Answer
$$\frac{1}{5}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 5} {{x^{ - 2}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_{ - \infty }^{ - 5} {{x^{ - 2}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 5} {{x^{ - 2}}} dx \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 5} {{x^{ - 2}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right)_a^{ - 5} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)_a^{ - 5} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{x}} \right)_a^{ - 5} \cr
& {\text{property of integrals}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{x}} \right)_{ - 5}^a \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 388}}} \right) \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{a} - \frac{1}{{ - 5}}} \right) \cr
& {\text{evaluate the limit when }}b \to \infty \cr
& = \frac{1}{\infty } + \frac{1}{5} \cr
& = \frac{1}{5} \cr} $$
