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# L FINAL - Version 038 L FINAL EXAM Neitzke(56585 This...

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Version 038 – L FINAL EXAM – Neitzke – (56585) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine f ( x ) so that lim n → ∞ n summationdisplay k =1 6 n parenleftBig 2 + k n parenrightBig 2 = integraldisplay 1 0 f ( x ) dx . 1. f ( x ) = 6(1 + 2 x 2 ) 2. f ( x ) = 2(6 + x ) 2 3. f ( x ) = 6(2 + x ) 2 correct 4. f ( x ) = 2(1 + 6 x ) 2 5. f ( x ) = 2 + 6 x 2 6. f ( x ) = 6(2 + x 2 ) Explanation: As a Riemann Integral, integraldisplay 1 0 f ( x ) dx = lim n → ∞ n summationdisplay k =1 f ( x k ) 1 n where x k is a sample point in the subinterval bracketleftBig k - 1 n , k n bracketrightBig of [0 , 1] . For the given sum n summationdisplay k =1 6 n parenleftBig 2 + k n parenrightBig 2 , therefore, f ( x ) = 6(2 + x ) 2 . 002 10.0 points Continuous functions f, g are known to have the properties integraldisplay 6 2 f ( x ) dx = 16 , integraldisplay 6 2 g ( x ) dx = 32 respectively. Use these to find the value of the definite integral I = integraldisplay 6 2 (8 f ( x ) - 3 g ( x ) + 6) dx. 1. I = 52 2. I = 50 3. I = 54 4. I = 58 5. I = 56 correct Explanation: Because of the linearity of integration, I = 8 integraldisplay 6 2 f ( x ) dx - 3 integraldisplay 6 2 g ( x ) dx + 6 integraldisplay 6 2 1 dx = I 1 + I 2 + I 3 . Since I 1 = 128 , I 2 = - 96 , I 3 = 24 , it thus follows that I = 56 . 003 10.0 points Determine g ( x ) when g ( x ) = integraldisplay 5 x 2 t 2 sec t dt . 1. g ( x ) = - 4 x sec 2 x 2. g ( x ) = 4 x sec x tan x 3. g ( x ) = 2 x 2 sec x 4. g ( x ) = 2 x 2 tan x 5. g ( x ) = 4 x sec 2 x

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Version 038 – L FINAL EXAM – Neitzke – (56585) 2 6. g ( x ) = - 2 x 2 sec x correct 7. g ( x ) = - 2 x 2 tan x 8. g ( x ) = - 4 x sec x tan x Explanation: By Properties of integrals and the Funda- mental Theorem of Calculus, d dx parenleftBig integraldisplay a x f ( t ) dt parenrightBig = d dx parenleftBig - integraldisplay x a f ( t ) dt parenrightBig = - f ( x ) . When g ( x ) = integraldisplay 5 x f ( t ) dt , f ( t ) = 2 t 2 sec t , therefore, g ( x ) = - 2 x 2 sec x . 004 10.0 points Find the value of lim x 0 e 3 x - 1 6 x + ln(4 x + 1) . 1. limit does not exist 2. limit = 3 10 correct 3. limit = 3 5 4. limit = 3 11 5. limit = 2 5 6. limit = 1 2 Explanation: The limit in question is of the form lim x 0 f ( x ) g ( x ) where f, g are differentiable functions and lim x 0 f ( x ) = 0 , lim x 0 g ( x ) = 0 . L’Hospital’s Rule can thus be applied: lim x 0 e 3 x - 1 6 x + ln(4 x + 1) = lim x 0 3 e 3 x lim x 0 parenleftBig 6 + 4 4 x + 1 parenrightBig . Consequently, limit = 3 10 . 005 10.0 points Find the value of f ( π ) when f ( t ) = 5 3 cos 1 3 t - 2 sin 2 3 t and f ( π 2 ) = 4. 1. f ( π ) = - 3 2 + 7 2 3 2. f ( π ) = 1 2 - 7 2 3 3. f ( π ) = 3 2 - 5 2 3 4. f ( π ) = - 1 2 + 7 2 3 5. f ( π ) = - 3 2 + 5 2 3 correct 6. f ( π ) = 1 2 - 5 2 3 Explanation: The function f must have the form f ( t ) = 5 sin 1 3 t + 3 cos 2 3 t + C
Version 038 – L FINAL EXAM – Neitzke – (56585) 3 where the constant C is determined by the condition f parenleftBig π 2 parenrightBig = 5 sin π 6 + 3 cos π 3 + C = 4 .

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L FINAL - Version 038 L FINAL EXAM Neitzke(56585 This...

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