Question:
The value of $\int_{-1}^{1} \mathrm{x}^{2} \mathrm{e}^{\left[\mathrm{x}^{3}\right]} \mathrm{dx}$, where $[\mathrm{t}]$ denotes the greatest integer $\leq \mathrm{t}$, is :
Correct Option: , 3
Solution:
$I=\int_{-1}^{0} x^{2} \cdot e^{-1} d x+\int_{0}^{1} x^{2} d x$
$\therefore I=\left.\frac{x^{3}}{3 e}\right|_{-1} ^{0}+\left.\frac{x^{3}}{3}\right|_{0} ^{1}$
$\Rightarrow I=\frac{1}{3 e}+\frac{1}{3}$