What is the most general way to think about Integrals? Given a single-variable scalar function, $f : \mathbb{R} \to \mathbb{R}$ 
The "area under the curve" (of the graph of the function $f$ in $\mathbb{R^2}$) is given by
$$\int_{a}^{b} f(x) \ dx = Area$$

Given a multi-variable scalar function, $g: \mathbb{R^2} \to \mathbb{R}$
The "volume under the curve" (of the graph of the function $g$ in $\mathbb{R^3}$) is given by 
$$\int\int_{a}^{b} g(x, y) \ dx \ dy = Volume$$

But when thinking about integrals in this sense as translating from $\ Lines \to Areas \to Volumes$, (i.e. taking the Riemann Sum of an infinite number of infinitesimal line segments to get an area, or taking the Riemann Sum of an infinite number of infinitesimal areas to get a volume) it's a pretty "applied" approach. I mean you need a to plot the function $f$ as a graph for explanations such as "area under the curve" or "volume under the curve" to make any sort of sense. 
I'm sure that this way of thinking about Integrals breaks down at some point.
For example, what would the double integral of the single-variable function I gave in the first example "represent".
$$\int\int_{a}^{b} f(x) \ dx^2 = \ ???$$
The graph of $f$ can be plotted in $\mathbb{R^2}$, by means of a vector-function (correct me if I'm wrong here), however the graph of $f$ only exists in $\mathbb{R^2}$, and "Volumes" only have any meaning in $\mathbb{R^3}$ so there can be no way that the double integral of a single-variable scalar function could represent a "volume under the curve"

My Question
Is there a more "Pure" Mathematical approach to thinking about integrals? Because I'm sure that this more "applied" way of thinking about Integrals cannot be the most general. What would be the most general, and pure mathematical way to think about Integrals, and specifically Multiple Integrals?

If you have spotted any gaps in my understanding, please feel free to comment below, as an undergraduate student, majoring in Pure Mathematics, I'm always looking to improve.
 A: Of course this is by no means a complete answer (which would litterally take volumes, and actually does), or even an accurate one, but just a little piece of intuition about the difference between Riemann integration and Lebesgue integration, as my teacher used to explain my class. You can easily find rigourous details in any book on measure theory.
Measure theory gives a very satisfactory account of integration in very general situations. As the name suggests, there is still a strong correlation with notions of area/volume (which are subsumed in the general term "measure").
But in some sense the Lebesgue integral reverses the point of view of Riemann integral. With Riemann integrals, basically you have a natural notion of "measure" on the domain (say $X$) of the function, and what you do is divide $X$ in "small" pieces $X_i$, and take someting like $\sum_i f(x_i)\mu(X_i)$ where $x_i\in X_i$ and $\mu(X_i)$ is the "measure" of $X_i$. Then you take a limit for $X_i$ of size converging to zero.
In Lebesgue integral, you divide the codomain (say $Y$) of the function in small pieces $Y_i$, and take $\sum_i \mu(f^{-1}(Y_i))y_i$ with $y_i\in Y_i$. 
So the point of view is on the one hand "the approximate value of $f$ on a given set of size $V_i$ is $y_i$", and on the other hand "the set of variables on which $f$ has given approximate value $y_i$ has size $V_i$", and in both cases you take $\sum_i y_iV_i$ (and then some kind of limit). In a sense these points of view are dual to one another.
