/ 2 dx 3. I = lim t → ∞ integraldisplay t 2 xe x/ 2 dx 4. I = lim t → ∞ integraldisplay 3 2 xe x/t dx 5. I = lim t → ∞ integraldisplay 3+ t 2 xe x/ 2 dx Explanation: The integral I = integraldisplay 3 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 3 2 xe x/ 2 dx . 002(part2of3)10.0points (ii) Compute the value of I t . 1. I t = 20 e 3 2 4( t + 2) e t/ 2 correct 2. I t = 4(3 t ) { e 3 2 e t/ 2 } 3. I t = 20 e 3 2 + 4( t + 2) e t/ 2 4. I t = 20 e 3 2 + 4( t + 2) e t/ 2 5. I t = 5 e 3 2 ( t + 2) e t/ 2 Explanation: To evaluate I t = integraldisplay t 3 2 xe x/ 2 dx we integrate by parts. Then I t = bracketleftBig 4 xe x/ 2 bracketrightBig t 3 + 4 integraldisplay t 3 e x/ 2 dx = 4 bracketleftBig ( x + 2) e x/ 2 bracketrightBig t 3 . Consequently, I t = 20 e 3 2 4( t + 2) e t/ 2 . 003(part3of3)10.0points (iii) Determine if lim t → ∞ I t exists, and if it does, find its value. 1. I = 20 e 3 2 2. I = 4 e 3 2 3. I = 20 e 3 2 correct 4. I does not exist 5. I = 4 e 3 2 Explanation: By L’Hospital’s Rule, lim x → ∞ x m e x = 0 for any power m of x . It follows that lim t → ∞ ( t + 2) e t/ 2 = 0 . Consequently, lim t → ∞ I t exists, and I = 20 e 3 2 .
pacheco (jnp926) – Homework 4 – staron – (52840) 2 004 10.0points Determine if the improper integral I = integraldisplay 0 4 (4 x + 3) 2 dx is convergent, and if it is, find its value. 1. I = 5 12 2. I = 1 4 3. I is not convergent 4. I = 1 2 5. I = 1 3 correct 6. I = 7 12 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 = integraldisplay t 0 4 (4 x + 3) 2 dx , exists. To evaluate I t , set u = 4 x + 3, Then du = 4 dx , while x = 0 = u = 3 x = t = u = 4 t + 3 . In this case, I t = 1 4 integraldisplay 4 t +3 3 4 u 2 du = 1 4 bracketleftBig 4 u bracketrightBig 4 t +3 3 , so that lim t → ∞ I t = lim t → ∞ parenleftBig 1 3 1 4 t + 3 parenrightBig = 1 3 . Consequently, the integral is convergent and I = 1 3 . . 005 10.0points Determine if I = integraldisplay 3 x 3 x 2 2 dx converges, and if it does, compute its value. 1. I = 7 2 / 3 4 2. I = 7 2 / 3 3. I = 3 · 7 2 / 3 2 4. I = 3 · 7 2 / 3 4 5. I does not converge correct 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