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Unformatted text preview: Version 010 – EXAM 2 – Odell – (58340) 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 4 x + 6 ln x exists, and if it does, find its value. 1. none of the other answers 2. limit =∞ 3. limit = 0 correct 4. limit = 10 5. limit = ∞ 6. limit = 4 Explanation: Use of L’Hospital’s Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 4 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 ) = 4 + 6 x , so lim x →∞ f ′ ( x ) g ′ ( x ) = lim x →∞ 2 ln x 4 x + 6 . We need to apply L’Hospital once again, for then lim x →∞ 2 ln x 4 x + 6 = lim x →∞ 2 x 4 = 0 . Consequently, the limit exists and lim x →∞ (ln x ) 2 4 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 7 1 f ( x ) dx . 1. I ≈ 88 3 2. I ≈ 89 3 correct 3. I ≈ 30 4. I ≈ 29 5. I ≈ 86 3 Explanation: Simpson’s Rule estimates the integral I = integraldisplay 7 1 f ( x ) dx Version 010 – EXAM 2 – Odell – (58340) 2 by I ≈ 1 3 braceleftBig f (1) + 4 f (2) + 2 f (3) + 4 f (4) + 2 f (5) + 4 f (6) + f (7) bracerightBig , taking n = 6. Reading off the values of f from its graph we thus see that I ≈ 89 3 . keywords: definite integral, graph, Simpson’s rule 003 10.0 points Evaluate the definite integral I = integraldisplay 1 6 xe 3 x dx. 1. I = 2 3 e 3 2. I = 2 3 parenleftBig 3 e 3 + 1 parenrightBig 3. I = 2 3 parenleftBig 2 e 3 + 1 parenrightBig correct 4. I = 2 parenleftBig 2 e 3 + 1 parenrightBig 5. I = 2 e 3 6. I = 2 parenleftBig 3 e 3 + 1 parenrightBig Explanation: After integration by parts, I = bracketleftBig 2 xe 3 x bracketrightBig 1 2 integraldisplay 1 e 3 x dx = bracketleftBig 2 xe 3 x 2 3 e 3 x bracketrightBig 1 . Consequently, I = 2 3 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 = 7 4 π 2. I = 5 4 π 3. I = 3 4 π 4. I = π correct 5. I = 3 2 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 . Consequently, I = 4 bracketleftBig tan − 1 u bracketrightBig 1 = π since tan − 1 1 = π 4 . Version 010 – EXAM 2 – Odell – (58340) 3 keywords: definite integral, trig substitution, inverse tan integral 005 10.0 points Determine the integral I = integraldisplay π/ 2 (2 cos θ + cos 3 θ ) dθ ....
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This note was uploaded on 04/15/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.
 Fall '08
 Cepparo

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