Exam01vr32

# Exam01vr32 - Cheung, Anthony – Exam 1 – Due: Oct 10...

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Unformatted text preview: 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 ) = 2 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. 1. A ≈ 13 5 2. A ≈ 8 3 3. A ≈ 27 10 4. A ≈ 79 30 5. A ≈ 77 30 correct 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 ≈ 1 + 2 3 + 1 2 + 2 5 = 77 30 . 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 5 on the interval [4 , 8]. 1. area = lim n → ∞ n X i = 1 ‡ 4 + 7 i n · 5 4 n 2. area = lim n → ∞ n X i = 1 ‡ 4 + 4 i n · 5 4 n correct 3. area = lim n → ∞ n X i = 1 ‡ 4 + 5 i n · 5 4 n 4. area = lim n → ∞ n X i = 1 ‡ 4 + 5 i n · 5 5 n 5. area = lim n → ∞ n X i = 1 ‡ 4 + 7 i n · 5 5 n 6. area = lim n → ∞ n X i = 1 ‡ 4 + 4 i n · 5 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 5 , [ a, b ] = [4 , 8] , and x * i = x i , we see that area = lim n → ∞ n X i = 1 ‡ 4 + 4 i n · 5 4 n . keywords: area, limit Riemann sum, cubic function 003 (part 1 of 1) 10 points A function h has graph Cheung, Anthony – Exam 1 – Due: Oct 10 2006, 11:00 pm – Inst: David Benzvi 2 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. 2- 2- 4 2- 2 correct 2. 2- 2- 4 2- 2 3. 2- 2- 4 2- 2 4. 2- 2- 4 2- 2 5. 2- 2- 4 2- 2 6. 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 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 ( x ) = h ( x ) on (- 3 , 3), so f ( x ) will be increasing on...
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## This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.

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Exam01vr32 - Cheung, Anthony – Exam 1 – Due: Oct 10...

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