There is this note about Quadratic Interpolation by Simpsons Rule that I don't quite understand how to get the LHS.
Simpsons Rule by the Method of Undetermined Coefficients
We seek an approximation $$I\approx A_0f(-h)+A_1 f(0)+A_2 f(h)$$
Indeed, one expects the coefficients to be proportional to the length of the interval, but not to change if the $x$-variable is simply translated. So it is reasonable to guess that $A_i=ha_i$ gives true constants $a_i$. Anyway, this will be confirmed in the calculation.
Exactness for quadratics means exactness for $f(x)=1,x,x^2$ and inserting each of these in turn gives $$2h=h(a_0+a_1+a_2)$$ $$0=h^2(-a_0+a_2)$$ $$\frac{2}{3}h^3=h^3(a_0+a_2)$$
The solution is $a_0=a_2=1/3, a_1=1/4$.
My questions are:
1. Why is the coefficients are proportional to the length of the interval?
2. How do they get the LHS: $2h,0,\frac{2}{3}h^3$?
Thanks for the help!
For 1:
Letting $x = hy$, so $dx = h dy$, $\int_{-h}^h f(x) dx =\int_{-1}^1 f(hy) hdy =h\int_{-1}^1 f(hy) dy $.
For 2:
$\int_{-h}^h x^k dx =\dfrac{x^{k+1}}{k+1}\big|_{-h}^h =\dfrac{h^{k+1}-(-h)^{k+1}}{k+1} =(2h, 0, (2h^3)/3, 0) \text{ for } k=(0, 1, 2, 3) $.
Correspondingly,
$\begin{array}\\ a_0(-h)^k+a_0(0^k)+a_2h^k &=a_0+a_1+a_2 \text{ for } k=0,\\ &=-a_0h+a_2h =h(a_2-a_0)\text{ for } k=1,\\ &=a_0h^2+a_2h^2 =h^2(a_0+a_2) \text{ for } k=2,\\ \end{array} $
But these need to be multiplied by $h$ as shown in the answer to 1.