Prove variance in Uniform distribution (continuous) I read in wikipedia article, variance is $\frac{1}{12}(b-a)^2$ , can anyone prove or show how can I derive this?

By the definition of the variance, $\operatorname{Var} X = \mathbb{E}[X^2] - (\mathbb{E} X)^2$.
Since here $\mathbb{E} X = \frac{1}{b-a}\int_{[a,b]}x dx = \frac{a+b}{2}$, and $\mathbb{E} X^2 = \frac{1}{b-a}\int_{[a,b]}x^2 dx = \frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3}$, it follows that
$$
\operatorname{Var} X = \frac{a^2+ab+b^2}{3} - \frac{a^2+2ab+b^2}{4} = \frac{a^2-2ab+b^2}{12} =\frac{(b-a)^2}{12} 
$$
