proof of Simpson’s rule

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:

$${\int }_a^{b}f(x)𝑑x\approx \frac{h}{3}(f(x_0)+4f(x_1)+f(x_2)).$$