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HW_2_Answers - Miller Kierste Homework 2 Due 3:00 am Inst...

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Miller, Kierste – Homework 2 – Due: Jan 30 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Rewrite the sum n 2+ 1 9 · 2 o + n 4+ 2 9 · 2 o + . . . + n 12+ 6 9 · 2 o using sigma notation. 1. 6 X i = 1 n 2 i + i 9 · 2 o correct 2. 6 X i = 1 n i + 2 i 9 · 2 o 3. 9 X i = 1 n 2 i + i 9 · 2 o 4. 6 X i = 1 2 n i + i 9 · 2 o 5. 9 X i = 1 2 n i + i 9 · 2 o 6. 9 X i = 1 2 n i + 2 i 9 · 2 o Explanation: The terms are of the form n 2 i + i 9 · 2 o , with i = 1 , 2 , . . . , 6. Consequently, in sigma notation the sum becomes 6 X i = 1 n 2 i + i 9 · 2 o . keywords: Stewart5e, summation notation, 002 (part 1 of 1) 10 points Estimate the area, A , under the graph of f ( x ) = 4 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. Correct answer: 5 . 133 . Explanation: With four equal subintervals and right end- points as sample points, A n f (2) + f (3) + f (4) + f (5) o 1 since x i = x * i = i + 1. Consequently, A 5 . 133 . keywords: Stewart5e, area, rational function, Riemann sum, 003 (part 1 of 1) 10 points Cyclist Joe accelerates as he rides away from 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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Miller, Kierste – Homework 2 – Due: Jan 30 2007, 3:00 am – Inst: Gary Berg 2 1. 47 ft < distance < 61 ft 2. 47 ft < distance < 59 ft 3. 47 ft < distance < 63 ft 4. 49 ft < distance < 61 ft 5. 49 ft < distance < 63 ft 6. 45 ft < distance < 61 ft correct 7. 45 ft < distance < 63 ft 8. 49 ft < distance < 59 ft 9. 45 ft < distance < 59 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 increasing, the best possible lower estimate occurs taking left hand endpoints as sample points and the area of the rectangles shown in 1 2 3 4 5 4 8 12 16 20 On the other hand, the best upper estimate will occur taking right hand endpoints and the area of the rectangles shown in 1 2 3 4 5 4 8 12 16 20 Consequently, reading off values from the graphs to compute the height of the rect- angles, we see that 45 ft < distance < 61 ft . keywords: 004 (part 1 of 1) 10 points Stewart Section 5.1, Example 3(a), page 321 Decide which of the following regions has area = lim n → ∞ n X i = 1 π 4 n tan 4 n without evaluating the limit. 1. n ( x, y ) : 0 y tan x, 0 x π 8 o 2. n ( x, y ) : 0 y tan 2 x, 0 x π 4 o 3. n ( x, y ) : 0 y tan 3 x, 0 x π 8 o 4. n ( x, y ) : 0 y tan 2 x, 0 x π 8 o 5. n ( x, y ) : 0 y tan 3 x, 0 x π 4 o 6. n ( x, y ) : 0 y tan x, 0 x π 4 o correct Explanation:
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Miller, Kierste – Homework 2 – Due: Jan 30 2007, 3:00 am – Inst: Gary Berg 3 The area under the graph of y = f ( x ) on an interval [ a, b ] is given by the limit lim n → ∞ n X i = 1 f ( x i ) Δ x when [ a, b ] is partitioned into n equal subin- tervals [ a, x 1 ] , [ x
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