EXAM 1 - Portillo, Tom Exam 1 Due: Oct 10 2006, 11:00 pm...

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Portillo, Tom – 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 – fnd 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 ) = 3 x on [1 , 5] by dividing [1 , 5] into Four equal subintervals and using right endpoints. 1. A 15 4 2. A 77 20 correct 3. A 39 10 4. A 79 20 5. A 19 5 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 3 2 + 1 + 3 4 + 3 5 = 77 20 . keywords: Stewart5e, area, rational Function, Riemann sum 002 (part 1 oF 1) 10 points ±ind an expression For the area oF the region under the graph oF f ( x ) = x 2 on the interval [1 , 6]. 1. area = lim n → ∞ n X i = 1 1 + 8 i n · 2 6 n 2. area = lim n → ∞ n X i = 1 1 + 6 i n · 2 5 n 3. area = lim n → ∞ n X i = 1 1 + 5 i n · 2 5 n correct 4. area = lim n → ∞ n X i = 1 1 + 5 i n · 2 6 n 5. area = lim n → ∞ n X i = 1 1 + 6 i n · 2 6 n 6. area = lim n → ∞ n X i = 1 1 + 8 i n · 2 5 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 2 , [ a, b ] = [1 , 6] , and x * i = x i , we see that area = lim n → ∞ n X i = 1 1 + 5 i n · 2 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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Portillo, Tom – Exam 1 – Due: Oct 10 2006, 11:00 pm – Inst: David Benzvi 2 2 4 - 2 2 4 on ( - 3 , 4). If f ( x ) = Z x - 2 h ( t ) dt, ( x ≥ - 2) , which of the following is the graph of f on ( - 3 , 4)? 1. 2 4 - 2 2 - 2 2. 2 4 - 2 2 - 2 3. 2 4 - 2 2 - 2 4. 2 4 - 2 2 - 2 5. 2 4 - 2 2 - 2 correct 6. 2 4 - 2 2 - 2 Explanation: Since f ( x ) is deFned only for x ≥ - 2, there will be no graph of f on the interval ( - 3 , - 2). This already eliminates two of the possible
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Portillo, Tom – Exam 1 – Due: Oct 10 2006, 11:00 pm – Inst: David Benzvi 3 graphs. On the other hand, f ( - 2) = Z - 2 - 2 h ( x ) dx = 0 , eliminating two more graphs. Finally, by the Fundamental Theorem of Calculus, f 0 ( x ) = h ( x ) on ( - 2 , 4), 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 we thus have to look at the sign of h on an interval and the slope of a possible graph of f on that interval. Consequently, the
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This note was uploaded on 02/18/2010 for the course MATH 408L taught by Professor Benzvi during the Spring '06 term at North Texas.

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EXAM 1 - Portillo, Tom Exam 1 Due: Oct 10 2006, 11:00 pm...

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