Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {{e^x}} dx \cr
& {\text{definition of improper integral}} \cr
& \int_{ - \infty }^0 {{e^x}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{e^x}} dx \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left. {\left( {{e^x}} \right)} \right|_a^0 \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {{e^0} - {e^a}} \right) \cr
& {\text{simplify}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {1 - {e^a}} \right) \cr
& {\text{evaluate the limit}} \cr
& = 1 - {e^{ - \infty }} \cr
& = 1 - 0 \cr
& = 1 \cr} $$
