Editing Riemann sum (section)

=== Midpoint rule ===
[[Image:MidRiemann2.svg|thumb|right|Middle Riemann sum of {{math|''x'' ↦ ''x''<sup>3</sup>}} over [0, 2] using 4 subintervals]]
For the midpoint rule, the function is approximated by its values at the midpoints of the subintervals. This gives {{math|''f''(''a'' + Δ''x''/2)}} for the first subinterval, {{math|''f''(''a'' + 3Δ''x''/2)}} for the next one, and so on until {{math|''f''(''b'' − Δ''x''/2)}}. Summing the resulting areas gives
<math display="block">S_\mathrm{mid} = \Delta x\left[f\left(a + \tfrac{\Delta x}{2}\right) + f\left(a + \tfrac{3\Delta x}{2}\right) + \dots + f \left(b - \tfrac{\Delta x}{2}\right)\right].</math>

The error of this formula will be
<math display="block">\left\vert\int_a^b f(x)\, dx - S_\mathrm{mid}\right\vert \le \frac{M_2(b - a)^3}{24n^2},</math>
where <math>M_2</math> is the maximum value of the [[absolute value]] of <math>f^{\prime\prime}(x)</math> over the interval. This error is half of that of the trapezoidal sum; as such the middle Riemann sum is the most accurate approach to the Riemann sum.
==== Generalized midpoint rule ====
A generalized midpoint rule formula, also known as the enhanced midpoint integration, is given by
<math display="block">\int_0^1 f(x)\,dx
= 2\sum_{m=1}^M {\sum_{n=0}^\infty {\frac{1}{{\left(2M\right)^{2n+1}}\left({2n+1}\right)!}{{\left. f^{(2n)}(x) \right|}_{x=\frac{m-1/2}{M}}}}}\,\,,</math>
where <math>f^{(2n)}</math> denotes even derivative.

For a function <math> g(t) </math> defined over interval <math> (a,b) </math>, its integral is
<math display="block">\int_a^b g(t) \, dt = \int_0^{b-a} g(\tau+a) \, d\tau= (b-a) \int_0^1 g((b-a)x+a) \, dx.</math>
Therefore, we can apply this generalized midpoint integration formula by assuming that <math> f(x) = (b-a) \, g((b-a)x+a) </math>. This formula is particularly efficient for the numerical integration when the integrand <math> f(x) </math> is a highly oscillating function.