cal3 - choi (kc24547) HW03 BERG (56525) 1 This print-out...

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Unformatted text preview: choi (kc24547) HW03 BERG (56525) 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 4 n parenleftBig 6 + 5 n parenrightBig 2 + 4 n parenleftBig 6 + 10 n parenrightBig 2 + . . . + 4 n parenleftBig 6 + 5 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i = 1 5 n parenleftBig 6 + 4 i n parenrightBig 2 2. n summationdisplay i = 1 4 i n parenleftBig 6 + 5 i n parenrightBig 2 3. n summationdisplay i = 1 5 i n parenleftBig 6 + 4 i n parenrightBig 2 4. n summationdisplay i = 1 4 n parenleftBig 6 i + 5 i n parenrightBig 2 5. n summationdisplay i = 1 5 n parenleftBig 6 i + 4 i n parenrightBig 2 6. n summationdisplay i = 1 4 n parenleftBig 6 + 5 i n parenrightBig 2 correct Explanation: The terms are of the form 4 n parenleftBig 6 + 5 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i = 1 4 n parenleftBig 6 + 5 i n parenrightBig 2 . 002 10.0 points Estimate the area, A , under the graph of f ( x ) = 1 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. Correct answer: 1 . 283. 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 1 . 283 . 003 10.0 points Decide which of the following regions has area = lim n n summationdisplay i = 1 4 n sin i 4 n without evaluating the limit. 1. braceleftBig ( x, y ) : 0 y sin x, x 8 bracerightBig 2. braceleftBig ( x, y ) : 0 y sin 3 x, x 8 bracerightBig 3. braceleftBig ( x, y ) : 0 y sin 2 x, x 4 bracerightBig 4. braceleftBig ( x, y ) : 0 y sin 3 x, x 4 bracerightBig 5. braceleftBig ( x, y ) : 0 y sin x, x 4 bracerightBig correct 6. braceleftBig ( x, y ) : 0 y sin 2 x, x 8 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 choi (kc24547) HW03 BERG (56525) 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 x = ( b- a ) /n . When the area is given by A = lim n n summationdisplay i = 1 4 n sin i 4 n , therefore, we see that f ( x i ) = sin i 4 n , x = 4 n , where in this case x i = i 4 n , f ( x ) = sin x, [ a, b ] = bracketleftBig , 4 bracketrightBig . Consequently, the area is that of the region under the graph of y = sin x on the interval [0 , / 4]. In set-builder notation this is the region braceleftBig ( x, y ) : 0 y sin x, x 4 bracerightBig ....
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cal3 - choi (kc24547) HW03 BERG (56525) 1 This print-out...

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