Answer
$$ - 1$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 1} {\frac{2}{{{x^3}}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_{ - \infty }^{ - 1} {\frac{2}{{{x^3}}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {\frac{2}{{{x^3}}}} dx \cr
& = 2\mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {{x^{ - 3}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = 2\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{{{x^{ - 2}}}}{{ - 2}}} \right]_a^{ - 1} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left[ {{x^{ - 1}}} \right]_a^{ - 1} \cr
& {\text{property of integrals}} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{x}} \right]_1^a \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{a} - \frac{1}{1}} \right) \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{a} - 1} \right) \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{1}{{\left( { - \infty } \right)}} - 1 \cr
& = - 1 \cr} $$