M 408L HW03 - kodeih (njk359) – HW03 – Henry –...

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Unformatted text preview: kodeih (njk359) – HW03 – Henry – (54974) 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 5 n parenleftBig 6 + 2 n parenrightBig 2 + 5 n parenleftBig 6 + 4 n parenrightBig 2 + . . . + 5 n parenleftBig 6 + 2 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i = 1 5 i n parenleftBig 6 + 2 i n parenrightBig 2 2. n summationdisplay i = 1 5 n parenleftBig 6 i + 2 i n parenrightBig 2 3. n summationdisplay i = 1 5 n parenleftBig 6 + 2 i n parenrightBig 2 correct 4. n summationdisplay i = 1 2 i n parenleftBig 6 + 5 i n parenrightBig 2 5. n summationdisplay i = 1 2 n parenleftBig 6 i + 5 i n parenrightBig 2 6. n summationdisplay i = 1 2 n parenleftBig 6 + 5 i n parenrightBig 2 Explanation: The terms are of the form 5 n parenleftBig 6 + 2 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i = 1 5 n parenleftBig 6 + 2 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 ≈ 53 2. area ≈ 57 3. area ≈ 55 4. area ≈ 54 5. area ≈ 56 correct 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 ≈ 56 , 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 3 n tan iπ 3 n kodeih (njk359) – HW03 – Henry – (54974) 2 without evaluating the limit. 1. braceleftBig ( x, y ) : 0 ≤ y ≤ tan x, ≤ x ≤ π 6 bracerightBig 2. braceleftBig ( x, y ) : 0 ≤ y ≤ tan x, ≤ x ≤ π 3 bracerightBig correct 3. braceleftBig ( x, y ) : 0 ≤ y ≤ tan 3 x, ≤ x ≤ π 3 bracerightBig 4. braceleftBig ( x, y ) : 0 ≤ y ≤ tan 2 x, ≤ x ≤ π 3 bracerightBig 5. braceleftBig ( x, y ) : 0 ≤ y ≤ tan 3 x, ≤ x ≤ π 6 bracerightBig 6. braceleftBig ( x, y ) : 0 ≤ y ≤ tan 4 x, ≤ x ≤ π 6 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 3 n tan iπ 3 n , therefore, we see that f ( x i ) = tan iπ 3 n , Δ x = π 3 n , where in this case x i = iπ 3 n , f ( x ) = tan x, [ a, b ] = bracketleftBig , π 3 bracketrightBig ....
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This note was uploaded on 09/25/2010 for the course M 54980 taught by Professor Henry during the Fall '10 term at University of Texas at Austin.

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M 408L HW03 - kodeih (njk359) – HW03 – Henry –...

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