 Solution

Suppose an approximation has an error expansion $A(h)=I+c{h}^p+O(h^q)$ with $q>p$. With refinement factor $r>1$, combining the two approximations eliminates the leading term: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 $R_{k,j}=(4^j{R}_{k,j-1}-R_{k-1,j-1})/(4^j-1)$, 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 isThe leading power is two, so one extrapolation step givesThis 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.