Answer
The improper integral converges to $2e$.
Work Step by Step
We are given that $I=\int_{-2}^{+\infty} e^{-0.5x} \ dx$
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 $\int_{-2}^{+\infty} e^{-0.5x} \ dx=\lim\limits_{a \to +\infty}\int_{-2}^{a} e^{-0.5x} \ dx$
or, $=-2 \lim\limits_{a \to +\infty}[e^{-0.5x}]_{-2}^{a}$
or, $=- 2 \lim\limits_{a \to +\infty}(e^{-a} -e)$
or, $=-2[0-e]$
or, $=2e$
Therefore, the given improper integral converges to $2e$.
