Composite trapezoidal error bound ($|T_n-I|\le \frac{(b-a{)}^3}{12n^2}{\max }_{[a,b]}|f^{\prime \prime }|$)

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.