# Can someone explain to me the jump between steps in the bottom two lines of this proof (not yet totally complete)?

Suppose we have a region $$G$$ that is bounded by the straight lines $$x=a$$, $$x=b$$, $$y=c$$ and by an arc $$y = f(x)$$ (which lies above $$y=c$$) where $$a \leq x \leq b$$. If $$f$$, $$P(x,y)$$ and $$Q(x,y)$$ are all continuously differentiable in a region which contains the region $$G$$ and its boundary, denoted as $$\partial G$$, then:

$$\iint_G \Bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Bigg) dA = {\oint}_{\partial G} \Big(P \cdot dx + Q \cdot dy\Big)$$

We can prove this lemma by showing the following integrals, which are taken by splitting up the above equality, are as such: $$$$\iint_G \Bigg(\frac{\partial Q}{\partial x}\Bigg) \cdot dA = \oint_{\partial G}Q \cdot dy$$$$ $$$$-\iint_G \Bigg(\frac{\partial P}{\partial y}\Bigg) \cdot dA = \oint_{\partial G}P \cdot dx$$$$ First, we shall prove the bottom equation. First we need to manipulate the double integral using the bounds given and then apply the Fundamental Theorem of Calculus. This will give us the following: \begin{align} -\iint_G \Bigg(\frac{\partial P}{\partial y}\Bigg) \cdot dA &= - \int_{a}^{b} \int_{c}^{f(x)} \frac{\partial P}{\partial y} \cdot dy \cdot dx \label{eqn:27} \\ & = - \int_{a}^{b} \Big(P[x,f(x)] - P[x,c]\Big) \cdot dx \label{eqn:28} \end{align} Letting $${\overset{\rightharpoonup}{C_3}}$$ be the same as $$y = f(x)$$ and $${\overset{\rightharpoonup}{C_1}}$$ be the same as $$y=c$$ we can rewrite the line above as: $$\int_{\overset{\rightharpoonup}{C_3}}P(x,y) \cdot dx + \int_{\overset{\rightharpoonup}{C_1}}P(x,y) \cdot dx$$

How does: $$- \int_{a}^{b} \Big(P[x,f(x)] - P[x,c]\Big) \cdot dx = \int_{\overset{\rightharpoonup}{C_3}}P(x,y) \cdot dx + \int_{\overset{\rightharpoonup}{C_1}}P(x,y) \cdot dx$$ ???

Apologies I couldn't get the anti clockwise arrow on the integrals for the contours.