Variance of a Uniform Distribution

Compute $E[X^2]$ by integrating $x^2$ against the constant density.

Use $\text{Var}(X)=E[X^2]-(E[X]{)}^2$.

1. 1. Compute the second moment

   $E[X^2]={\int }_a^b{x}^2\cdot \frac{1}{b-a}dx=\frac{a^2+ab+b^2}{3}$.

2. 2. Subtract the squared mean

   Using $E[X]=\frac{a+b}{2}$, the variance simplifies to a familiar closed form.

   $$\text{Var}(X)=\frac{a^2+ab+b^2}{3}-\frac{(a+b{)}^2}{4}=\frac{(b-a{)}^2}{12}$$

Answer: (b-a)^2/12