Answer
\[ = \frac{{{e^{16}} - 1}}{4}\]
Work Step by Step
\[\begin{gathered}
\int_{ - 2}^2 {{e^{4x + 8}}dx} \hfill \\
\hfill \\
set\,\,u = 4x + 8{\text{ then}}\,{\text{ }}du = 4dx \hfill \\
\frac{1}{4}du = dx \hfill \\
\hfill \\
= \int_{}^{} {{e^{4x + 8}}\,} \,dx = \frac{1}{4}\int {{e^u}} du \hfill \\
integrate \hfill \\
= \frac{1}{4}{e^u} + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{ - 2}^2 {{e^{4x + 8}}dx} = \frac{1}{4}\,\,\left[ {{e^{4x + 8}}} \right]_{ - 2}^2 \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \frac{1}{4}\,\,\left[ {{e^{4\,\left( 2 \right) + 8}} - {e^{4\,\left( { - 2} \right) + 8}}} \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{1}{4}\,\left( {{e^{16}} - {e^0}} \right) \hfill \\
\hfill \\
= \frac{{{e^{16}} - 1}}{4} \hfill \\
\end{gathered} \]
