M 408L exam 1

# M 408L exam 1 - Version 075 EXAM 1 Cepparo(58400 This print-out should have 19 questions Multiple-choice questions may continue on the next column

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Version 075 – EXAM 1 – Cepparo – (58400) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points When 2 4 6 8 10 2 4 2 4 is the graph oF a Function f , use rectangles to estimate the defnite integral I = i 10 0 | f ( x ) | dx by subdividing [0 , 10] into 10 equal subin- tervals and taking right endpoints oF these subintervals. 1. I 24 2. I 25 3. I 21 4. I 23 5. I 22 correct Explanation: The defnite integral I = i 10 0 | f ( x ) | dx is the area between the graph oF f and the interval [0 , 10]. The area is estimated using the gray-shaded rectangles in 2 4 6 8 10 2 4 2 4 Each rectangle has base-length 1; and it’s height can be read o± From the graph. Thus Area = 4 + 3 + 2 + 1 + 1 + 3 + 4 + 3 + 1 . Consequently, I 22 . 002 10.0 points ²ind 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 s i = 1 p 4 + 5 i n P 5 5 n 2. area = lim n →∞ n s i = 1 p 4 + 7 i n P 5 4 n 3. area = lim n →∞ n s i = 1 p 4 + 7 i n P 5 5 n 4. area = lim n →∞ n s i = 1 p 4 + 4 i n P 5 5 n 5. area = lim n →∞ n s i = 1 p 4 + 5 i n P 5 4 n 6. area = lim n →∞ n s i = 1 p 4 + 4 i n P 5 4 n correct

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Version 075 – EXAM 1 – Cepparo – (58400) 2 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 s 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 s i = 1 p 4 + 4 i n P 5 4 n . 003 10.0 points Express the limit lim n →∞ n s i =1 2 x i sin x i Δ x as a deFnite integral on the interval [3 , 7]. 1. limit = i 3 7 2 x sin x dx 2. limit = i 3 7 2 sin x dx 3. limit = i 7 3 2 x dx 4. limit = i 7 3 2 x sin x dx correct 5. limit = i 7 3 2 sin x dx 6. limit = i 3 7 2 x dx Explanation: By deFnition, the deFnite integral I = i b a f ( x ) dx of a continuous function f on an interval [ a, b ] is the limit I = lim n →∞ n s i = 1 f ( x i ) Δ x of the Riemann sum n s i = 1 f ( x i ) Δ x formed when the interval [ a, b ] is divided into n subintervals of equal width Δ x and x i is any sample point in the i th subinterval [ x i 1 , x i ]. In the given example, f ( x ) = 2 x sin x, [ a, b ] = [3 , 7] . Thus limit = i 7 3 2 x sin x dx . 004 10.0 points If F ( x ) = i x 0 5 e 12 sin 2 θ dθ , Fnd the value of F ( π/ 4).
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## This note was uploaded on 03/24/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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M 408L exam 1 - Version 075 EXAM 1 Cepparo(58400 This print-out should have 19 questions Multiple-choice questions may continue on the next column

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