General definition of area in calculus In Cartesian coordinate system, I know that we define the area as:
\begin{equation}
A_1= \iint dy\ dx
\end{equation}
But I already know that the area under a certain curve is:
\begin{equation}
A_2= \int f(x) \ dx = \int y \ dx
\end{equation}
or even between two curves so $(y_1-y_2)$ isnted of $y$.
In some books, I configure this relation:
\begin{equation}
A_3 = \int x dy 
\end{equation}
What is then the difference between the 3 forms of area?
 A: The general definition of area of a subset of $\mathbb{R}^n$ is known as Lebesgue measure. This notion is a quite deep concept usually introduced in a second year course of analysis.
This Lebesgue measure is a function $\mathcal{L}^n : \mathcal{B} \subset \mathcal{P}(\mathbb{R}^n) \to [0, +\infty]$ which satisfies some properties. This is the definition of area and in your particular case, $n=2$ and $\mathcal{L}^2(A) = \int_A 1$. The last $2$ integral are only a particular case for a set of the plane $\mathbb{R}^2$ of the form $graph_{\leq}(f)=\{ (x,y) : x \in dom(f), 0 \leq y \leq f(x)\}$ for $f: [a,b] \to [0, +\infty)$. To obtain that $\int_{graph_{\leq}(f)} 1 = \int_a^bf(x)dx$ we need the Fubini-Tonelli's theorem.
Hope this will help you.
A: While Filippo Giovanni's answer is fully correct, it is probably a bit high level for someone doing a calculus course. Here's a simpler, though less general way to look at it.
The ancient Greeks already had a way to define a general area, called the method of exhaustion. The general idea is to take the figure whose area $A$ you want to find, and inscribe a simpler figure with known area $A_-$, and circumscribe a simpler figure with known area $A_+$. The circumscribed area is guaranteed to be larger than $A$, and the inscribed area is guaranteed to be less than $A$, that is, $A_-\leq A\leq A_+$. So we have at least an interval in which the actual area has to be.
Now if you find a sequence of such in- and circumscribed areas which both converge to the same area, the sandwich principle tells us that this area must be $A$. Archimedes did this in order to find the area of a circle, like this:
https://upload.wikimedia.org/wikipedia/commons/thumb/c/c9/Archimedes_pi.svg/1280px-Archimedes_pi.svg.png
He in- and circumscribed regular polygons, whose areas he knew. And he knew that as the number of vertices increased, these areas would converge to $\pi r^2$ (which in itself is something that requires proof, but it's true)
Now when integrating in calculus, we're actually using the method of exhaustion, too! The Riemann integral of a function can be defined using upper- und lower sums, where the lower sum is essentially an inscribed area made up of rectangular areas and the upper sum a circumscribed area, also made up of rectangular areas (some additional care is required for negative functions, but you get the gist). As we make finer upper and lower sums, they converge to what we call the integral, which is then the area under the graph of the function.
The other two methods of calculating areas using integrals will at some point, when going down the rabbit hole of how they are actually defined, go back in some way to the method of exhaustion, but applied in a different way.
A: 
\begin{equation} A_1= \iint dy\ dx \end{equation}

Here, the unspecified integration domain appears to be a lamina (let's call it $L$) on the $x$-$y$ plane; as such, $A_1$ gives the area of of $L.$
More generally, if $f$ is the density (mass per unit area) function of $L,$ then \begin{align}\iint_L f(x,y) &\:\mathrm dy\,\mathrm dx. \tag1\end{align} gives the mass of $L.$

\begin{equation} A_2= \int f(x) \ dx = \int y \ dx \end{equation}

Here, the unspecified integration domain appears to be a straight line segment on the $x$-axis; as such, $A_2$ gives the signed area between some interval in the $x$-axis and the curve $y=f(x).$

\begin{equation} A_3 = \int x dy  \end{equation}

$A_3,$ once the integration interval is specified, gives the signed area between that interval in the $y$-axis and the curve $x=f(y).$
Similarly, expression $(1)$ above can be interpreted as the signed volume between the lamina $L$ and the surface $z=f(x,y).$ This is an alternative to the ‘mass’ interpretation.
Read more here.
A: they all effectively represent the same thing: If I have a curve I represent by the equation $y=f(x)$ and I want the area under this curve you are integrating between $y=0$ and $y=f(x)$, then between whatever your limits are on the $x$ axis so:
$$\text{Area}=\int\limits_{x=0}^{x=X}\int\limits_{y=0}^{y=f(x)}dydx=\int_0^Xf(x)dx$$. The way I like to think of it is that we are integrating $1$, which represents the uniform plane we are integrating over, and our bounds are the four functions.
