Finding the CDF when the PDF is given and the median Let $X$ be the annual income to a randomly selected person in a population group. It is normal to assume that $X$ is Partero-distributed, that means that $X$ has the PDF of
$$
f_X(x)=\begin{cases}\theta k^\theta x^{-\theta -1}& \text{for} ~~x>k\\0 & \text{otherwise,}\end{cases}
$$
show that the CDF is given by
$$
F_X(x)=\begin{cases}1-k^\theta x^{-\theta} & \text{for} ~~x>k\\ 0 & \text{otherwise}\end{cases}  
$$
and find the median of the annual income
Me:
I know that 
$$
\int^x_{-\infty}{f_X(x)dx}=F_X(x)
$$ 
since $x$ is not defined under $k$ we get $\int_k^x{f(x)}dx$ if we integrate this with respect to $x$ and apply the limits we would get 
$$
F_X(x)=1-k^{\theta}x^{\theta} ~~\text{for}~~~ x>k
$$
now that that's done. I have to find the median. and this is where i'm stuck. Do I find the median of the CDF or PDF?
$$
F_X(x)=\frac{1}{2}\Longrightarrow \int^M_? 1-k^{\theta}x^{\theta}=\frac{1}{2}
$$
and what under limit am i supposed to use? 

The idea is
$$
\int_{-\infty}^m f_X(x){\rm d}x = \frac{1}{2} = F(m)
$$
So the problem becomes
$$
\frac{1}{2} = 1 - \left(\frac{k}{m}\right)^{\theta}
$$
and from there
$$
\frac{k}{m} = 2^{-1/\theta}
$$
So the median is
$$
m = 2^{1/\theta}k
$$
