The Simpson's rule for approximation of integral is based upon $$ \int_{a-h}^{a+h} f(x)dx \approx \frac {h}{3}(f(a-h)+4f(h)+f(a+h)).$$ The coefficients are usually found by integrating the quadratic function passing through the points $(a-h,f(a-h)) ,(a,f(a)),(a+h,f(a+h))$.
What if we try to find the coefficients by the requirement that the formula yields exact answer for $f(x)=1, f(x)=x,f(x)=x^2 $ on the interval $[-1,1]$ ?
We have \begin{align}\int _{-1}^{1} f(x) dx &= \alpha f(-1)+\beta f(0)+\gamma f(1)\\ f(x)&=1\implies \alpha +\beta+\gamma =2\\ f(x)&=x \implies \alpha=\gamma\\ f(x)&=x^2 \implies \alpha+\gamma =\frac {2}{3}\end{align} Solving for the coefficients we get $$\alpha =\gamma =1/3$$ and $$\beta =4/3$$ Note that we have $h=1$ so the coefficients match with the Simpson's rule.
your arguments are ok , Simpson even integrates x^3 and so all polynoms of grade <=3 exakt. so is there a question? trula