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SampleRiemannSum

# SampleRiemannSum - 1 and x 2 is(x 1 x 2 3 Find the height...

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Example of finding a Riemann Sum for finite “n” using a table rather than sigma notation. Use a Riemann Sum to approximate the definite integral dx. Use 3 rectangles of equal length and use the midpoints as sample points. 1. Determine the 3 subintervals which will be the bases of the 3 rectangles. Each rectangle will have length ∆x = . Here b = 11, a = 2 and n = 3. So we get ∆x = . The means our subintervals will need to have length 3. Note, since all of our rectangles are the same length, we used the notation ∆x for the lengths. If they were of different lengths, as we did in class today, we would use the notation ∆X i rather ∆x than so we could distinguish between them. 2. Find the sample point in each subinterval. Note: the sample points may be the left endpoints, the right endpoints, or the midpoints, or any other point, but in this example we are using midpoints. The midpoint of x
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Unformatted text preview: 1 and x 2 is (x 1 + x 2 ). 3. Find the height of each rectangle. Evaluate the sample point in f(x) to determine the height of the rectangle. 4. Find the area of each rectangle. Multiply the height of the rectangle by the length to get the area. 5. Use the areas of all the rectangles to approximate the area under the curve. Total the areas of all the rectangles to get an approximation of the area under the curve y = x 2 , over the interval [2, 11]. Our approximate area is 434.25. Subinterval Sample Pt X i * Height of Rectangle f(X i *) Length of Rectangle ∆x Area of Rectangle f(X i *)∆x [2 , 5 ] X 1 * = .5(2 + 5 ) X 1 * = 3.5 f(3.5) = (3.5) 2 3 (3.5) 2 (3) = 36.75 [5, 8] X 2 * = .5(5 + 8) X 2 * = 6.5 f(6.5) = (6.5) 2 3 (6.5) 2 (3) = 126.75 [8 11] X 3 * = .5(8 + 11) X 3 * = 9.5 f(9.5) = (9.5) 2 3 (9.5) 2 (3) =270.75 TOTAL 434.25...
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