Answer
$$\frac{1}{{10}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 10} {{x^{ - 2}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_{ - \infty }^{ - 10} {{x^{ - 2}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 10} {{x^{ - 2}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{{{x^{ - 1}}}}{{ - 1}}} \right]_a^{ - 10} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{x}} \right]_a^{ - 10} \cr
& {\text{property of integrals}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{x}} \right]_{ - 10}^a \cr
& {\text{evaluate}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{a} - \frac{1}{{ - 10}}} \right) \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{a} + \frac{1}{{10}}} \right) \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{1}{{ - \infty }} + \frac{1}{{10}} \cr
& = \frac{1}{{10}} \cr} $$
