# Hw03 - badruddin (ssb776) – HW03 – Radin – (57410) 1...

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Unformatted text preview: badruddin (ssb776) – HW03 – Radin – (57410) 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 Decide which of the following regions has area = lim n →∞ n summationdisplay i = 1 π 3 n tan iπ 3 n without evaluating the limit. 1. braceleftBig ( x, y ) : 0 ≤ y ≤ tan 3 x, ≤ x ≤ π 3 bracerightBig 2. braceleftBig ( x, y ) : 0 ≤ y ≤ tan x, ≤ x ≤ π 6 bracerightBig 3. braceleftBig ( x, y ) : 0 ≤ y ≤ tan x, ≤ x ≤ π 3 bracerightBig correct 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 2 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 , b ] each of length Δ x = ( b- a ) /n . If 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 . But in this case x i = iπ 3 n , f ( x ) = tan x, [ a, b ] = bracketleftBig , π 3 bracketrightBig . Consequently, the area is that of the region under the graph of y = tan x on the interval [0 , π/ 3]. In set-builder notation this is the region braceleftBig ( x, y ) : 0 ≤ y ≤ tan x, ≤ x ≤ π 3 bracerightBig . 002 10.0 points Estimate the area under the graph of f ( x ) = 3 sin x between x = 0 and x = π 3 using five approx- imating rectangles of equal widths and right endpoints as sample points. 1. area ≈ 1 . 747 2. area ≈ 1 . 687 3. area ≈ 1 . 767 correct 4. area ≈ 1 . 707 5. area ≈ 1 . 727 Explanation: An estimate for the area, A , under the graph of f on [0 , b ] with [0 , b ] partitioned in n equal subintervals [ x i − 1 , x i ] = bracketleftBig ( i- 1) b n , ib n bracketrightBig and right endpoints x i as sample points is A ≈ braceleftBig f ( x 1 ) + f ( x 2 ) + . . . + f ( x n ) bracerightBig b n . For the given area, f ( x ) = 3 sin x, b = π 3 , n = 5 , badruddin (ssb776) – HW03 – Radin – (57410) 2 and x 1 = 1 15 π, x 2 = 2 15 π, x 3 = 1 5 π, x 4 = 4 15 π, x 5 = 1 3 π . Thus A ≈ 3 braceleftBig sin( 1 15 π ) + . . . + sin( 1 3 π ) bracerightBig π 15 . After calculating these values we obtain the estimate area ≈ 1 . 767 for the area under the graph. 003 10.0 points Cyclist Joe brakes as he approaches a stop sign. His velocity graph over a 5 second period (in units of feet/sec) is shown in 1 2 3 4 5 4 8 12 16 20 Compute best possible upper and lower es- timates for the distance he travels over this period by dividing [0 , 5] into 5 equal subinter- vals and using endpoint sample points....
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