= Composite trapezoidal error bound
{title2=$|T_n-I|\leq\frac{(b-a)^3}{12n^2}\max_{[a,b]}|f''|$}
If $f$ is twice continuously differentiable on $[a,b]$, take a uniform bound $|f''|\leq M$. On one panel $[u,v]$, two <integrations by parts> give the signed <quadrature error>
$$
\frac{v-u}{2}(f(u)+f(v))-\int_u^v f(t)\,dt=\frac12\int_u^v(t-u)(v-t)f''(t)\,dt.
$$
The 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.
Back to article page
