Fundamental theorem of calculus for double integrals If we have a function $f(x)$ that is difficult to integrate, it's well known from the fundamental theorem of calculus that if we can find an antiderivative $F(x)$ such that $F'(x)=f(x)$, then $\int_{a}^{b} f(x) dx=F(b)-F(a)$.
Suppose instead we have a function $f(x,y)$ that we're not sure how to integrate and we want to calculate $I=\int_{c}^{d} \int_{a}^{b} f(x,y) dxdy$. However, we have functions $F(x,y)$ and $G(x,y)$ such that $\frac{\partial F(x,y)}{\partial x}=f(x,y)$ and $\frac{\partial G(x,y)}{\partial y}=F(x,y)$ (and therefore $\frac{\partial^2 G(x,y)}{\partial x\partial y}=f(x,y)$), then is there a way to express $I$ as evaluations of $G(x,y)$ and $F(x,y)$?
Integrating with respect to $x$ first, we have $I=\int_{c}^{d} F(b,y)-F(a,y) dy$. Next, I would like to claim $\int_{c}^{d} F(b,y)dy=G(b,d)-G(b,c)$ and $-\int_{c}^{d} F(a,y)dy=-G(a,d)+G(a,c)$ and so $I=G(b,d)-G(b,c)-G(a,d)+G(a,c)$. Is this right?
It seems obviously true but I'm a little uneasy about for example $\int_{c}^{d} F(b,y)dy=G(b,d)-G(b,c)$ since I feel like this could be false if $F(x,y)$ is badly behaved at $x=b$, or is the existence of a partial derivative or (additionally) continuity of $F$ at $x=b$ enough to ensure this always hold?
EDIT: corrected a typo where $a$ and $b$ were the wrong way round.
 A: -For the last part:
If you define the function $\phi(t) = G(b,t)$ then $\phi'(t) = \frac{\partial}{\partial t} G(b,t) = F(b,t)$, if you assume continuity of F, then the fundamental theorem of calculus applies and
$$
\int_{c}^d F(b,t) \, dt = \int_c^d \phi'(t) \, dt = \phi(d) - \phi(c) = G(b,d) - G(b,c).
$$
We used continuity of $\phi'$ here.
To answer your full question: It is true if you have continuity of the second partial derivatives of $G$, otherwise you can run into some issues. Maybe the integral does not exist, or maybe it doesn't converge absolutely and you can't use Fubini's theorem, maybe the second partial derivatives are not symmetric.
A: You can do something of that sort if you use differential forms. In the language of forms:
$$ I = \int_D f(x,y) dx \wedge dy = \int_{\partial D} P(x,y) dy$$
Where $dP(x,y) = f(x,y) dx$. For instance, if we had $f(x,y)=1$, then $P(x,y)$ can be taken as $P(x,y)= x$. We have:
$$ I = \int_D (1) dx \wedge dy = \int_{\partial D} x dy$$
When we integrate the two form over the domain, we get the area, and that is equal to the $x dy$ integrating over the bounding loop.
