I saw a YouTube video that messed up all my intuition. They were calculating the integral of something like -1+x^dx which in the end he solved by taking the limit as dx->0 and he got the solution. But everyone in the comments was saying complex stuff like like “well to be rigorous x^dx can be thought of an element of C*(R,R), the exterior algebra on R”. Huh. When did dx get any different from lim x->0.
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dx can be interpreted in a couple of different ways. They all produce the same (or nearly the same, up to some technicalities about domain) results, so these are at the end of the day just different ways to formalize the same ideas.
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One way to think of dx is, as you roughly say, limit of delta-x as delta-x -> 0. This intuition can get you surprisingly far with two rules:
* lim delta-x -> 0 is dx
* dx^2 is “0” (or more properly, infinitesimally smaller than dx)
So for example, suppose we want to compute the area of a polar figure. To do that, we need to take two close-by points on the polar curve and compute the area of the wedge between them. This ends up being the combination of a circular wedge of area (delta-theta/2pi) * pi * r^(2), plus an approximately triangular shape of area (1/2) (r * delta theta) (delta r).
The first term gets us (1/2) r^2 delta-theta, and the second term gets us 1/2 * r * delta-theta * delta-r. If we then think about this in the limit, the first term is (1/2) r^2 d(theta), and the second term ends up with a d(theta) times a dr -> two d’s multiplied -> effectively zero.
So we get, correctly, the area element (1/2) r^2 d(theta).
In this approach, you’re thinking of your integrals as the Riemann integrals you probably learned first: you have a bunch of bars, you make them infinitely thin, and you add them up.
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But there’s another, deeper, way to think about what dx represents, and it turns out to be a way that generalizes better to more complex math.
That way is to think about dx as assigning a “weight” to each segment of the thing you’re integrating over. So for example, dx assigns a weight of (b-a) to every interval [a,b] of the real line. And then your integral over an interval is simply the “height” of the function f(x) times the weight of each tiny interval dx. Similarly, you write dA, and what you’re really doing is assigning a weight (b-a)(d-c) to each rectangle in R^(2).
Up until now, we’ve thought of integrals as having two basic inputs: the set you’re integrating over (typically an interval) and the function you’re integrating. But with this formulation, an integral has *three* basic inputs: the set you’re integrating over, the function you’re integrating, **and the weight function dx you’re using**.
Formally speaking, we think of such integrals as regular integrals of some special product between the function and the weight function, more technically called a *distribution* or a *measure* (depending on exactly what you’re doing; the concepts are equivalent-ish). And because functions form a vector space, where you can add functions or multiply them by constants, and because the product of a function and a distribution or measure produces a real number, the measure is (by defintion) a function from vectors to scalars, which is what we call a *functional*.
The reason that this is useful is that it’s more generalizable. For example, if you’ve had some differential equations, you may have encountered the Dirac delta function, a function with the property that:
* delta(x) = 0 everywhere except x = 0 and
* integral(a,b) delta(x) dx = 1 if the interval contains 0, and 0 otherwise.
No actual function has this property. So the Dirac delta is not, strictly speaking, a function. However, it *is* a *distribution*, and what we more properly mean by this definition is that:
* integral(a,b) f(x) delta(x) = f(x) if the interval contains 0 and integral(a,b) f(x) delta(x) = 0 otherwise
And *this* idea is well-defined.
With me so far?
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