Answer
The improper integral converges to $-1$.
Work Step by Step
We are given that $I=\int_{-\infty }^{0} e^{-x} \ dx$
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 $\int_{-\infty }^{0} e^{-x} \ dx=\lim\limits_{a \to -\infty}\int_{a}^{0} e^{-x} \ dx$
or, $= -\lim\limits_{a \to -\infty}[e^{-x}]_{a}^{0}$
or, $= -\lim\limits_{a \to -\infty} [e^{-0}-e^{-a}]$
or, $=-(1-0)$
or, $=-1$
Therefore, the given improper integral converges to $-1$.
