Let $X$ be a continuous random variable with PDF,
$f_X(x) = \begin{cases} {ke^{-2x}} & \text{$x > 0$} \\ 0 & \text{otherwise}\end{cases}$
a) Find $k$.
b) Find the mean of $X$.
c) Find the variance of $X$.
c) Find the Cumulative Distribution Function (CDF) of $X$.
a) $k = 2$. No problems there.
b) I am getting $E(X) = \frac{1}{2}$.
c) This is where things get a little crazy. Is it me or is the integral required to find the variance a little complex?
$\int_{0}^\infty (x - \frac{1}{2})^2 \ (2e^{-2x}) \ dx$
I'm struggling to evaluate this! Is this the correct integral?
edit:: It all came to me and I solved it ($var(X) = \frac{1}{4}$). Currently working on part d)...
d)
$F(x) = \begin{cases} {0} & \text{$x \leq 0$} \\ {-e^{-2x} + 1} & \text{$x > 0$}\end{cases}$