Simpson Rule special case of Gauss quadrature Prove that the Simpson Rule for integrating the function $f$ on the interval $[a,b]$: $$I_2(f)=\frac{b-a}{6}(f(a)+4f(\frac{a+b}{2})+f(b))$$ is actually the Gauss quadrature for the weight $g=1$ and $3$ nodes.
I can not get why when we have $3$ nodes in the Gauss quadrature the nodes have to be the ones from the formula above! Can anyone give me some hints on how to prove this?

Start from the definition of a Gauss quadrature:
$$\int_{-1}^1 f(x)\ dx \approx \sum_{i=1}^3 g_i f(x_i).$$
We perform a change of variables to change the range of integration, so
$$\int_a^b f(x)\ dx = \frac{b-a}{2}\int_{-1}^1 f\left(\frac{b-a}{2}z+\frac{b+a}{2}\right)\ dz.$$
Apply gaussian quadrature to the new integral on the right.
$$\frac{b-a}{2}\int_{-1}^1 f\left(\frac{b-a}{2}z+\frac{b+a}{2}\right)\ dz \approx \frac{b-a}{2}\sum_{i=1}^3 g_i f\left(\frac{b-a}{2}z_i+\frac{b+a}{2}\right).$$
With a three-point quadrature, our points are $x_1 = a, x_2 = \frac{a+b}{2}, x_3 = b$. These correspond to $z_1 = -1, z_2 = 0, z_3 = 1$.
Plugging these in, we find
$$\begin{align*}
\int_a^b f(x)\ dx \approx \frac{b-a}{2}& \left[g_1 f\left(\frac{b-a}{2}\cdot -1
+ \frac{a+b}{2}\right)\right. \\
&+\left.g_2 f\left(\frac{b-a}{2}\cdot 0
+ \frac{a+b}{2}\right)\right. \\
&+ \left.g_3 f\left(\frac{b-a}{2}\cdot 1
+ \frac{a+b}{2}\right)\right].
\end{align*}$$
The right-hand side simplifies to $$\frac{b-a}{2}\left[ g_1 f(a) + g_2 f\left(\frac{a+b}{2}\right) + g_3 f(b)\right].$$
Now you can see the function arguments are in the right form, and now all you have to do is compute the weights, which is straightforward since $g(x) = 1$. You can use $g_i = \int_\alpha^\beta g(x) \prod_{\substack{1 \le j \le 3 \\ j \neq i}} \frac{x-x_j}{x_i-x_j}\ dx$. But note that we've already done the change of variables to shift $[a,b]$ to $[-1,1]$ in the quadrature, so $\alpha = -1$ and $\beta = 1$. It is very straightforward to compute the integrals.
For $g_1$, we have
$$g_1 = \int_{-1}^1 \frac{x-0}{-1-0}\cdot \frac{x-1}{-1 - 1}\ dx = \frac12\left[\left.\frac{x^3}{3}\right|_{-1}^1 - \left.\frac{x^2}{2}\right|_{-1}^1\right] = \frac12\left[\frac13+\frac13-\left(\frac12-\frac12\right)\right] = \frac13.$$
You should find that $g_2$ and $g_3$ are similar. The result follows.
