Use Green's theorem to prove: if the vertices of the polygon in counterclockwise order are $(x_i,y_i)$, $i=1(1)n$ then the area of the polygon is $$A=\frac 12 \sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$ $x_{n+1}=x_1, y_{n+1}=y_1$.
Now the statement of Green's theorem: ${\displaystyle \int_{\partial D}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dx\,dy}.$
It is clear that we have to take $L=-y,M=x$ the confusion occurs while calculating ${\displaystyle \int_{(x_i,y_i)}^{(x_{i+1},y_{i+1})}(L\,dx+M\,dy)}.$ I am telling that I got the answer by taking this path $L_i:(tx_{i+1}+(1-t)x_{i},ty_{i+1}+(1-t)y_{i})$.
I attempted to calculate the integration over the path "$L_i^1(x_{i},y_{i})$ to $(x_{i+1},y_{i})$" and then $L_i^2:(x_{i+1},y_{i})$ to $(x_{i+1},y_{i+1})$. I think that the integration
$${\displaystyle \int_{(x_i,y_i)}^{(x_{i+1},y_{i+1})}(L\,dx+M\,dy):=\int_{L_i}(L\,dx+M\,dy)=\int_{L_i^1}(L\,dx+M\,dy)+ \int_{L_i^2}(L\,dx+M\,dy)}$$ because the the integration is independent of path but
$$\int_{L_i^1}(-y\,dx+x\,dy)+ \int_{L_i^2}(-y\,dx+x\,dy)=x_{i+1}y_{i+1}+x_iy_i-2x_{i+1}y_i$$ and $$\int_{L_i}(-y\,dx+x\,dy)=x_{i}y_{i+1}-x_{i+1}y_i.$$
Can you please help me, why aren't they same?
The issue is that the vector field $\vec{F} = \left<-y , x \right>$ isn't conservative.
We can show this using Clairaut's theorem: if there was some $f$ such that $f_x = -y$ and $f_y = x$ then $f_{xy} = -1$, but $f_{yx} = 1$. Therefore there can be no $f$ such that $\vec{F} = \nabla f$ and $\vec{F}$ is not conservative.
Also, this is to be expected because if the integral were path-independent, then your total area would always have to equal zero.