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HW03-solutions

# HW03-solutions - kim(tk5895 HW03 Henry(54974 10 1 This...

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kim (tk5895) – 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 4 n parenleftBig 5 + 2 n parenrightBig 2 + 4 n parenleftBig 5 + 4 n parenrightBig 2 + . . . + 4 n parenleftBig 5 + 2 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i =1 2 i n parenleftBig 5 + 4 i n parenrightBig 2 2. n summationdisplay i =1 4 n parenleftBig 5 i + 2 i n parenrightBig 2 3. n summationdisplay i =1 4 n parenleftBig 5 + 2 i n parenrightBig 2 correct 4. n summationdisplay i =1 2 n parenleftBig 5 i + 4 i n parenrightBig 2 5. n summationdisplay i =1 2 n parenleftBig 5 + 4 i n parenrightBig 2 6. n summationdisplay i =1 4 i n parenleftBig 5 + 2 i n parenrightBig 2 Explanation: The terms are of the form 4 n parenleftBig 5 + 2 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i =1 4 n parenleftBig 5 + 2 i n parenrightBig 2 . 002 10.0 points The graph of a function f on the interval [0 , 10] is shown in -1 0 1 2 3 4 5 6 7 8 9 10 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 56 2. area 54 3. area 57 4. area 53 5. area 55 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 55 , 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 tan 2 n

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kim (tk5895) – HW03 – Henry – (54974) 2 without evaluating the limit. 1. braceleftBig ( x, y ) : 0 y tan x, 0 x π 2 bracerightBig correct 2. braceleftBig ( x, y ) : 0 y tan 3 x, 0 x π 2 bracerightBig 3. braceleftBig ( x, y ) : 0 y tan x, 0 x π 4 bracerightBig 4. braceleftBig ( x, y ) : 0 y tan 3 x, 0 x π 4 bracerightBig 5. braceleftBig ( x, y ) : 0 y tan 2 x, 0 x π 2 bracerightBig 6. braceleftBig ( x, y ) : 0 y tan 4 x, 0 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 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 tan 2 n , therefore, we see that f ( x i ) = tan 2 n , Δ x = π 2 n , where in this case x i = 2 n , f ( x ) = tan x, [ a, b ] = bracketleftBig 0 , π 2 bracketrightBig . Consequently, the area is that of the region under the graph of y = tan x on the interval [0 , π/ 2]. In set-builder notation this is the region braceleftBig ( x, y ) : 0 y tan x, 0 x π 2 bracerightBig .
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HW03-solutions - kim(tk5895 HW03 Henry(54974 10 1 This...

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