The value of

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 :

  1. (1) $\frac{\mathrm{e}+1}{3}$

  2. (2) $\frac{e-1}{3 e}$

  3. (3) $\frac{\mathrm{e}+1}{3 \mathrm{e}}$

  4. (4) $\frac{1}{3 \mathrm{e}}$


Correct Option: , 3

Solution:

$I=\int_{-1}^{0} x^{2} \cdot e^{-1} d x+\int_{0}^{1} x^{2} d x$

$\therefore I=\left.\frac{x^{3}}{3 e}\right|_{-1} ^{0}+\left.\frac{x^{3}}{3}\right|_{0} ^{1}$

$\Rightarrow I=\frac{1}{3 e}+\frac{1}{3}$

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