Answer
$$\frac{1}{{12}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {5{e^{60x}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_{ - \infty }^0 {5{e^{60x}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {5{e^{60x}}} dx \cr
& {\text{write as}} \cr
& = \frac{5}{{60}}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{e^{60x}}\left( {60} \right)} dx \cr
& {\text{integrate by using }}\int {{e^u}} du \cr
& = \frac{1}{{12}}\mathop {\lim }\limits_{a \to - \infty } \left( {{e^{60x}}} \right)_a^0 \cr
& {\text{evaluate}} \cr
& = \frac{1}{{12}}\mathop {\lim }\limits_{a \to - \infty } \left( {{e^{60\left( 0 \right)}} - {e^{60a}}} \right) \cr
& = \frac{1}{{12}}\mathop {\lim }\limits_{a \to - \infty } \left( {1 - {e^{60a}}} \right) \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{1}{{12}}\left( {1 - {e^{ - \infty }}} \right) \cr
& = \frac{1}{{12}}\left( {1 - 0} \right) \cr
& = \frac{1}{{12}} \cr} $$
