Question:
The value of $\int_{-1}^{1} x^{2} e^{\left[x^{3}\right]} d x$, where $[t]$ denotes the greatest integer $\leq t$, is :
Correct Option: , 3
Solution:
$I=\int_{-1}^{1} x^{2} e^{\left[x^{3}\right]} d x$
$=\int_{-1}^{0} x^{2} e^{\left[x^{3}\right]} d x+\int_{0}^{1} x^{2} e^{\left[x^{3}\right]} d x$
$=\int_{-1}^{0} x^{2} e^{-1} d x+\int_{0}^{1} x^{2} e^{0} d x$
$=\frac{1}{e} \times\left.\frac{x^{3}}{3}\right|_{-1} ^{0}+\left.\frac{x^{3}}{3}\right|_{0} ^{1}$
$=\frac{1}{\mathrm{e}} \times\left(0-\left(\frac{-1}{3}\right)\right)+\frac{1}{3}$
$=\frac{1}{3 \mathrm{e}}+\frac{1}{3}=\frac{1+\mathrm{e}}{3 \mathrm{e}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.