Answer
$\displaystyle \frac{\pi}{8}$
Work Step by Step
Formula 113.
$\displaystyle \int\sqrt{2au-u^{2}}du=\frac{u-a}{2}\sqrt{2\mathrm{a}u-u^{2}}+\frac{a^{2}}{2}\cos^{-1}(\frac{a-u}{a})+C$
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$\displaystyle \int_{0}^{1}\sqrt{x-x^{2}}dx=\int_{0}^{1}\sqrt{2(\frac{1}{2})x-x^{2}}dx=$
Use Formula 113, $a=\displaystyle \frac{1}{2}, u=x$
=$\left|\dfrac{x-\frac{1}{2}}{2}\sqrt{2(\frac{1}{2})x-x^{2}}+\frac{(\frac{1}{2})^{2}}{2}\cos^{-1}(\dfrac{\frac{1}{2}-x}{\frac{1}{2}})\right|_{0}^{1}$
$=\left|\dfrac{2x-1}{4}\sqrt{x-x^{2}}+\frac{1}{8}\cos^{-1}(1-2x)\right|_{0}^{1}$
$= [\displaystyle \frac{2-1}{4}\sqrt{1-1}+\frac{1}{8}\cos^{-1}(1-2)] -[\frac{0-1}{4}\sqrt{0-0}+\frac{1}{8}\cos^{-1}(1-0)]$
$= [0+\displaystyle \frac{1}{8}\cos^{-1}(-1)] -[0+\frac{1}{8}\cos^{-1}(1)]$
$=\displaystyle \frac{1}{8}\cdot\pi-0$
$=\displaystyle \frac{\pi}{8}$
