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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$
✓

Answer

Let $\text{I}=\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\sqrt{\frac{1}{4}-\big(\text{x}-\frac{1}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\frac{\big(\text{x}-\frac{1}{2}\big)^2}{\frac{1}{4}}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)^2}\text{ dx}$
Let $\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)=\sin\text{u}$
$\Rightarrow2\text{dx}=\cos\text{u du}$
$\therefore\ \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\sqrt{1-\sin^2\text{u}}\cos\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\cos^2\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\Big(\frac{\cos2\text{u}+1}{2}\Big)\text{du}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\sin2\text{u}}{2}+\text{u}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\pi}{2}+\frac{\pi}{2}\Big]$
$\Rightarrow\text{I}=\frac{\pi}{8}$

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