Transformation of Variables (Finding The Distribution) Let $V \sim f(v) = Tv^2,$ such that $-2<v<1$ and $T$ is a constant. I want to know the distribution of $W=V^2$ and the value $T$. My attempt is that $V = \pm \sqrt{ W }.$ We know that
$$\int_{-2}^1  Kx^2 = 1  \quad \Rightarrow
3K  =1, \quad K = 1/3$$
Then $V\sim f(v) = \frac{1}{3}v^2, \ -2<v<1$ and $$F_V(v) = \int_{-2}^y \frac{1}{3} v^2 dy
 + \int_{y}^1 \frac{1}{3} v^2 dy = \frac{y^3+8}{9}+\frac{1-y^3}{9} = 1$$
Then,
\begin{align*}
F_V (W \le w) & = P(V \le \pm \sqrt{w})\\
& = P(-\sqrt{w} \le V \le \sqrt{w})\\
& = F_V(\sqrt{w}) - F_V(-\sqrt{w})\\
f_V(w) & = \frac12 f_V (w^{-1/2}) - \frac12 f_V (-w^{-1/2})\\
& = \frac12\left(\frac13(w^{-1}) \right) + \frac12\left(\frac13(w^{-1}) \right)\\
& \stackrel{?}{=}  \frac{1}{3}w
\end{align*}
Could that be wrong?
 A: The value of the constant is right. I continue later with your approach. First of all we need the cdf of $V$, since we work with inequalities. For this purpose we integrate the pdf from $-2$ (lower bound) to $v$:
$$\int_{-2}^v \frac13t^2 \ dt=\left[\frac19t^3\right]_{-2}^v=\frac{v^3}{9}+\frac{8}{9}$$
Thus the cdf is
$$F_V(v)=\begin{cases} 0, \  v\leq -2 \\ \frac{v^3}{9}+\frac{8}{9}, \  -2< v \leq 1 \\ 1, \  v>1  \end{cases}$$
Since the interval is not symmetric around 0 we make a case distinction:
First case: $v\in [-1,1) \Rightarrow w\in [0,1)$. Then we have $v=w^{1/2} $ for $v\in [0,1)$ and $v=-w^{1/2}$ for $v\in [-1,0)$
Second case: $v\in [-2,-1] \Rightarrow w\in [1,4]$
At the first case we can use your interval:
$$F_V(\sqrt{w}) - F_V(-\sqrt{w})=\left(\frac{w^{\frac32}}{9}+\frac{8}{9}\right)-\left(\frac{\left(-w^{1/2}\right)^3}{9}+\frac{8}{9}\right)$$
$$F_W(w)=\left(\frac{w^{\frac32}}{9}+\frac{8}{9}\right)-\left(\frac{-w^{3/2}}{9}+\frac{8}{9}\right)=\frac{w^{\frac32}}{9}+\frac{8}{9}+\frac{w^{\frac32}}{9}-\frac89=\frac{2}{9}w^{\frac32}$$
At the second case $V$ is negative. Thus we have
$P((-V)^2<w)=P(-V<w^{\frac12})=P(V>-w^{\frac12})=1-F\left(-w^{\frac12} \right)$
Inserting the value into the cdf results in
$$F_W(w)=1-\left(\frac{\left(-w^{\frac12}\right)^3}{9}+\frac{8}{9}\right)=1-\left(\frac{-w^{\frac32}}{9}+\frac{8}{9}\right)=\frac{w^{\frac32}}{9}+\frac19$$
