We want to derive Simpson's rule for$$ \int_a^b f(x) \,dx.$$

We will use Newton and Cotes formulas for $n=2$ . In this case, $x_0=a$ , $x_2=b$ and $x_1 = (a+b)/2$ . We use Lagrange's interpolation formula to get a polynomial $p(x)$ such that $p(x_j)=f(x_j)$ for $j=0,1,2$ .

The corresponding interpolating polynomial is$$ p(x)=f(x_1)\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}+ f(x_2)\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)} +f(x_3)\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}.$$ and thus$$ \int_a^b f(x) \,dx\approx \int_a^b f(x_1)\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}+ f(x_2)\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)} +f(x_3)\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}\,dx.$$

Since integration is linear, we are concerned only with integrating each term in the sum. Now, taking $x_j = a + hj$ where $j=0,1,2$ and $h=|b-a|/2$ , we can rewrite the quotients on the last integral as$$ \int_a^b p(x)\, dx = hf(x_0)\int_0^2\frac{(t-1)(t-2)}{(0-1)(0-2)}\,dt + hf(x_1)\int_0^2\frac{(t-0)(t-2)}{(1-0)(1-2)}\,dt + hf(x_2)\int_0^2\frac{(t-0)(t-1)}{(2-0)(2-1)}\,dt.$$ and if we calculate the integrals on the last expression we get$$ \int_a^b p(x)\,dx=hf(x_0)\frac{1}{3} + hf(x_1)\frac{4}{3}+hf(x_2)\frac{1}{3},$$ which is Simpson's rule:$$ \int_a^b f(x)\,dx \approx \frac{h}{3}(f(x_0) + 4f(x_1) + f(x_2)).$$