Way to find CDF of a function with pdf similar to exponential distribution image link to the pdf of the function 
$$
f(x;\theta) = 
\begin{cases} 
  \frac{1}{4} e^{-\frac{1}{4} (x - \theta)} & \theta \lt x\\
   0 & \theta \gt x
\end{cases}
$$
How would you calculate the CDF of such a function? I am having trouble determining the domain after integrating the function. I know that, if $x$ must be bigger than $\theta$, then the CDF would be $0$ if $x$ smaller than theta. I got the integral to be $1-\exp[-(x-\theta)/4]$. Thank you in advance:)
 A: That is correct; $$F_X(x) = \begin{cases} 1 - e^{-(x-\theta)/4}, & x > \theta \\ 0, & x \le \theta. \end{cases}$$  You answered your own question.
A: Your way of phrasing your question makes me wonder if by "integrating the function" you mean just finding its antiderivatives. The cdf is
\begin{align}
F(x) & = \Pr(X\le x) = \int_{-\infty}^x f(t) \, dt \text{ where $f$ is the density function} \\[10pt]
& = \begin{cases}
\displaystyle\int_{-\infty}^x 0 \, dx & \text{if } x\le \theta, \\ \\
\displaystyle \left( \int_{-\infty}^\theta + \int_\theta^ x \right) f(t) \,dt = \int_{-\infty}^\theta 0\,dt + \int_\theta^x \frac 1 4 e^{-\frac 1 4 (t-\theta)} \, dx & \text{if }x>\theta,
\end{cases} \\[20pt]
& = \begin{cases} 0 & \text{if } x\le\theta, \\  \\
\displaystyle \int_\theta^x \frac 1 4 e^{-\frac 1 4 (t-\theta)} \, dt & \text{if } x>\theta.  \end{cases}
\end{align}
If $x>\theta$ then you have
$$
\int_\theta^x \frac 1 4 e^{-\frac 1 4 (t-\theta)} \,dt = \left. -e^{-\frac 1 4 (t-\theta)} \vphantom{\frac{\displaystyle\sum}1}\right|_{t\,:=\,\theta}^{t\,:=\,x} = -e^{-\frac 1 4 (x-\theta)} - (-1) = 1 - e^{-\frac 1 4(x-\theta)}.
$$
