Answer
$\frac{1}{2}$
Work Step by Step
$$\eqalign{
& \int_0^\infty {{e^{ - 2x}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_0^\infty {{e^{ - 2x}}} dx = \mathop {\lim }\limits_{t \to \infty } \int_0^t {{e^{ - 2x}}} dx \cr
& {\text{Integrating}}{\text{, use }}\int {{e^{ - ax}}} dx = - \frac{1}{a}{e^{ - ax}} + C \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ { - \frac{1}{2}{e^{ - 2x}}} \right]_0^t \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ {{e^{ - 2x}}} \right]_0^t \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ {{e^{ - 2t}} - {e^0}} \right] \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ {{e^{ - 2t}} - 1} \right] \cr
& = - \frac{1}{2}\left[ {\mathop {\lim }\limits_{t \to \infty } {e^{ - 2t}}} \right] - \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ { - 1} \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& \cr
& = 0 - \frac{1}{2}\left[ { - 1} \right] \cr
& = \frac{1}{2} \cr
& {\text{Therefore}}{\text{, the integral converges}} \cr} $$
