Not too hard. I presume you know how to determine an interpolating polynomial?
The idea is to find the parabola $ax^2+bx+c$ passing through $x_{i-1}$, $\dfrac{x_{i-1}+x_i}{2}$, and $x_i$, and then integrate that over $[x_{i-1},x_i]$. You could, for instance, set up the (Vandermonde) system of linear equations
$\begin{pmatrix}1&x_{i-1}&x_{i-1}^2\\1&\frac{x_{i-1}+x_i}{2}&\left(\frac{x_{i-1}+x_i}{2}\right)^2\\1&x_i&x_i^2\end{pmatrix}\begin{pmatrix}c\\b\\a\end{pmatrix}=\begin{pmatrix}f(x_{i-1})\\f\left(\frac{x_{i-1}+x_i}{2}\right)\\f(x_i)\end{pmatrix}$
since you know the three points where the parabola to be integrated passes through. You should be getting something like
$\displaystyle p(x)=\frac{2\left(f\left(x_{i-1}\right)-2f\left(\frac{x_{i-1}+x_i}{2}\right)+f\left(x_i\right)\right)}{\left(x_i-x_{i-1}\right)^2}\left(x-\frac{x_{i-1}+x_i}{2}\right)^2+\frac{f\left(x_i\right)-f\left(x_{i-1}\right)}{x_i-x_{i-1}}\left(x-\frac{x_{i-1}+x_i}{2}\right)+f\left(\frac{x_{i-1}+x_i}{2}\right)$
and evaluating $\int_{x_{i-1}}^{x_i}p(u)\;\mathrm du$ in the usual way should yield Simpson's rule.
To show that a linear combination of the trapezoidal and midpoint rules give Simpson's rule, consider the trapezoidal and midpoint approximations
$\begin{align*} \text{trap}&=\int_{x_{i-1}}^{x_i}\left(f\left(x_{i-1}\right)+\frac{f\left(x_i\right)-f\left(x_{i-1}\right)}{x_i-x_{i-1}}\left(u-x_{i-1}\right)\right)\;\mathrm du\\ &=\frac{x_i-x_{i-1}}{2}\left(f(x_{i-1})+f(x_i)\right)\\ \text{mid}&=\int_{x_{i-1}}^{x_i}f\left(\frac{x_{i-1}+x_i}{2}\right)\;\mathrm du\\ &=(x_i-x_{i-1})f\left(\frac{x_{i-1}+x_i}{2}\right) \end{align*}$
and figure out appropriate values of $p$ and $q$ such that
$\frac{x_i-x_{i-1}}{3}\left(\frac{f(x_{i-1})}{2}+2f\left(\frac{x_{i-1}+x_i}{2}\right)+\frac{f(x_i)}{2}\right)=p\;\text{trap}+q\;\text{mid}$