Example 7. Calculate the integral $I=\int_{\overparen{A B}}\left(x^{2}-2 x y\right) d x+\left(y^{2}-\right.$ $-2 x y) d y$, where $\overparen{A B}$ is the arc of a parabola connecting points $A(-1,1)$ and $B(1,1)$ and passing through the origin.
Solution. The arc $\overparen{A B}$ can be given by the function $y=x^{2}$, $x \in[-1,1]$. Then $d y=2 x d x$, and the line integral reduces to a definite integral:

$$
\begin{aligned}
I= & \overbrace{\overparen{A B}}\left(x^{2}-2 x y\right) d x+\left(y^{2}-2 x y\right) d y= \\
& =\int_{-1}^{1}\left[\left(x^{2}-2 x^{3}\right) d x+\left(x^{4}-2 x^{3}\right) 2 x d x\right]=\int_{-1}^{1}\left(x^{2}-2 x^{3}+2 x^{5}-4 x^{4}\right) d x
\end{aligned}
$$

The evaluation of the definite integral can be simplified by the fact that the integral over a symmetric interval relative to the origin of an odd function is zero, and that of an even function can be taken over the right half of the interval and then doubled, so that

$$
I=2 \int_{0}^{1}\left(x^{2}-4 x^{4}\right) d x=\left.2\left(\frac{x^{3}}{3}-\frac{4 x^{5}}{5}\right)\right|_{0} ^{1}=-\frac{14}{15}
$$

Answer. $I=-\frac{14}{15}$.