1286. Find the area of the figure bounded by the line

$$
y^{2}=x(x-1)^{2}
$$
Solution. The function $y^{2}=x(x-1)^{2}$ is even with respect to the variable $y$, therefore, the figure bounded by this line is symmetric with respect to the $O x$ axis (Fig. 81). Let's find the interval of integration. Let $y=0$, then $x_{1}=0$, $x_{2}=1$. Using formula (1), we get

$$
\frac{1}{2} S=\int_{0}^{1} V \bar{x}(x-1) d x=\int_{0}^{1}\left(x^{\frac{3}{2}}-x^{\frac{1}{2}}\right) d x=-\frac{4}{15}
$$

The negative sign indicates that the figure, the area of which has been found, is located below the $O x$ axis. This could have been anticipated, as the integrand $\sqrt{x}(x-1) \leqslant 0$ on the segment $[0,1]$. Therefore, the obtained result should be taken with the opposite sign, i.e., $\frac{1}{2} S=\frac{4}{15}$, hence the required area $S=\frac{8}{15}$.