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Unformatted text preview: Version 120 – EXAM 2 – Radin – (58415) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x →∞ (ln x ) 2 8 x + 6 ln x exists, and if it does, find its value. 1. limit = 14 2. limit = 8 3. limit =∞ 4. limit = 0 correct 5. limit = ∞ 6. none of the other answers Explanation: Use of L’Hospital’s Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 8 x + 6 ln x . Then f, g have derivatives of all orders and lim x →∞ f ( x ) = ∞ , lim x →∞ g ( x ) = ∞ . Thus L’Hospital’s Rule applies: lim x →∞ f ( x ) g ( x ) = lim x →∞ f ′ ( x ) g ′ ( x ) . But f ′ ( x ) = 2 ln x x , g ′ ( x ) = 8 + 6 x , so lim x →∞ f ′ ( x ) g ′ ( x ) = lim x →∞ 2 ln x 8 x + 6 . We need to apply L’Hospital once again, for then lim x →∞ 2 ln x 8 x + 6 = lim x →∞ 2 x 8 = 0 . Consequently, the limit exists and lim x →∞ (ln x ) 2 8 x + 6 ln x = 0 . 002 10.0 points The graph of a function f is shown in 2 4 6 8 10 2 4 6 8 Use Simpson’s Rule with n = 6 to estimate the integral I = integraldisplay 10 4 f ( x ) dx . 1. I ≈ 28 correct 2. I ≈ 86 3 3. I ≈ 85 3 4. I ≈ 83 3 5. I ≈ 82 3 Explanation: Simpson’s Rule estimates the integral I = integraldisplay 10 4 f ( x ) dx Version 120 – EXAM 2 – Radin – (58415) 2 by I ≈ 1 3 braceleftBig f (4) + 4 f (5) + 2 f (6) + 4 f (7) + 2 f (8) + 4 f (9) + f (10) bracerightBig , taking n = 6. Reading off the values of f from its graph we thus see that I ≈ 28 . keywords: definite integral, graph, Simpson’s rule 003 10.0 points Evaluate the definite integral I = integraldisplay 1 xe 3 x dx. 1. I = 1 3 e 3 2. I = 1 3 parenleftBig 2 e 3 + 1 parenrightBig 3. I = 1 9 parenleftBig 3 e 3 + 1 parenrightBig 4. I = 1 9 e 3 5. I = 1 9 parenleftBig 2 e 3 + 1 parenrightBig correct 6. I = 1 3 parenleftBig 3 e 3 + 1 parenrightBig Explanation: After integration by parts, I = bracketleftBig 1 3 xe 3 x bracketrightBig 1 1 3 integraldisplay 1 e 3 x dx = bracketleftBig 1 3 xe 3 x 1 9 e 3 x bracketrightBig 1 . Consequently, I = 1 9 parenleftBig 2 e 3 + 1 parenrightBig . 004 10.0 points Determine the integral I = integraldisplay π/ 2 4 cos θ 1 + sin 2 θ dθ . 1. I = 3 4 π 2. I = 1 2 π 3. I = 5 4 π 4. I = π correct 5. I = 1 4 π Explanation: Since d dθ sin θ = cos θ , the substitution u = sin θ is suggested. For then du = cos θ dθ , while θ = 0 = ⇒ u = 0 , θ = π 2 = ⇒ u = 1 , so that I = 4 integraldisplay 1 1 1 + u 2 du , which can now be integrated using the fact that d du tan − 1 u = 1 1 + u 2 ....
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This note was uploaded on 04/29/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin

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