HW 03 solution

# HW 03 solution - gutierrez(ig3472 – HW03 – Neitzke...

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Unformatted text preview: gutierrez (ig3472) – HW03 – Neitzke – (56585) 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 2 + 3 n parenrightBig 2 + 4 n parenleftBig 2 + 6 n parenrightBig 2 + . . . + 4 n parenleftBig 2 + 3 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i = 1 4 n parenleftBig 2 i + 3 i n parenrightBig 2 2. n summationdisplay i = 1 3 i n parenleftBig 2 + 4 i n parenrightBig 2 3. n summationdisplay i = 1 4 i n parenleftBig 2 + 3 i n parenrightBig 2 4. n summationdisplay i = 1 4 n parenleftBig 2 + 3 i n parenrightBig 2 correct 5. n summationdisplay i = 1 3 n parenleftBig 2 i + 4 i n parenrightBig 2 6. n summationdisplay i = 1 3 n parenleftBig 2 + 4 i n parenrightBig 2 Explanation: The terms are of the form 4 n parenleftBig 2 + 3 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i = 1 4 n parenleftBig 2 + 3 i n parenrightBig 2 . 002 10.0 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 ≈ braceleftBig f (2) + f (3) + f (4) + f (5) bracerightBig 1 since x i = x ∗ i = i + 1. Consequently, A ≈ 5 . 133 . 003 10.0 points Decide which of the following regions has area = lim n →∞ n summationdisplay i = 1 π 3 n sin iπ 3 n without evaluating the limit. 1. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 2 x, ≤ x ≤ π 3 bracerightBig 2. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 3 x, ≤ x ≤ π 6 bracerightBig 3. braceleftBig ( x, y ) : 0 ≤ y ≤ sin x, ≤ x ≤ π 3 bracerightBig correct 4. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 2 x, ≤ x ≤ π 6 bracerightBig 5. braceleftBig ( x, y ) : 0 ≤ y ≤ sin 3 x, ≤ x ≤ π 3 bracerightBig 6. braceleftBig ( x, y ) : 0 ≤ y ≤ sin x, ≤ x ≤ π 6 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 gutierrez (ig3472) – HW03 – Neitzke – (56585) 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 3 n sin iπ 3 n , therefore, we see that f ( x i ) = sin iπ 3 n , Δ x = π 3 n , where in this case x i = iπ 3 n , f ( x ) = sin x, [ a, b ] = bracketleftBig , π 3 bracketrightBig . Consequently, the area is that of the region under the graph of y = sin x on the interval [0 , π/ 3]. In set-builder notation this is the region braceleftBig ( x, y ) : 0 ≤ y ≤ sin x, ≤ x ≤ π 3 bracerightBig ....
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HW 03 solution - gutierrez(ig3472 – HW03 – Neitzke...

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