Calculating distribution function exponential distribution

Question

I have a question about the following example of a continuous random variable:

If $X$ has the exponential distribution with parameter $\lambda$, then

$ F_X(x)= \begin{cases} 0& \text{if }x\leq 0,\\ 1-e^{-\lambda x}& \text{if }x>0, \end{cases}$

with density function

$f_x(x)= \begin{cases} 0& \text{if }x\leq 0\\ \lambda e^{-\lambda x}& \text{if }x>0. \end{cases}$

Now, my textbook provides us with the following definition of continuous random variables:

A random variable $X$ is continuous if its distribution function $F_X$ may be written in the form

$F_X(x)=\mathbb P(X\leq x)=\int_{-\infty}^{x}f_X(u)\text{d}u\quad \text{for } x\in \mathbb R$,

for some non-negative function $f_X$.

So, I tried to write out $F_X(x)$ for some value $x<0$. What we'd get is:

\begin{align} F_X(x)=\int_{-\infty}^{x}f_X(u)\text{d}u=&\int_{-\infty}^{0}f_X(u)\text{d}u + \int_{0}^{x}f_X(u)\text{d}u\\ =&\int_{-\infty}^{0}0\text{d}u+\lim_{a\to 0}\int_{a}^{x}\lambda e^{-\lambda x}\text{d}u\\ =&\int_{0}^{x}\lambda e^{-\lambda x}\text{d}u= \left[-e^{-\lambda x}\right]_{o}^{x}=-e^{-\lambda x}. \end{align}

But how can I get '1'? I understand this is an integration constant, but as far as I know, integration constants are only involved in indefinite integrals?

Answer 1

$$ \mathrm{e}^{-0} = 1 $$ or more specifically $$ \left[-\mathrm{e}^{-\lambda x} \right]_0^x=-\mathrm{e}^{-\lambda x} - (-1) = 1 - \mathrm{e}^{-\lambda x} $$

Answer 2

A side note: You have mixed up $x$ and $u$. $f_X(u)$ must be written in terms of $u$, because $u$ is the integration variable. $x$ is the upper bound.

$$\int_{0}^{x}f_X(u)\, \text{d}u=\int_{0}^{x}\lambda e^{-\lambda \color{red} u}\, \text{d}u$$ $$=\left[ -e^{-\lambda \color{red} u} \right]_0^x=-e^{-\lambda x}-(-e^{-\lambda \cdot 0})=-e^{-\lambda x}+1=1-e^{-\lambda x}$$