Composite trapezoidal rule ($T_n(f)=h(f(a)/2+{\sum }_{j=1}^{n-1}f(a+jh)+f(b)/2)$)

On an equal-panel partition of an interval, with $h=(b-a)/n$, sum the trapezoidal rule over adjacent panels:There are $n+1$ distinct quadrature nodes. Each interior sample belongs to two panels and therefore has quadrature weight $h$, while each endpoint has quadrature weight $h/2$. Equivalently, the composite trapezoidal rule integrates the piecewise linear function obtained by linear interpolation exactly.^[ref]

For $f\in C^2[a,b]$ and $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$, summing panel quadrature errors givesfor some $\xi \in (a,b)$. Thus panel quadrature errors of order $O(h^3)$ accumulate to order $O(h^2)$, using big O notation.^[ref] On a nonuniform partition of an interval with widths $h_j$ and largest width $H$, the bound is $K{\sum }_j{h}_j^3/12\le K(b-a)H^2/12$.

For the polynomial $f(x)=x^2$ on $[0,1]$, direct summation yieldsBoth quadrature errors shrink by exactly four when $n$ doubles, with opposite signs. This exact relation belongs to this polynomial example; the trapezoidal rule is not always twice as inaccurate as the midpoint quadrature rule.

The left plot uses exact polynomial sums; the right adds a common sample bias $\delta =0.002$ to the composite trapezoidal rule. That mathematical model converges to $1/3+\delta$, showing why smaller panels alone do not ensure accuracy for noisy samples. It illustrates noise sensitivity of a quadrature rule, rather than measured data.

On a partition of an interval symmetric about zero, the composite trapezoidal rule gives opposite quadrature nodes equal quadrature weights. Samples of an odd function cancel pairwise, and a zero quadrature node contributes zero. For a continuous odd function, its exact integral over $[-L,L]$ is also zero. Adding any affine function preserves exactness by linearity of the integral and of the quadrature rule. For example, $1+\sin x$ is nonpolynomial but both the exact integral and every uniform-grid composite trapezoidal rule value on $[-1,1]$ are two.

On a uniform partition of $[a,b]$ with width $h$, the right Riemann sum and the composite trapezoidal rule have equal weights at every interior mesh point. They differ only by adding half a final-endpoint weight and removing half an initial-endpoint weight. This proves the displayed identity. For continuous $f$, both converge to its definite integral; their finite endpoint correction need not be confused with the smaller trapezoidal discretization error for a smooth integrand.

If $f$ is twice continuously differentiable on $[a,b]$, take a uniform bound $|f^{\prime \prime }|\le M$. On one panel $[u,v]$, two integrations by parts give the signed quadrature errorThe nonnegative kernel has integral $(v-u{)}^3/6$, so the absolute error is at most $M(v-u{)}^3/12$. Summing $n$ equal panels proves the displayed bound. The maximum exists on the closed interval; a supremum or any known uniform bound also suffices. The estimate certifies a sufficient panel count, which need not be the least count achieving a particular actual error.