Irvin Dukes

2021-12-12

What is midpoint Riemann sum?

Lindsey Gamble

I will assume that you know the general idea for a Riemann sum.
It is probably simplest to show an example:
For the interval: $\left[1,3\right]$ and for $n=4$
we find $\mathrm{△}x$ as always for Riemann sums:
$\mathrm{△}x=\frac{b-a}{n}=\frac{3-1}{4}=\frac{1}{2}$
Now the endpoints of the subintervals are:
$1,\frac{3}{2},2,\frac{5}{2},2$
The first four are left endpoint and the last four are right endpoints of subintervals.
The left Riemann sum uses the left endpoints to find the height of the rectangle. (And the right sum . . . )
The midpoint sum uses the midpoints of the subintervals:
$\left[1,\frac{3}{2}\right]\left[\frac{3}{2},2\right]\left[2,\frac{5}{2}\right]\left[\frac{5}{2},3\right]$
The midpoint of an interval is the average (mean) of the endpoints:
${m}_{1}=\frac{1}{2}\left(1+\frac{3}{2}\right)=\frac{5}{4}$
${m}_{2}=\frac{1}{2}\left(\frac{3}{2}+2\right)=\frac{7}{4}$
${m}_{3}=\frac{1}{2}\left(2+\frac{5}{2}\right)=\frac{9}{4}$
${m}_{4}=\frac{1}{2}\left(\frac{5}{2}+3\right)=\frac{11}{4}$
Now, whatever the function $f$ we get the sum:
$\mathrm{△}x\left(f\left({m}_{1}\right)+f\left({m}_{2}\right)+f\left({m}_{3}\right)+f\left({m}_{4}\right)\right)$
$=\frac{1}{2}\left(f\left(\frac{5}{4}\right)+f\left(\frac{7}{4}\right)+f\left(\frac{9}{4}\right)+f\left(\frac{11}{4}\right)\right)$

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