Cumulative Distribution Function of a Variable with Exponential Distribution

Question

Suppose we have a variable $X$ that has an exponential distribution with a probability density function:

$f(x) = 3e^{-3x}, x >0$

Then the cumulative distribution function is:

$\int_{-\infty}^{x} f(x)dx = \int_{-\infty}^0 f(x)dx + \int_0^x f(x)dx$

$= \int_0^x f(x) dx \space\space\space$ Since x > 0

$= (-e^{-3x})|_0^x$

$= -e^{-3x} - (-e^0)$

$= 1 - e^{-3x}$

Is this correct, that part I'm not sure about is disregarding the $\int_{-\infty}^0 f(x)dx$ because $x > 0$.

Answer 1

Stating what happens when $x\le0$ in the pdf definition explicitly could make things clearer.

$$f(x) = \begin{cases} 3e^{-3x} &, x > 0 \\ 0&, x\le 0\end{cases}$$

$\int_{-\infty}^{x} f(t)dt = \int_{-\infty}^0 f(t)dt + \int_0^x f(t)dt$

$= \int_{-\infty}^0 0 dt + \int_0^x 3e^{-3t}dt$

$= (-e^{-3t})|_0^x$

$= -e^{-3x} - (-e^0)$

$= 1 - e^{-3x}$