 Solution

Richardson extrapolation combines approximations with a known leading error power to cancel that error. If $A(h)=I+c{h}^p+O(h^{p+q})$, then $(2^{p}A(h)-A(2h))/(2^p-1)$ removes $c{h}^p$.

Let $x_j=a+jh$ for an even number $n$ of equal subintervals, with $x_n=b$. The composite trapezoidal rule isThe Simpson's rule isAssume an error expansion $T_h=I+c_2{h}^2+O(h^4)$, with the same coefficient $c_2$ on refined uniform meshes; $f\in C^4([a,b])$ is sufficient for this form. The coarser rule is $T_{2h}=h[f_0+f_n+2{\sum }_{\mathrm{even}}{f}_j]$. Direct coefficient collection givesThe leading $h^2$ errors cancel, leaving $S_h=I+O(h^4)$. The identity of the numerical integration weights is exact; the error-order conclusion additionally uses the smoothness and even-power error assumptions.