I'm new to stats and I facing problems in finding the median of a PDF. 

I have to find the median of this PDF

$$ f(x) =
\begin{cases}
cx^2, & \text{if 0 $\le$ $x$ $\le$ 3} \\
0, & \text{otherwise}
\end{cases}$$

**How I'm trying to solve it**

*First I found the value o $c$*

$$\int\_0^3 cx^2 dx = 1 \implies c = \frac{1}{26}$$

*Second*

Given that the median is a number $m$ such that $P(X\le m) =\frac{1}{2}$ I did:

$$\int\_0^m\frac{1}{26}x^2dx = \frac{1}{2}$$

I have two questions concerning my attempt:

1 - What I'm doing is correct? As I said, I'm new to stats and I'm sure if my approach is the correct one to find the median of the PDF

2 - How can I can integrate from zero to $m$? I have $x$ as a parameter in my integral and $m$ as its upper bond. I'm confused in how to solve that.