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HW03 solution - mao(tm23477 HW03 Radin(56570 This print-out...

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mao (tm23477) – HW03 – Radin – (56570) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rewrite the sum 5 n parenleftBig 4 + 2 n parenrightBig 2 + 5 n parenleftBig 4 + 4 n parenrightBig 2 + . . . + 5 n parenleftBig 4 + 2 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i =1 2 n parenleftBig 4 i + 5 i n parenrightBig 2 2. n summationdisplay i =1 5 n parenleftBig 4 i + 2 i n parenrightBig 2 3. n summationdisplay i =1 5 i n parenleftBig 4 + 2 i n parenrightBig 2 4. n summationdisplay i =1 5 n parenleftBig 4 + 2 i n parenrightBig 2 correct 5. n summationdisplay i =1 2 i n parenleftBig 4 + 5 i n parenrightBig 2 6. n summationdisplay i =1 2 n parenleftBig 4 + 5 i n parenrightBig 2 Explanation: The terms are of the form 5 n parenleftBig 4 + 2 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i =1 5 n parenleftBig 4 + 2 i n parenrightBig 2 . 002 10.0 points Estimate the area, A , under the graph of f ( x ) = 5 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. Correct answer: 6 . 417. Explanation: With four equal subintervals and right end- points as sample points, A braceleftBig f (2) + f (3) + f (4) + f (5) bracerightBig 1 since x i = x i = i + 1. Consequently, A 6 . 417 . 003 10.0 points Decide which of the following regions has area = lim n → ∞ n summationdisplay i =1 π 2 n cos 2 n without evaluating the limit. 1. braceleftBig ( x, y ) : 0 y cos 2 x, 0 x π 2 bracerightBig 2. braceleftBig ( x, y ) : 0 y cos 3 x, 0 x π 2 bracerightBig 3. braceleftBig ( x, y ) : 0 y cos x, 0 x π 4 bracerightBig 4. braceleftBig ( x, y ) : 0 y cos x, 0 x π 2 bracerightBig correct 5. braceleftBig ( x, y ) : 0 y cos 2 x, 0 x π 4 bracerightBig 6. braceleftBig ( x, y ) : 0 y cos 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 x
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mao (tm23477) – HW03 – Radin – (56570) 2 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 cos 2 n , therefore, we see that f ( x i ) = cos 2 n , Δ x = π 2 n , where in this case x i = 2 n , f ( x ) = cos x, [ a, b ] = bracketleftBig 0 , π 2 bracketrightBig . Consequently, the area is that of the region under the graph of y = cos x on the interval [0 , π/ 2]. In set-builder notation this is the region braceleftBig ( x, y ) : 0 y cos x, 0 x π 2 bracerightBig . 004 10.0 points Find an expression for the area of the region under the graph of f ( x ) = x 5 on the interval [3 , 8]. 1. area = lim n → ∞ n summationdisplay i =1 parenleftBig 3 + 6 i n parenrightBig 5 5 n 2. area = lim n → ∞ n summationdisplay i =1 parenleftBig 3 + 6 i n parenrightBig 5 6 n 3. area = lim n → ∞ n summationdisplay i =1 parenleftBig 3 + 5 i n parenrightBig 5 5 n correct 4. area = lim n → ∞ n summationdisplay i =1 parenleftBig 3 + 5 i n parenrightBig 5 6 n 5. area = lim
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