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प्रश्न
उत्तर
\[Let\, I = \int_{- 2}^1 \frac{\left| x \right|}{x} d x\]
\[\text{We have}, \]
\[\left| x \right| = \begin{cases}x&,& 0 \leq x \leq 1\\ - x&,& - 2 \leq x < 0\end{cases}\]
\[ \therefore \frac{\left| x \right|}{x} = \begin{cases}1&,& 0 \leq x \leq 1\\ - 1&,& - 2 \leq x < 0\end{cases}\]
\[\text{Therefore}, \]
\[I = \int_{- 2}^0 - 1dx + \int_0^1 1 dx\]
\[ = - \left[ x \right]_{- 2}^0 + \left[ x \right]_0^1 \]
\[ = 0 - 2 + 1 - 0\]
\[ = - 1\]
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