Answer
\[ - \frac{5}{2}\]
Work Step by Step
\[\begin{gathered}
\int_{1/2}^1 {\left( {{x^{ - 3}} - 8} \right)} dx \hfill \\
\hfill \\
{\text{Integrate using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= \left[ {\frac{{{x^{ - 2}}}}{{ - 2}} - 8x} \right]_{1/2}^1 \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
= \left[ { - \frac{1}{{2{x^2}}} - 8x} \right]_{1/2}^1 \hfill \\
\hfill \\
{\text{Use the fundamental theorem of calculus}} \hfill \\
\hfill \\
= \left[ { - \frac{1}{{2{{\left( 1 \right)}^2}}} - 8\left( 1 \right)} \right] - \left[ { - \frac{1}{{2{{\left( {1/2} \right)}^2}}} - 8\left( {1/2} \right)} \right] \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
\left( { - \frac{{17}}{2}} \right) - \left( { - 6} \right) \hfill \\
\hfill \\
- \frac{5}{2} \hfill \\
\hfill \\
\end{gathered} \]
