probability - Get CDF from given PDF

I have the following PFD:

$f_x(x) = \begin{cases} bx^2, & 0 < x < 2, \\ 5b-bx & 2 \le x < 3, \\ 0 & \text{otherwise}. \end{cases}$

I already calculated $b = \frac{6}{31}$.

However, I am unsure about the CDF:

$F_x(x) = \begin{cases} 0 & x \le 0, \\ 0 + \int_{0}^{x} bt^2 dt = bx^3 * \frac{1}{3} & 0 < x < 2, \\ 0 + \int_{0}^{2} bt^2dt + \int_{2}^{x}(5b-bt)dt = \frac{8}{3}b - \frac{b}{2}(x^2-10x+16) & 2 \le x < 3, \\ 1 & x \ge 3. \end{cases}$

Can this be true? I am unsure about how to get $F_x(x)$ for the interval $2\le x < 3$... A comment would be nice!

EDIT

$F_x(x)$ is correcty. To verify correctness $F_x(x)$ can be derivated piecewise. This should give $f_x(x)$ then.

Answer

I cannot comment since I'm a new user, but yes, that looks correct. To verify further you can try to get the PDF back by taking a piecewise derivative.