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Exam01 m408L - Cheung Anthony Exam 1 Due 11:00 pm Inst...

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Cheung, Anthony – Exam 1 – Due: Oct 10 2006, 11:00 pm – Inst: David Benzvi 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Estimate the area, A , under the graph of f ( x ) = 5 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. 1. A 79 12 2. A 25 4 3. A 19 3 4. A 77 12 correct 5. A 13 2 Explanation: With four equal subintervals and right end- points as sample points, A n f (2) + f (3) + f (4) + f (5) o 1 since x i = x * i = i + 1. Consequently, A 5 2 + 5 3 + 5 4 + 1 = 77 12 . keywords: Stewart5e, area, rational function, Riemann sum 002 (part 1 of 1) 10 points Find an expression for the area of the region under the graph of f ( x ) = x 4 on the interval [1 , 6]. 1. area = lim n → ∞ n X i = 1 1 + 8 i n · 4 5 n 2. area = lim n → ∞ n X i = 1 1 + 6 i n · 4 6 n 3. area = lim n → ∞ n X i = 1 1 + 5 i n · 4 5 n correct 4. area = lim n → ∞ n X i = 1 1 + 6 i n · 4 5 n 5. area = lim n → ∞ n X i = 1 1 + 8 i n · 4 6 n 6. area = lim n → ∞ n X i = 1 1 + 5 i n · 4 6 n Explanation: The area of the region under the graph of f on an interval [ a, b ] is given by the limit A = lim n → ∞ n X i = 1 f ( x * i ) Δ x when [ a, b ] is partitioned into n equal subin- tervals [ a, x 1 ] , [ x 1 , x 2 ] , . . . , [ x n - 1 , b ] each of length Δ x = ( b - a ) /n and x * i is an arbitrary sample point in [ x i - 1 , x i ]. Consequently, when f ( x ) = x 4 , [ a, b ] = [1 , 6] , and x * i = x i , we see that area = lim n → ∞ n X i = 1 1 + 5 i n · 4 5 n . keywords: area, limit Riemann sum, cubic function 003 (part 1 of 1) 10 points A function h has graph
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Cheung, Anthony – Exam 1 – Due: Oct 10 2006, 11:00 pm – Inst: David Benzvi 2 -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 - 2 - 4 on ( - 4 , 3). If f ( x ) = Z x - 3 h ( t ) dt, ( x ≥ - 3) , which of the following is the graph of f on ( - 4 , 3)? 1. -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 2 - 2 correct 2. -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 2 - 2 3. -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 2 - 2 4. -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 2 - 2 5. -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 2 - 2 6. -5 -4 -3 -2 -1 0 1 2 3 2 - 2 - 4 2 - 2 Explanation: Since f ( x ) is defined only for x ≥ - 3, there will be no graph of f on the interval ( - 4 , - 3). This already eliminates two of the possible
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Cheung, Anthony – Exam 1 – Due: Oct 10 2006, 11:00 pm – Inst: David Benzvi 3 graphs. On the other hand, f ( - 3) = Z - 3 - 3 h ( x ) dx = 0 , eliminating two more graphs. Finally, by the Fundamental Theorem of Calculus, f 0 ( x ) = h ( x ) on ( - 3 , 3), so f ( x ) will be increasing on any interval on which h > 0, and decreasing on any interval on which h < 0. To determine which of the remaining two possible graphs is the graph of f
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