Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^{ + \infty } {{e^{ - x}}} dx \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_a^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\,\,\int_0^{ + \infty } {{e^{ - x}}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_0^b {{e^{ - x}}} dx \cr
& {\text{Integrate}} \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ { - {e^{ - x}}} \right]_0^b \cr
& = - \mathop {\lim }\limits_{b \to + \infty } \left[ {{e^{ - x}}} \right]_0^b \cr
& = - \mathop {\lim }\limits_{b \to + \infty } \left( {{e^{ - b}} - {e^0}} \right) \cr
& = - \mathop {\lim }\limits_{b \to + \infty } \left( {{e^{ - b}} - 1} \right) \cr
& \cr
& {\text{calculate the limit when }}b \to + \infty \cr
& = - \left( {{e^{ - \infty }} - 1} \right) \cr
& {\text{simplifying}} \cr
& = - \left( {0 - 1} \right) \cr
& = 1 \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}1 \cr} $$
