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What is the intuition/idea behind integration of a function with respect to another function? Say $$int f(x)d(g(x)) ;;;;;?$$
or may be a more particular example $$int x^2d(x^3)$$

My concern is not at the level of problem solving. To solve we could simply substitute $u=x^3$ and then $x^2=u^{2/3}$. My concern is rather about what meaning physical/geometrical does this impart?

ADDED if you ask what kind of meaning I seek, integration of a function w.r.t a variable gives the area under the curve and above x-axis.

You shouldn't think of $g$ as a function, it's a change of variable. It's a way to parametrize the integral, with no geometrical or physical sense.

ADDENDUM : in fact, both in mathematics and in physics, a major concern has been to develop theories in which the significant results do not depend on the choice of such a paremetrization. For example, in maths, the object that is "natural" to integrate are differential forms and not functions because the integral will not depend on the choice of the parametrization.

ADDENDUM 2 : What I mean by parametrization. Consider the following problem : you have a segment of curve $C$ and you wish to calculate its length (the simplest example : $C$ is a straight segment). By definition, the length is $int_C |gamma'(t)| dt$, where $gamma : [0,1] rightarrow mathbb{R}^3$ is a parametrization of the curve. (once again, you may replace $ mathbb{R}^3$ by $mathbb{R}$ if you prefer, in which case $C$ is a straight segment). But there are many different parametrizations $gamma$ (think of it as a point moving along the curve : you may move at different speeds, but you will still move along the curve). The number $|gamma'(t)|$ represents the speed of the parametrization at the time $t$. The number $L = int_C |gamma'(t)| dt$ is independent of the parametrization, as it should be since it represents the length. You can see this using the change of variable formula.

There's no geometric interpretation of $int f(x) mbox{d}(g(x))$ as far as I'm concern. However, it does have a meaning. Say, you want to take the differentiation of $sin(x)$ w.r.t x, and $sin (x)$, they give 2 different results:

• $frac{mbox{d} sin x}{mbox{d}x} = cos x$


• $frac{mbox{d} sin x}{mbox{d} sin x} = 1$

Another example is to take the derivatives of $x^4$ w.r.t x, and $x^2$ respectively.

• $frac{mbox{d} x^4}{mbox{d}x} = 4x^3$


• $frac{mbox{d} x^4}{mbox{d} x^2} = frac{mbox{d} (x^2)^2}{mbox{d} x^2} = 2x^2$

An antiderivative of $f$ w.r.t $x$ is some function $F$, such that $dfrac{mbox{d}F}{mbox{d}x} = f$

So:

• $int mbox{d}(x^2)$ is some function, such that its derivative w.r.t $x^2$ is 1, this family of function is, of course, $x^2 + C$.


• $int x^5mbox{d}(x^5)$ is some function, such that its derivative w.r.t $x^5$ is $x^5$, this family of function is, of course, $dfrac{x^{10}}{2} + C$, since: $frac{1}{2}dfrac{mbox{d}(x^{10} + C)}{mbox{d}(x^5)} = frac{1}{2}dfrac{mbox{d}((x^5)^2)}{mbox{d}(x^5)} + 0 = x^5$

When solving problems, you just need to remember that $int f(x)mbox{d}(g(x)) = int f(x).g'(x) mbox{d}x$, e.g, when you take some function out of d, you have to differentiate it, and vice versa, when you put some function into d, you'll have to integrate it, like this:

• $int x^5 mbox{d}(x^2) = int 2x^6 mbox{d}(x) = dfrac{2x^7}{7} + C$. (Take $x^2$ out of d, we differentiate it, and have $2x$)


• $int sin^2 x cos x mbox{d}x = int sin^2 x mbox{d}(sin x) = frac{sin ^ 3 x}{3} + C$ (Put $cos x$ into d, we have to integrate it to get $sin x$).

I found this relevant discussion on math overflow.

Visualisation of Riemann-Stieltjes Integral