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homework 2 solutions

# homework 2 solutions - Mcbiles Emily Homework 2 Due 3:00 am...

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Mcbiles, Emily – Homework 2 – Due: Sep 12 2006, 3:00 am – Inst: David Fonken 1 This print-out should have 20 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 4+ 1 9 · 2 o + n 8+ 2 9 · 2 o + . . . + n 28+ 7 9 · 2 o using sigma notation. 1. 9 X i = 1 n 4 i + i 9 · 2 o 2. 7 X i = 1 4 n i + i 9 · 2 o 3. 9 X i = 1 4 n i + i 9 · 2 o 4. 7 X i = 1 n i + 4 i 9 · 2 o 5. 9 X i = 1 4 n i + 4 i 9 · 2 o 6. 7 X i = 1 n 4 i + i 9 · 2 o correct Explanation: The terms are of the form n 4 i + i 9 · 2 o , with i = 1 , 2 , . . . , 7. Consequently, in sigma notation the sum becomes 7 X i = 1 n 4 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 Decide which of the following regions has area = lim n → ∞ n X i = 1 π 5 n sin 5 n without evaluating the limit. 1. n ( x, y ) : 0 y sin 3 x, 0 x π 5 o 2. n ( x, y ) : 0 y sin x, 0 x π 10 o 3. n ( x, y ) : 0 y sin x, 0 x π 5 o correct 4. n ( x, y ) : 0 y sin 3 x, 0 x π 10 o 5. n ( x, y ) : 0 y sin 2 x, 0 x π 10 o 6. n ( x, y ) : 0 y sin 2 x, 0 x π 5 o Explanation:

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Mcbiles, Emily – Homework 2 – Due: Sep 12 2006, 3:00 am – Inst: David Fonken 2 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 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 X i = 1 π 5 n sin 5 n , therefore, we see that f ( x i ) = sin 5 n , Δ x = π 5 n , where in this case x i = 5 n , f ( x ) = sin x, [ a, b ] = h 0 , π 5 i . Consequently, the area is that of the region under the graph of y = sin x on the interval [0 , π/ 5]. In set-builder notation this is the region n ( x, y ) : 0 y sin x, 0 x π 5 o . keywords: limit Riemann sum, area, trig func- tion 004 (part 1 of 1) 10 points Find an expression for the area of the region under the graph of f ( x ) = x 3 on the interval [3 , 10]. 1. area = lim n → ∞ n X i = 1 3 + 10 i n · 3 8 n 2. area = lim n → ∞ n X i = 1 3 + 8 i n · 3 7 n 3. area = lim n → ∞ n X i = 1 3 + 7 i n · 3 7 n correct 4. area = lim n → ∞ n X i = 1 3 + 10 i n · 3 7 n 5. area = lim n → ∞ n X i = 1 3 + 8 i n · 3 8 n 6. area = lim n → ∞ n X i = 1 3 + 7 i n · 3 8 n Explanation: The area of the region under the graph of f on an interval [ a, b ] is given by the limit A = 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 1 , x 2 ] , . . . , [ x n - 1 , b ] each of length Δ x = ( b - a ) /n and x * i is an arbitrary sample point in [ x i - 1 , x i ].
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