I am interested in deriving the distribution for $Y=|X|$ such that $f_X(x) = T |x|,$ $-1< x<2$, $T$ is a constant. I started by finding $T$ and deriving the CDF of $Y$ as follows:
$$\int_{-1}^2 T |x| dx = 2T + \frac{T}{2} =1, \quad \Rightarrow T = \frac25.$$
\begin{align*} F_Y (Y \le y) = P(|X| \le y) = P(-y \le X \le y) = F_X(y) - F_X(-y) \end{align*}
\begin{align*} F_X(x) = \int_{-1}^x \frac25 |t| dt = \frac{x|x| + 1}{5} = \begin{cases} \frac{x^2 + 1}{5}, & -1 \le x \le 1 \\\\ \frac{-x^2 + 1}{5}, & -1 < x \le 1 \end{cases}\\ \end{align*}
$$F_X(y) - F_X(-y) = \left(\frac{y|y| + 1}{5} \right) + \left( \frac{y|y| + 1}{5} \right) = \frac{2}{5}y|y| + 1.$$
I don't trust my work. I need the correct solution. This is self-study question.