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प्रश्न
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उत्तर
\[\left| x \right| = \begin{cases} - x &,& - 1 < x < 0\\ x &,& 0 < x < 1\end{cases}\]
\[ \therefore x\left| x \right| = \begin{cases} - x^2 &,& - 1 < x < 0\\ x^2 &,& 0 < x < 1\end{cases}\]
\[Now\, \int_{- 1}^1 x\left| x \right| d x\]
\[ = \int_{- 1}^0 - x^2 dx + \int_0^1 x^2 dx\]
\[ = - \int_{- 1}^0 x^2 dx + \int_0^1 x^2 dx\]
\[ = - \left[ \frac{x^3}{3} \right]_{- 1}^0 + \left[ \frac{x^3}{3} \right]_0^1 \]
\[ = - \left( 0 + \frac{1}{3} \right) + \left( \frac{1}{3} - 0 \right)\]
\[ = 0 - \frac{1}{3} + \frac{1}{3} - 0\]
\[ = 0\]
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