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प्रश्न
पर्याय
π/2
π/4
π/6
π/8
उत्तर
\[\pi\]\8
\[Let, I = \int_0^1 \sqrt{x\left( 1 - x \right)} d x\]
\[ = \int_0^1 \sqrt{x - x^2} d x\]
\[ = \int_0^1 \sqrt{\frac{1}{4} - \left( x^2 - x + \frac{1}{4} \right)} d x\]
\[ = \int_0^1 \sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2} dx\]
\[ = \left[ \frac{\left( x - \frac{1}{2} \right)}{2}\sqrt{x - x^2} + \frac{1}{2} \times \frac{1}{4} \sin^{- 1} \left( 2x - 1 \right) \right]_0^1 \]
\[ = \frac{1}{8} \left[ \sin^{- 1} \left( 1 \right) - \sin^{- 1} \left( - 1 \right) \right]_0^1 \]
\[ = \frac{1}{8}\left[ \frac{\pi}{2} + \frac{\pi}{2} \right]\]
\[ = \frac{\pi}{8}\]
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