= Solution
{c}
<Richardson extrapolation> combines approximations with a known leading error power to cancel that error. If $A(h)=I+ch^p+O(h^{p+q})$, then $(2^pA(h)-A(2h))/(2^p-1)$ removes $ch^p$.
Let $x_j=a+jh$ for an even number $n$ of equal subintervals, with $x_n=b$. The <composite trapezoidal rule> is
$$
T_h=h\left[\frac{f_0+f_n}{2}+\sum_{j=1}^{n-1}f_j\right].
$$
The <Simpson's rule> is
$$
S_h=\frac h3\left[f_0+f_n+
4\sum_{\substack{1\leq j<n\\j\ {\rm odd}}}f_j+
2\sum_{\substack{2\leq j<n\\j\ {\rm even}}}f_j\right].
$$
Assume an error expansion $T_h=I+c_2h^2+O(h^4)$, 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 $T_{2h}=h[f_0+f_n+2\sum_{\rm even}f_j]$. Direct coefficient collection gives
$$
\boxed{\frac{4T_h-T_{2h}}3=S_h.}
$$
The 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.
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