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प्रश्न
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
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उत्तर
We have,
\[\left| 2x - 1 \right| = \begin{cases} - \left( 2x - 1 \right)&,& 0 \leq x \leq \frac{1}{2}\\ 2x - 1&,& \frac{1}{2} \leq x \leq 1\end{cases}\]
\[ \therefore \int_0^1 \left| 2x - 1 \right| d x\]
\[ = \int_0^\frac{1}{2} - \left( 2x - 1 \right) dx + \int_\frac{1}{2}^1 \left( 2x - 1 \right) dx\]
\[ = \left[ - x^2 + x \right]_0^\frac{1}{2} + \left[ x^2 - x \right]_\frac{1}{2}^1 \]
\[ = \frac{- 1}{4} + \frac{1}{2} + 1 - 1 - \frac{1}{4} + \frac{1}{2}\]
\[ = \frac{1}{2}\]
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