Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {8{e^{ - 8x}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_0^\infty {8{e^{ - 8x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_0^b {8{e^{ - 8x}}} dx \cr
& {\text{write as}} \cr
& = - \mathop {\lim }\limits_{b \to \infty } \int_1^b {8{e^{ - 8x}}} \left( { - 8} \right)dx \cr
& {\text{integrating by using }}\int {{e^u}} du = {e^u} + C \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 8x}}} \right]_0^b \cr
& {\text{ using ftc}} \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 8b}} - {e^{ - 8\left( 0 \right)}}} \right] \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left( {{e^{ - 8b}} - 1} \right) \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = - \left( {{e^{ - \infty }} - 1} \right) \cr
& = - \left( {0 - 1} \right) \cr
& = 1 \cr} $$
