Answer
\[ - \frac{{32}}{3}\]
Work Step by Step
\[\begin{gathered}
\int_{ - 2}^2 {\left( {{x^2} - 4} \right)} dx \hfill \\
\hfill \\
{\text{Use }}\frac{1}{{{x^{1/2}}}} = {x^{ - 1/2}} \hfill \\
\hfill \\
\int_{ - 2}^2 {\left( {{x^2} - 4} \right)} dx \hfill \\
\hfill \\
{\text{Use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ }}} \hfill \\
\hfill \\
\left[ {\frac{{{x^3}}}{3} - 4x} \right]_{ - 2}^2 \hfill \\
\hfill \\
{\text{Fundamental Theorem of calculus}} \hfill \\
\hfill \\
\left[ {\frac{{{{\left( 2 \right)}^3}}}{3} - 4\left( 2 \right)} \right] - \left[ {\frac{{{{\left( { - 2} \right)}^3}}}{3} - 4\left( { - 2} \right)} \right] \hfill \\
\hfill \\
{\text{Simplify and subtract}} \hfill \\
\hfill \\
\left( {\frac{8}{3} - 8} \right) - \left( { - \frac{8}{3} + 8} \right) \hfill \\
\hfill \\
- \frac{{16}}{3} - \frac{{16}}{3} \hfill \\
\hfill \\
- \frac{{32}}{3} \hfill \\
\end{gathered} \]
