Answer
$$\frac{1}{{64}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 4} {\frac{3}{{{x^4}}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_{ - \infty }^{ - 4} {\frac{3}{{{x^4}}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 4} {\frac{3}{{{x^4}}}} dx \cr
& = 3\mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 4} {{x^{ - 4}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = 3\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{{{x^{ - 3}}}}{{ - 3}}} \right]_a^{ - 4} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left[ {{x^{ - 3}}} \right]_a^{ - 4} \cr
& {\text{property of integrals}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{x^3}}}} \right]_{ - 4}^a \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{{{a^3}}} - \frac{1}{{{{\left( { - 4} \right)}^3}}}} \right) \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{{{a^3}}} + \frac{1}{{64}}} \right) \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{1}{{{{\left( { - \infty } \right)}^3}}} + \frac{1}{{64}} \cr
& = \frac{1}{{64}} \cr} $$
