Answer
$$\frac{6}{e}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {6{e^{ - x}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_1^\infty {6{e^{ - x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {6{e^{ - x}}} dx \cr
& = - 6\mathop {\lim }\limits_{b \to \infty } \int_1^b {{e^{ - x}}} \left( { - 1} \right)dx \cr
& {\text{integrating by using }}\int {{e^u}} du = {e^u} + C \cr
& = - 6\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - x}}} \right]_1^b \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 388}}} \right) \cr
& = - 6\mathop {\lim }\limits_{b \to \infty } \left( {{e^{ - b}} - {e^{ - 1}}} \right) \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = - 6\left( {{e^{ - \infty }} - {e^{ - 1}}} \right) \cr
& = - 6\left( {0 - {e^{ - 1}}} \right) \cr
& = 6{e^{ - 1}} \cr
& = \frac{6}{e} \cr} $$
