Do people agree on what dx is in Riemann integration?

Question

I've been rather curious about what the official rigorous understanding of the dx term in a Riemann integral is. Of course, we could use the formal definition of a Riemann integral but I have seen on Wikipedia other proposed ways of defining the differential of some function (we'll call x), dx. The classic high school teacher approach is to say it's a "very small change in x", or rather a small change in a linear approximation of x. However, as someone who studies differential geometry, I am wondering how this can be interpreted as the exterior derivative of a function x, in which dx is a 1-form. Do we then reinterpret $\int{f(x)}\,dx$ as being the integral of a 1-form? And how then can we consider a 1-form to be a "small change in x"?

Answer 1

The title doesn't really match the question, but perhaps it suffices to say that $d x$ is a 1-form which is the exterior derivative of the function $x$, i.e. the identity function on $\mathbb{R}^1$.

This function takes $x$ to $x$, hence the name $x$; compare e.g. when we write $\sin(x)$ for the function that takes $x$ to $\sin(x)$. (These are both abuses of notation; the latter should just be called $\sin$ when we're talking about the function, while the former should be called ${\rm id}$ or something.)

Answer 2

This question has been asked a million times on this site in one form or another. But I suppose it can’t hurt to answer it again since it is so natural a question but also so typically badly answered.

The short answer to your title question is yes: there is agreement about what the $dx$ means — but only once you get precise about exactly what kind of integral you’re talking about. The one-dimensional baby’s-first-integral in $\mathbb{R}$ can be interpreted in many different ways that only come apart once you move to integration in more general spaces.

If you’re thinking about $\int_{[a,b]}f dx$ as giving the integral of a density $f$ over an unoriented interval $[a,b]$, then $dx$ generally denotes a measure. Example: if $f$ measures mass per unit length over a rod, then the integral gives the mass of the rod. This generalizes easily to higher-dimensional Euclidean spaces: the “area element” $dA$ and the “volume element” $dV$ are really measures. In differential geometry you want to integrate such things not just over subsets of Euclidean space but over abstract manifolds. In that case you don’t have a measure, but you do have the differential geometric analog of a measure called a density.

If, however, you’re thinking about $\int_a^bfdx$ as giving the integral of $f$ over the oriented interval $[a,b]$, then $dx$ generally denotes a differential form. Example: if $f$ measures the rate of change of some quantity as a function of $x$, then the integral gives the total change in the quantity as $x$ ranges from $a$ to $b$ — that’s the fundamental theorem, which is just a special case of Stokes theorem, which is a theorem that depends crucially on orientations and thus on differential forms. There is a connection between differential forms and densities, however. Densities are in some sense the natural objects to integrate: they’re defined whether you have an orientation or not. You can only integrate a differential form if you have an orientation, by contrast — and indeed the whole point of the orientation is that it provides enough information to turn the differential form into a density, which can then be integrated.

If you reject all this seeming abstract nonsense and insist on an answer in the case of the most familiar, calculus 101 one-dimensional Riemann integral of a bounded real function $f$ on $[a,b]$, the answer is that the $dx$ is literally meaningless, a mere piece of suggestive notation. The value of the integral, if it exists, is solely a function of $f$ and of $[a,b]$; the best notation to emphasize this is $\int_{[a,b]}f$, with no $dx$ at all.