Official solution
\[I = \int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}dx} \]
Here -1 \[ \le \] x \[ \le \] 1 then -1 \[ \le \] x3 \[ \le \] 1
Integer between -1 and 1 is 0. So integration will be divided into two parts, -1 to 0 and 0 to 1.
\[ = \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}dx} + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}dx} \]
\[ = \int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}dx} + \int\limits_0^1 {{x^2}{e^0}dx} \]
= \[{1 \over e} \times \left[ {{{{x^3}} \over 3}} \right]_{ - 1}^0 + \left[ {{{{x^3}} \over 3}} \right]_0^1\]
\[ = {1 \over e} \times \left( {0 - \left( {{{ - 1} \over 3}} \right)} \right) + {1 \over 3}\]
\[ = {1 \over {3e}} + {1 \over 3} = {{1 + e} \over {3e}}\]