Answer
The improper integral converges to $1$.
Work Step by Step
We are given that $I=\int_{0}^{+\infty} e^{-x} \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $\int_{0}^{+\infty} e^{-x} \ dx=\lim\limits_{a \to +\infty}\int_{0}^{a} e^{-x} \ dx$
or, $=\lim\limits_{a \to +\infty}[e^{-x}]_0^{a}$
or, $=- \lim\limits_{a \to +\infty}(e^{-a} -\dfrac{1}{e^0})$
or, $=-[0-1]$
or, $=1$
Therefore, the given improper integral converges to $1$.
