7
$\begingroup$

The first explanation I heard for the $\mathrm{d}x$ - it just shows by which variable we are integrating. Which made sense because $(F(x)+C)^\prime=f(x)$, not $f(x)\mathrm{d}x$. Now, some time later, the $\mathrm{d}x$ has become a source of confusion again. If there's $\int \frac{x}{x^2+1} \mathrm{d}x$, then why can we solve it like that: $\int \frac{1}{x^2+1} x \mathrm{d}x= \frac{1}{2}\int\frac{1}{x^2+1} 2 x \mathrm{d} x=\frac{1}{2}\int \frac{1}{x^2+1} \mathrm{d}(x^2+1)$ ? The other parts seem more or less normal but the transition from $\int\frac{x}{x^2+1} \mathrm{d}x$ to $\int \frac{1}{x^2+1} x \mathrm{d}x$ seems very strange.

It works but why does it? If $\mathrm{d}x$ just shows by which variable we are integrating $f(x)$ then we cannot treat it as if $f(x)$ were multiplied by it. And on the other hand, if $f(x)$ IS actually multiplied by $\mathrm{d}x$ then why can we do it? I know there's simple explanation for it when we calculate the definite integral, that we break up some line or surface or volume into infinitely little pieces and then add up those infinitely little pieces to get the whole thing, so it makes sense.

But why do we treat $\mathrm{d}x$ in the indefinite integral as if $f(x)$ were multiplied by it? Thanks.

  • 2
    "Don't treat it as multiplication", I would say. Think of $\int$ and $\mathrm dx$ (assuming $x$ is your dummy variable) as a compound operator: $\int$ indicates that you're integrating, and $\mathrm dx$ merely indicates the variable you're integrating with respect to in the function enclosed between $\int$ and $\mathrm dx$.2011-11-06
  • 1
    @LiisiKerik If you are at terms with $\mathrm{d} x$ in the definite integral $\int_a^b f(x) \mathrm{d}x$, think of the indefinite integral $\int f(x) \mathrm{d} x$ as a definite integral with variable upper, or lower, bound, $\int f(x) \mathrm{d} x = \int_a^x f(y) \mathrm{d} y + C$.2011-11-06
  • 0
    Related: http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio2011-11-06
  • 0
    @J.M. - The problem is, in the example I gave dx is NOT treated as a dummy variable. It is treated as if the other stuff which is being integrated were multiplied by it and for some reason the result is correct. I'd just like to know why.2011-11-06
  • 0
    ...and I don't recommend that abuse of notation you speak of. :)2011-11-06
  • 0
    @J.M. - I didn't like it also at first. But unfortunately the "abuse" was suggested by our math teacher and gives a correct result - we get (ln|x^2+1|)/2, which, if we take a derivative, is [1/(x^2+1)]*2x/2= x/(x^2+1)2011-11-06
  • 0
    Your teacher's being somewhat slovenly, I'd say. The proper way of going about it is to make a substitution: "let $u=x^2+1$, and $\mathrm du=2x\,\mathrm dx$"...2011-11-06
  • 4
    If you are interested, it is possible to understand $f d x$ in isolation, without the $\int$ sign, as some object called "differential form", however, you'll need much background.2011-11-06
  • 0
    @Liisi: If you agree that $\frac{x}{x^2 + 1} = \frac{1}{x^2 + 1} x$, then under _any_ reasonable definition, we must have $\frac{x}{x^2 + 1} \, \mathrm{d}x = \frac{1}{x^2 + 1} x \, \mathrm{d}x$, since we are literally just substituting one for the other. There's no magic here.2011-11-06

2 Answers 2