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# solution2_pdf - So(dds785 – HW03 – Radin –(54915 1...

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Unformatted text preview: So (dds785) – HW03 – Radin – (54915) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rewrite the sum 3 n parenleftBig 5 + 6 n parenrightBig 2 + 3 n parenleftBig 5 + 12 n parenrightBig 2 + . . . + 3 n parenleftBig 5 + 6 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i = 1 3 n parenleftBig 5 i + 6 i n parenrightBig 2 2. n summationdisplay i = 1 6 n parenleftBig 5 i + 3 i n parenrightBig 2 3. n summationdisplay i = 1 6 i n parenleftBig 5 + 3 i n parenrightBig 2 4. n summationdisplay i = 1 3 i n parenleftBig 5 + 6 i n parenrightBig 2 5. n summationdisplay i = 1 6 n parenleftBig 5 + 3 i n parenrightBig 2 6. n summationdisplay i = 1 3 n parenleftBig 5 + 6 i n parenrightBig 2 correct Explanation: The terms are of the form 3 n parenleftBig 5 + 6 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i = 1 3 n parenleftBig 5 + 6 i n parenrightBig 2 . 002 10.0 points The graph of a function f on the interval [0 , 10] is shown in 2 4 6 8 10 2 4 6 8 Estimate the area under the graph of f by dividing [0 , 10] into 10 equal subintervals and using right endpoints as sample points. 1. area ≈ 49 2. area ≈ 52 3. area ≈ 50 4. area ≈ 51 correct 5. area ≈ 53 Explanation: With 10 equal subintervals and right end- points as sample points, area ≈ braceleftBig f (1) + f (2) + . . . f (10) bracerightBig 1 , since x i = i . Consequently, area ≈ 51 , reading off the values of f (1) , f (2) , . . . , f (10) from the graph of f . 003 10.0 points Decide which of the following regions has area = lim n →∞ n summationdisplay i = 1 π 2 n sin iπ 2 n So (dds785) – HW03 – Radin – (54915) 2 without evaluating the limit. 1. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 3 x, ≤ x ≤ π 2 bracerightBig 2. braceleftBig ( x, y ) : 0 ≤ y ≤ sin x, ≤ x ≤ π 4 bracerightBig 3. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 4 x, ≤ x ≤ π 4 bracerightBig 4. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 2 x, ≤ x ≤ π 2 bracerightBig 5. braceleftBig ( x, y ) : 0 ≤ y ≤ sin x, ≤ x ≤ π 2 bracerightBig correct 6. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 3 x, ≤ x ≤ π 4 bracerightBig Explanation: The area under the graph of y = f ( x ) on an interval [ a, b ] is given by the limit lim n →∞ n summationdisplay 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 , x n ] each of length Δ x = ( b- a ) /n . When the area is given by A = lim n →∞ n summationdisplay i = 1 π 2 n sin iπ 2 n , therefore, we see that f ( x i ) = sin iπ 2 n , Δ x = π 2 n , where in this case x i = iπ 2 n , f ( x ) = sin x, [ a, b ] = bracketleftBig , π 2 bracketrightBig ....
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## This note was uploaded on 10/25/2010 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.

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solution2_pdf - So(dds785 – HW03 – Radin –(54915 1...

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