875. Compute the line integrals:
1) $\oint_{-i} 2 x d x-(x+2 y) d y \quad$ and
2) $\oint_{+l} y \cos x d x+\sin x d y$

along the perimeter of the triangle with vertices $A(-1 ; 0), B(0 ; 2)$ and $C(2 ; 0)$.
Solution. 1) Here (Fig. 186) the integration line (closed) consists of three segments lying on different lines (with different equations). Accordingly, the line integral along the broken line $A B C A$ is calculated as the sum of integrals taken over the segments $A B, B C$, and $C A$.

By formulating the equation of the line $A B$,

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-304.jpg?height=261&width=341&top_left_y=1138&top_left_x=848)

Fig. 186 $y-2 x=2$, and based on this equation, we transform the line integral over the segment $A B$ into an ordinary integral with the variable $x$:

$$
\begin{gathered}
x=2 x+2, \quad d y=2 d x, \quad \int_{A B}=-8 \int_{x_{A}=-1}^{x_{B}=0}(x+1) d x= \\
=-\left.4(x+1)^{2}\right|_{-1} ^{0}=-4
\end{gathered}
$$

Similarly, calculating the line integral over the segments $B C$ and $C A$, we get

$$
\begin{aligned}
& x=2-y, d x=-d y, \quad \int_{B C}==\int_{u_{R}=2}^{4 / c^{=0}}(y-6) d y=\left.\frac{(y-6)^{2}}{2}\right|_{2} ^{0}=10 ; \\
& y=0, \quad d y=0, \int_{C A}=2 \int_{x_{C}=2}^{x_{A}=-1} x d x=\left.x^{2}\right|_{2} ^{-1}=-3
\end{aligned}
$$

Therefore,

$$
\oint_{A B C A}=\int_{A B}+\int_{B C}+\int_{C A}=-4+10-3=3
$$

2) Here the integrand is a complete differential of a function of two variables, since $(y \cos x)_{y}^{\prime}=(\sin x)_{x}^{\prime}=\cos x$. Therefore, this line integral, taken over the perimeter of the given triangle, is equal to zero. It will also be equal to zero over any other closed contour.

Calculate the line integrals: