Let C be a smooth simple closed curve in the xy-plane with the property that lines parallel to the axes cut it at no more than two points. Let R be the region enclosed by C and suppose that M, N, and their first partial derivatives are continuous at every point of some open region containing C and R. We want to prove the circulation-curl form of Green’s Theorem:
$$\oint_{C} Mdx+Ndy = \iint\frac{ \partial N}{\partial x} - \frac{\partial M}{\partial y}dxdy$$ The boundary curve $C$ is made up of $C_1$, the graph of y = $ƒ_1(x)$, and $C_2$ , the graph of $y$ = $ƒ_2(x)$.
The link above shows C made up of two directed pairs, $C_1 : y=f_1(x), a \leq x \leq b$, $C_2:$ $ b \geq x \geq a $
For any x between a and b, we can integrate $ \frac {\partial M}{\partial y}$ with respect to y from $y=f_1(x)$ to $ y=f_2(x)$ and obtain
$$\int_{f{_1}(x)}^{f{_2}(x)} \frac{\partial M}{\partial y} = M(x,y) \Big|_{y=f_1{(x)}}^{y=f_2(x)} = M(x,f_2(x))-M(x,f_1(x))$$.
We can then integrate this with respect to $x$ from $a$ to $b$:
$$\int_a^b\int_{f_1(x)}^{f_2(x)}\frac{\partial M}{\partial y}dydx = \int_{a}^{b} (M(x,f_2(x))-M(x,f_1(x))dx$$ ...
Please open to link shaded into blue to see the picture.
Please help me understand why we're integrating $\frac{\partial M}{\partial{y}}$ with the bounds $y=f_1(x)$ and $y=f_2(x)$. One case is if $ x=a$ or $ x=b$, then $f_1(x)=f_2(x)$ and what else would happen? The other case is if x $\neq a $and $x\neq b$.
Does the value of $M(x,f_2(x))-M(x,f_1(x))$ doesn't depend on the bounds of the integral $f_2(x)$ and $f_1(x)$?. I assume the upper and lower bounds $y=y_2(x)$ and $y=y_1(x)$ respectively are real numbers.