= Solution
Suppose an approximation has an error expansion $A(h)=I+ch^p+O(h^q)$ with $q>p$. With refinement factor $r>1$, combining the two approximations eliminates the leading term:
$$
\boxed{R(h)=\frac{r^pA(h/r)-A(h)}{r^p-1}=I+O(h^q).}
$$
If further expansion terms are known, repeat the elimination at their successive powers. For an even-power trapezium expansion and repeated halving, this gives the Romberg recurrence $R_{k,j}=(4^jR_{k,j-1}-R_{k-1,j-1})/(4^j-1)$, starting from $R_{k,0}=T(2^{-k})$.
For the stated cubic the one-panel trapezium value is $T(1)=[0^3+1^3]/2=1/2$. The two-panel value is
$$
T(1/2)=\frac14\left[0+2\left(\frac12\right)^3+1\right]=\frac5{16}.
$$
The leading power is two, so one extrapolation step gives
$$
\boxed{\frac{4T(1/2)-T(1)}3=\frac{4(5/16)-1/2}3=\frac14.}
$$
This equals the exact integral $[x^4/4]_0^1=1/4$; for a cubic the trapezium error has only the leading $h^2$ term, which this combination removes exactly.
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