 Composite midpoint quadrature rule

Apply the midpoint quadrature rule on each of $n$ equal panels of a partition of an interval, with $h=(b-a)/n$:Each quadrature node is a panel midpoint and each quadrature weight is $h$. For $f\in C^2[a,b]$, put $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$. Summing the panel quadrature errors givesfor some $\xi \in (a,b)$. Each panel contributes at most $K{h}^3/24$; there are $n=(b-a)/h$ panels, explaining the composite midpoint rule error of order $O(h^2)$ in big O notation. On a nonuniform partition of an interval with panel widths $h_j$ and largest width $H$, the corresponding bound is $K{\sum }_j{h}_j^3/24\le K(b-a)H^2/24$.

For $f(x)=x^2$ on $[0,1]$, summing the sampled polynomial exactly yieldsDoubling $n$ divides the absolute error by exactly four. For a general $C^2$ function, halving $h$ divides the displayed error bound by four, but the actual absolute error need not follow that ratio: cancellation can make it smaller or zero. Continuous functions still give convergence as the largest panel width tends to zero, by uniform continuity; without the bounded second derivative, the second-order rate is not guaranteed. Noise sensitivity of a quadrature rule explains a separate limitation when samples are inexact.