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M408L_HW03

# M408L_HW03 - hyun(hh7953 HW03 gogolev(57440 This print-out...

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hyun (hh7953) – HW03 – gogolev – (57440) 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 4 n tan 4 n without evaluating the limit. 1. braceleftBig ( x, y ) : 0 y tan x, 0 x π 4 bracerightBig correct 2. braceleftBig ( x, y ) : 0 y tan 3 x, 0 x π 8 bracerightBig 3. braceleftBig ( x, y ) : 0 y tan 2 x, 0 x π 8 bracerightBig 4. braceleftBig ( x, y ) : 0 y tan x, 0 x π 8 bracerightBig 5. braceleftBig ( x, y ) : 0 y tan 2 x, 0 x π 4 bracerightBig 6. braceleftBig ( x, y ) : 0 y tan 3 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 ) 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 4 n tan 4 n , therefore, we see that f ( x i ) = tan 4 n , Δ x = π 4 n . But in this case x i = 4 n , f ( x ) = tan x, [ a, b ] = bracketleftBig 0 , π 4 bracketrightBig . Consequently, the area is that of the region under the graph of y = tan x on the interval [0 , π/ 4]. In set-builder notation this is the region braceleftBig ( x, y ) : 0 y tan x, 0 x π 4 bracerightBig . 002 10.0 points Estimate the area under the graph of f ( x ) = 3 sin x between x = 0 and x = π 4 using five approx- imating rectangles of equal widths and right endpoints as sample points. 1. area 1 . 023 2. area 1 . 063 3. area 1 . 083 4. area 1 . 043 correct 5. area 1 . 103 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 4 , n = 5 ,

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hyun (hh7953) – HW03 – gogolev – (57440) 2 and x 1 = 1 20 π, x 2 = 1 10 π, x 3 = 3 20 π, x 4 = 1 5 π, x 5 = 1 4 π . Thus A 3 braceleftBig sin( 1 20 π ) + . . . + sin( 1 4 π ) bracerightBig π 20 . After calculating these values we obtain the estimate area 1 . 043 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. 1. 40 ft < distance < 62 ft 2. 42 ft < distance < 58 ft 3. 42 ft < distance < 62 ft 4. 42 ft < distance < 60 ft 5. 38 ft < distance < 62 ft 6. 38 ft < distance < 58 ft 7. 40 ft < distance < 58 ft correct 8. 40 ft < distance < 60 ft 9. 38 ft < distance < 60 ft Explanation: The distance Joe travels during the 5 sec- ond period is the area under the velocity graph and above [0 , 5]. Since Joe’s speed is decreasing, the best possible lower estimate occurs taking right hand endpoints as sample points and the area of the rectangles shown in 1 2 3 4 5 4
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M408L_HW03 - hyun(hh7953 HW03 gogolev(57440 This print-out...

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