probability density function for a random variable For a given probability density function $f(x)$, how do I find out the probability density function for say, 
$Y = x^2$?
$$f(x)=\begin{cases}cx&,0<x<2\\2c&,2<x<5\\0&,\text{otherwise}\end{cases}$$

Let's call the probability density for the random variables $X$ and $Y$ as $f_X$ and $f_Y$, respectively.
$f_Y$ has the property that $Pr[a\le Y \le b] = \int_a^b f_Y(y) dy$.
You want to know what this $f_Y$ is using $f_X$.
Since you want $Y = X^2$, note that
$Pr[a\le Y \le b] = Pr[a\le X^2 \le b]$
Let's assume that $a$ and $b$ are both positive (Q: Why is it okay to do this?), so 
$\displaystyle Pr[a\le X^2 \le b] = Pr[\sqrt{a} \le X \le \sqrt{b}] = \int_{\sqrt{a}}^{\sqrt{b}} f_X(x) dx$. 
(Question: Why am I only taking the positive square root?)
But this isn't quite what you want. You want the integral in terms of $y$. And also, the bounds should be from $a$ to $b$, but this will be fixed once we perform a change of variables.
Use the fact that $y = x^2$. Then, $\sqrt{y} = x$, so
$$Pr[a\le Y \le b] = \int_{a}^{b} \dfrac{f_X(\sqrt{y})}{2\sqrt{y}} dy$$
So $f_Y(y) = \dfrac{f_X(\sqrt{y})}{2\sqrt{y}}$
