Answer
\[ = \frac{{{e^9} - 1}}{3}\]
Work Step by Step
\[\begin{gathered}
\int_{ - 1}^2 {{x^2}{e^{{x^3} + 1}}dx} \hfill \\
\hfill \\
set\,\,u = {x^3} + 1\,\,\,{\text{which implies that}}\,\,\,du = 3{x^2}dx \hfill \\
\hfill \\
{\text{Changing limits of integration}} \hfill \\
\hfill \\
x = - 1\,\,\,implies\,\,u = 0 \hfill \\
x = 2\,\,\,\,\,implies\,\,u = 9 \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
= \int_0^9 {{e^u}du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \frac{1}{3}\,\,\left[ {{e^u}} \right]_0^9 \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \frac{1}{3}\,\left( {{e^9} - {e^0}} \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{{{e^9} - 1}}{3} \hfill \\
\end{gathered} \]
