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.
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