Answer
$$\frac{1}{3}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {{e^{3x}}} dx \cr
& \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_{ - \infty }^b {f\left( x \right)} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \int_{ - \infty }^0 {{e^{3x}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{e^{3x}}dx} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{e^{3x}}dx} \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{e^{3x}}\left( 3 \right)dx} \cr
& \cr
& {\text{Integrate}} \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left[ {{e^{3x}}} \right]_a^0 \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left[ {{e^{3\left( 0 \right)}} - {e^{3a}}} \right] \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left( {1 - {e^{3a}}} \right) \cr
& \cr
& {\text{calculate the limit when }}a \to - \infty \cr
& = \frac{1}{3}\left( {1 - {e^{3\left( { - \infty } \right)}}} \right) \cr
& = \frac{1}{3}\left( {1 - 0} \right) \cr
& = \frac{1}{3} \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\frac{1}{3} \cr} $$
