Use Simpsons rule with n = 4 to estimate the integral $\int_0^2\sqrt{4-x^2}dx$. Notice that the integral gives you an approximation for ${\pi}$ and therefore demonstrate without evaluating this integral that it is indeed equal to $\pi$.
Getting equally-spaced points at 0, 0.5, 1, 1.5 and 2. Letting $f(x) = \sqrt{2-x^2}$. I get the Simpson expression as,
$$\frac{1}{6}(2 + 4 \sqrt{ 4-0.5^2} +2\sqrt{4-1^2}+4\sqrt{4-1.5^2}+0) = 3.0836$$
The problem is i'm not sure how to demonstrate that this integral is equal to ${\pi}$, without evaluating it. I'm thinking along the lines of rearranging the function so it is the equation of a circle.