Paragraph

Then for the meanĀ 1
\begin{align*} \mu & = \int_{-1}^1 x \cdot \frac{3}{4} \cdot (1-x^2) dx\\ & = \int_{-1}^1 \frac{3}{4} \cdot (x-x^3) dx\\ & = \frac{3}{4} \cdot (x^2/2-x^4/4) \big |_{-1}^1\\ & = \frac{3}{4} \cdot [(1/2)-(1/4)] - [(1/2) - (1/4)]\\ & = 0 \end{align*}
as expected since the probability function is symmetric about x=0.
in-context