If the probability density function of a random variable X is given by 𝑓(𝑥) = {2𝑘𝑥𝑒 − 𝑥2 , 𝑥 > 0 0, 𝑥 ≤ Determine (i)k (ii)distribution function.

$f(x)=\{\begin{array}{ll}2kx{e}^{-x^2}& x>0\\ 0& x\le 0\end{array}$

(a)

$\displaystyle\int_{-\infin}^{\infin}f(x)dx=1$

$\displaystyle\int_{0}^{\infin} 2kxe^{-x^2}dx=\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t} 2kxe^{-x^2}dx$

$=k\lim\limits_{t\to \infin}[-e^{-x^2}]\begin{matrix} t \\ 0 \end{matrix}=k=1$

$k=1$

$f(x)= \begin{cases} 2xe^{-x^2} &x>0\\ 0 &x\le0 \end{cases}$

(b)

$F(x)=\displaystyle\int_{-\infin}^{x}f(y)dy$

If $x\le0, F(x)=0.$

If $x>0$

$F(x)=\displaystyle\int_{0}^{x}2ye^{-y^2} dy=[-e^{-y^2}]\begin{matrix} x \\ 0 \end{matrix}=1-e^{-x^2}$

$F(x)= \begin{cases} 0 &x\le0\\ 1-e^{-x^2} &x>0 \end{cases}$