Question

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\sqrt{x\left(1-x\right)}\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^1\sqrt{\frac{1}{4}-{\left(x-\frac{1}{2}\right)}^2}dx \\ \Rightarrow I=\frac{1}{2}{\int }_0^1\sqrt{1-\frac{{\left(x-{\displaystyle \frac{1}{2}}\right)}^2}{{\displaystyle \frac{1}{4}}}}dx \\ \Rightarrow I=\frac{1}{2}{\int }_0^1\sqrt{1-{\left(\frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}\right)}^2}dx \\ \mathrm{Let}\left(\frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}\right)=\sin u \\ \Rightarrow 2dx=\cos udu \\ \therefore I=\frac{1}{4}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{1-{\sin }^{2}u}\cos udu \\ \Rightarrow I=\frac{1}{4}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\cos }^{2}udu \\ \Rightarrow I=\frac{1}{4}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{\cos 2u+1}{2}\right)du \\ \Rightarrow I=\frac{1}{8}{\left[\frac{\sin 2u}{2}+u\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=\frac{1}{8}\left[\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow I=\frac{\mathrm{\pi }}{8} \end{gathered}$$