Answer
$$\frac{3}{5}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 1} {{x^{ - 8/3}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_{ - \infty }^{ - 1} {{x^{ - 8/3}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {{x^{ - 8/3}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{{{x^{ - 8/3 + 1}}}}{{ - 8/3 + 1}}} \right]_a^{ - 1} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{{{x^{ - 5/3}}}}{{ - 5/3}}} \right]_a^{ - 1} \cr
& = - \frac{3}{5}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{x^{5/3}}}}} \right]_a^{ - 1} \cr
& {\text{property of integrals}} \cr
& = \frac{3}{5}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{x^{5/3}}}}} \right]_{ - 1}^a \cr
& {\text{evaluate}} \cr
& = \frac{3}{5}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{a^{5/3}}}} - \frac{1}{{{{\left( { - 1} \right)}^{5/3}}}}} \right] \cr
& = \frac{3}{5}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{a^{5/3}}}} + 1} \right] \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{3}{5}\left[ {\frac{1}{{{{\left( { - \infty } \right)}^{5/3}}}} + 1} \right] \cr
& = \frac{3}{5} \cr} $$
