If $f : \mathbb R^2 \to \mathbb R$ is an integrable function, then do we have $$ \int f dxdy = \int f dydx $$ or $$ \int f dxdy = -\int f dydx? $$ (I am leaving the domain of integration as it does not matter for the question, but suppose we have some compactum $[a,b]\times[c,d]$ as this make integration theory easier).
To give some context. I am currently reading about differential forms and integration, and if we have a $2$-form $\omega = f\cdot (dx \wedge dy)$, then $$ \int f dx\wedge dy = -\int f dy \wedge dx $$ and the notation $\int f dx \wedge dy$ could be understood as shorthand for $\int f dxdy$. So then the second interpretation would be true.
On the other side, the "ordinary" integral as interpreted as the Riemann-integral where $dx$ and $dy$ are sometimes understood as the "infinitesimal" sides of the rectangles in which the domain of integrations gets partitioned. But then $dxdy = dydx$, or $\Delta x \cdot \Delta y = \Delta y \cdot \Delta x$. I know this language is a little bit "hand-wavy" as it is not clear what infinitesimal objects are, and so some authors leave them out of the integral expression and just write $\int f$, but at least this is the intuitive way and in some way resembles what is done in Riemann integration.
Also, another argument in favour of the first (i.e. interchanging does not changes sign) is the Fubini theorem and multiple integrals (see wikipedia), i.e. if Fubini applies we have $$ \int f dxdy = \int \left(\int f dx\right) dy $$ and $$ \int \left(\int f dx \right) dy = \int \left(\int f dy\right) dx $$ which would yield $\int f dxdy = \int f dydx$ by comparing those expressions.
So there are arguments to support both interpretations, but what is the right interpretation?
As far as i know $\int f dx \wedge dy$ is shorthand for something like $\int f\circ g * (\frac{\partial g}{\partial x}-\frac{\partial g}{\partial y})dxdy$ where g is a parametrisation of the integrationdomain. This makes $dx\wedge dy$ alternating, so $dx\wedge dy=-dy\wedge dx$.