int cal final exam

# int cal final exam - Version 084 M408L FINAL EXAM...

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Version 084 – M408L FINAL EXAM – Radin – (57410) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points For each n the interval [3 , 9] is divided into n subintervals [ x i 1 , x i ] o± equal length Δ x , and a point x i is chosen in [ x i 1 , x i ]. Express the limit lim n →∞ n s i = 1 (5 x i sin x i ) Δ x as a defnite integral. 1. limit = i 3 9 5 x dx 2. limit = i 9 3 5 x dx 3. limit = i 3 9 5 sin x dx 4. limit = i 9 3 5 x sin x dx correct 5. limit = i 3 9 5 x sin x dx 6. limit = i 9 3 5 sin x dx Explanation: By defnition, the defnite integral I = i b a f ( x ) dx o± a continuous ±unction f on an interval [ a, b ] is the limit I = lim n →∞ n s i = 1 f ( x i ) Δ x o± the Riemann sum n s i = 1 f ( x i ) Δ x ±ormed when [ a, b ] is divided into n subinter- vals [ x i 1 , x i ] o± equal length Δ x and x i is some point in [ x i 1 , x i ]. In the given example, f ( x ) = 5 x sin x, [ a, b ] = [3 , 9] . Thus limit = i 9 3 5 x sin x dx . 002 10.0 points Determine i± lim x 0 sin 1 (4 x ) tan 1 (3 x ) exists, and i± it does, fnd its value. 1. limit = 3 2. limit = 0 3. limit = 4 3 correct 4. limit = 3 4 5. limit does not exist 6. limit = 4 Explanation: Since the limit has the ±orm 0 / 0, it is in- determinate and L’Hospital’s Rule can be ap- plied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) with f ( x ) = sin 1 (4 x ) , g ( x ) = tan 1 (3 x ) , and f ( x ) = 4 1 16 x 2 , g ( x ) = 3 1 + 9 x 2 .

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Version 084 – M408L FINAL EXAM – Radin – (57410) 2 In this case, f ( x ) g ( x ) = 4 3 p 1 + 9 x 2 1 16 x 2 P −→ 4 3 as x 0. Consequently, the limit exists and limit = 4 3 . 003 10.0 points Determine g ( x ) when g ( x ) = i x 0 r 5 6 t 2 dt . 1. g ( x ) = 6 x 5 6 x 2 2. g ( x ) = r 5 6 x 2 correct 3. g ( x ) = r 5 6 x 2 4. g ( x ) = x 5 6 x 2 5. g ( x ) = 6 x r 5 6 x 2 6. g ( x ) = x 5 6 x 2 Explanation: By the Fundamental Theorem of Calculus, d dx p i x a f ( t ) dt P = f ( x ) . When g ( x ) = i x 0 f ( t ) dt , f ( t ) = r 5 6 t 2 , therefore, g ( x ) = r 5 6 x 2 . 004 10.0 points Find f (1) when f ′′ ( x ) = 9 x and f ( 1) = 5 , f ( 1) = 4 . 1. f (1) = 4 2. f (1) = 7 correct 3. f (1) = 6 4. f (1) = 8 5. f (1) = 5 Explanation: When f ′′ ( x ) = 9 x the most general anti- derivative of f ′′ is f ( x ) = 9 2 x 2 + C where the arbitrary constant C is determined by the condition f ( 1) = 4. For f ( 1) = 4 = f ( x ) = 9 2 x 2 1 2 .
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int cal final exam - Version 084 M408L FINAL EXAM...

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