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