Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 2} {\frac{1}{{{x^2}}}} dx \cr
& {\text{Using the definition of improper integrals see page 1053}} \cr
& \underbrace {\int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{M \to - \infty } } \int_M^b {f\left( x \right)} dx}_ \Downarrow \cr
& \int_{ - \infty }^{ - 2} {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{M \to - \infty } \int_M^{ - 2} {\frac{1}{{{x^2}}}} dx \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{M \to - \infty } \left[ { - \frac{1}{x}} \right]_M^{ - 2} \cr
& = - \mathop {\lim }\limits_{M \to - \infty } \left[ {\frac{1}{x}} \right]_M^{ - 2} \cr
& = - \mathop {\lim }\limits_{M \to - \infty } \left[ { - \frac{1}{2} - \frac{1}{M}} \right] \cr
& {\text{Evaluate the limit when }}M \to - \infty \cr
& = - \left[ { - \frac{1}{2} - \frac{1}{{ - \infty }}} \right] \cr
& = - \left[ { - \frac{1}{2} - 0} \right] \cr
& = \frac{1}{2} \cr} $$
