MCQ
$\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx}$ equals:
- A$\frac{\pi}{2}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- D$\frac{\pi}{8}$
Solution:
$\text{I}=\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx} $
$\text{I}= \int\limits^1_0\sqrt{\text{x}-\text{x}^2}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}+\text{x}-\text{x}^2+\frac{1}{4}}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}-\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$\text{I}=\Bigg[\frac{\text{x}-\frac{1}{2}}{2}\sqrt{\text{x}(1-\text{x})}+\frac{1}{2}\times\frac{1}{4}\sin^{-1}(2\text{x}-1)\Bigg]^1_0$
$\text{I}=0+\frac{1}{8}\big(\sin^{-1}(1)-\sin^{-1}(-1)\big)$
$\text{I}= \frac{1}{8}\Big(\frac{\pi}{2}-\Big(\frac{\pi}{2}\Big)\Big)$
$\text{I}= \frac{\pi}{8}$
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Then the triangle $P Q R$ has $S$ as its
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