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प्रश्न
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
उत्तर
We have,
\[\left| \sin2\pi x \right| = \begin{cases}\left( \sin2\pi x \right),& 0 \leq x \leq \frac{1}{2}\\ - \left( \sin2\pi x \right),& \frac{1}{2} \leq x \leq 1\end{cases}\]
\[ \therefore \int_0^1 \left| \sin2\pi x \right| d x = \int_0^\frac{1}{2} \sin2\pi x dx + \int_\frac{1}{2}^1 - \sin2\pi x dx\]
\[ = \left[ \frac{- \cos2\pi x}{2\pi} \right]_0^\frac{1}{2} + \left[ \frac{\cos2\pi x}{2\pi} \right]_\frac{1}{2}^1 \]
\[ = \frac{1}{2\pi} + \frac{1}{2\pi} + \frac{1}{2\pi} + \frac{1}{2\pi}\]
\[ = \frac{2}{\pi}\]
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