A continuous random variable X has probability density function given
by
$$f(x) = \begin{cases}\alpha x^{2} &\mbox{if } 1\leqslant x \leqslant 3\\ 0 &\mbox{otherwise.} \end{cases}$$
i) Find the value of $\alpha$
ii) Find the cumulative distribution function $F(x) = P(X\leqslant x)$ for the random variable x.
iii) Hence, or otherwise, find $P(0\leqslant X \leqslant 2)$.
I have got no idea about how to find alpha.
all I know is $f(x)=$$\frac{d}{dx}F(x)$, even if I integrate the $\alpha x^{2}$ , I get $\frac{\alpha x^{3}} {3}$, still cannot find $\alpha$.
If part i) is found, so do part ii) and iii)
would someone explain in details please?
Thank you
Best Answer
$$\int\limits_{\color{red}1}^{\color{green}3}\alpha{x^2}=1\implies\frac{\alpha\cdot\color{green}3^3}{3}-\frac{\alpha\cdot\color{red}1^3}{3}=1\implies\alpha=\frac{3}{26}$$