The RiemannSum(f(x), x = a..b, method = midpoint, opts) command calculates the midpoint Riemann sum of f(x) from a to b. The first two arguments (function expression and range) can be replaced by a definite integral.

If the independent variable can be uniquely determined from the expression, the parameter x need not be included in the calling sequence.

Given a partition $P=\left(a=x_0,x_1,...,x_N=b\right)$ of the interval $\left(a\,b\right)$, the midpoint Riemann sum is defined as:

where the chosen point of each subinterval $\left(x_{i-1}\,x_i\right)$ of the partition is the midpoint $\frac{\left(x_{i-1}+x_i\right)}{2}$.

By default, the interval is divided into $10$ equal-sized subintervals.

For the options opts, see the RiemannSum help page.

This integration method can be applied interactively, through the ApproximateInt Tutor.

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Calculus1}\right]\right)\:$

$\mathrm{RiemannSum}\left(\sin \left(x\right)\,x=0...5.0\,\mathrm{method}=\mathrm{midpoint}\right)$

$0.7238544375$

$\mathrm{RiemannSum}\left(x\left(x-2\right)\left(x-3\right)\,x=0..5\,\mathrm{method}=\mathrm{midpoint}\,\mathrm{output}=\mathrm{plot}\right)$

$\mathrm{RiemannSum}\left(\tan \left(x\right)-2x\,-1..1\,\mathrm{method}=\mathrm{midpoint}\,\mathrm{output}=\mathrm{plot}\,\mathrm{partition}=20\,\mathrm{boxoptions}=\left[\mathrm{filled}=\left[\mathrm{color}=''Burgundy''\right]\right]\right)$

$\mathrm{exact}≔\mathrm{int}\left(\ln \left(x\right)\,x=1..100\right)$

$\mathrm{exact}≔-99+200\ln \left(2\right)+200\ln \left(5\right)$

$\mathrm{evalf}\left(\mathrm{exact}\right)$

$361.5170185$

$\mathrm{RiemannSum}\left(\ln \left(x\right)\,x=1..100\,\mathrm{method}=\mathrm{midpoint}\,\mathrm{outline}=\mathrm{true}\,\mathrm{output}=\mathrm{animation}\right)$

Left Riemann Sum

Lower Riemann Sum

Right Riemann Sum

Upper Riemann Sum