Answer
$$\displaystyle{\int_0^{\frac{1}{2}}x\sqrt{1-4x^2}dx=\frac{1}{12}}\\$$
Work Step by Step
$\displaystyle{I=\int_0^{\frac{1}{2}}x\sqrt{1-4x^2}dx}\\
\displaystyle{I=\int_0^{\frac{1}{2}}x\sqrt{1-{\left(2x\right)}^2}dx}\\$
$\displaystyle \left[\begin{array}{ll} 2x=\sin\theta & 4 x^2=\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{1}{2}\cos\theta & dx=\frac{1}{2}\cos\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_0^{\frac{\pi}{2}}\frac{1}{2}\sin\theta\sqrt{1-\sin^2\theta}\times\frac{1}{2}\cos\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi}{2}}\frac{1}{2}\sin\theta\cos\theta\times\frac{1}{2}\cos\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi}{2}}\frac{1}{4}\sin\theta\cos^2\theta\ d\theta}\\
\displaystyle{I=\frac{1}{4}\int_0^{\frac{\pi}{2}}\frac{1}{3}\times3\sin\theta\cos^2\theta\ d\theta}\\
\displaystyle{I=\frac{1}{12}\int_0^{\frac{\pi}{2}}3\sin\theta\cos^2\theta\ d\theta}\\
\displaystyle{I=\frac{1}{12}\left[-\cos^3\theta\right]_0^{\frac{\pi}{2}}}\\
\displaystyle{I=-\frac{1}{12}\cos^3\left(\frac{\pi}{2}\right)\ +\frac{1}{12}\cos^3\left(0\right)}\\
\displaystyle{I=0 +\frac{1}{12}}\\
\displaystyle{I=\frac{1}{12}}\\
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