The reason the dx is not used in integration

Question

Integration problems end with dx. But, nothing is done with that dx. It just sits there. For example - when solving the problem $\int x^3\ dx$, the fact that the dx is there does nothing to the problem. Is the reason for this that by definition when taking an integral we are saying dx is getting infinitesimally small? To the point where it basically doesn't exist?

Answer 1

Mostly, the "$dx$" serves as a means to specify the variable in which the integration is being done. Indeed, often, you have more than one variable in the integrand, and in those cases, without a "$dx$" to specify the variable in which we are integrating, things would get confusing, to say the least.

e.g.:

What is

$$\int x^2y $$?

Without specifying with respect to what variable we are integrating, the above expression is ill-defined, i.e. it doesn't have one unique answer.

Answer 2

The $dx$ term is there for a good reason: it tells us how to measure the domain! This can extend very broadly, but recall that a (Riemann) integral such as $$\int_0^1f(x)\,dx$$ is, more or less, defined to be the sum of a rectangular approximation to the region under the curve $f$ - more precisely, we divide the unit interval into small intervals $[c_0,c_1],\,[c_1,c_2],\ldots, [c_{n-1},c_n]$ where $c_0=0$ and $c_n=1$ and then approximate the area above the interval $[a,b]$ as $f(x)\cdot (b-a)$ for some $x\in [a,b]$. Note that this has the same form as before: the term $f(x)\,dx$ transforms by choosing $x$ from the interval and changing $dx$ to be the length of the interval - so it's actually, from this perspective, specifying a actual multiplication.

This turns out to be fairly useful even - you can change how you measure intervals and get integrals that mean different things! A silly example would be to take $d(2x)$ where you simply measure $[a,b]$ as having "weight" $(2b-2a)$ in the limiting sums* - which simply doubles the integral. However, you could also integrate with respect to other things - for instance integrating against $d(x^2)$ measures intervals as having weight $b^2-a^2$, and you can discover that the relation that $d(x^2)=2x\,dx$ lets you replace weird terms at the end with more meaningful things - more generally, a $u$-substitution is an implementation of the rule by noting $$df(x)=f'(x)\,dx$$ where we get to manipulate these differential terms in clever ways to simply the questions.

While $dx$ might not seem to do much when you're only ever integrating functions on the real line according to the usual notion of area, it's actually really important in that it situates that kind of integration in a more general context. Rather than vanishing, it tells you how to break an interval into lots of small pieces, each with some weight, and how to sum these back up to represent the whole.

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(*In general, assigning intervals weights based on the differences of a function evaluated at each endpoint gives you Riemann-Stieltjes integration, which is not the most common way to interpret differentials formally; usually, differentials like $dx$ are more explicitly defined to mean a way to assign little pieces of the domain a weight - either by assigning weights to each subset of the domain (as is done in measure theory) or by relating the differential to the standard ones in a coordinate system (as is done in differential geometry))

Answer 3

You do do something with the $dx$. In the most basic calculus exercises, it may be relatively inert, calculation-wise. But in more advanced math (like differential geometry or measure theory), the symbol $dx$ has an actual, concrete meaning (although what that meaning is may differ from field to field). And even in regular calculus, you use it explicitly when doing substitutions:

Using $x^2 = u$, we get $\frac{du}{dx} = 2x$, which is to say $2x\,dx = du$ (as far as basic calculus is concerned, this is an abuse of notation, but it works). This gives us: $$\int_a^b2x^3dx = \int_{a^2}^{b^2} u\,du$$

Answer 4

$'dx'$ tells with respect to what you are integrating. Suppose we want to calculate area of circle with radius R and Outside surface area of a cylinder with radius R and height H. $ A= \int_0^R 2 \pi r \: dr = \pi R^2 $. Now outside surface area of the cylinder is $C=\int_0^H 2 \pi R \: dh = 2\pi RH$. So it is very important. In your example $\int x^3 dx = \frac{x^4}{4}$+constant, while $\int x^3 dt = x^3 t$. So unless $dx$ or $dt$ or some "$d..$" is there, integration is not defined.