The problem is $f(x)=1/3$ the range of $X$ is $(-1,2)$ zero elsewhere find the pdf of $Y=X^2$.
I tried using the pdf of $Y$ using transformation formula (not cumulative method)
and about the setting range... jacobian part $X=+-\sqrt {y}$ so divided it to part when for $-1< x<0$ and $0< x<2$ but when it becomes about $y$, there are $2$ overlapped range part for $-1< x<0,0< x<1$ both becomes $ 0< y<1$,
so i calculate is for when $-1< x<0$ and
multiple $2$ for the $0< x<1$ part... is it correct logic?
and i continued to finish the question the answer becomes $1/3\sqrt {y }, 0< y<1$ and $1/6\sqrt {y} ,1< y<4$ and $0$ elsewhere.
Is it correct?
and one more... how can i solve this with a cdf method?