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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
MCQ
$\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx}$ equals:
  • A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • ✓
    $\frac{\pi}{8}$
✓

Answer

Correct option: D.
$\frac{\pi}{8}$
$\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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