Integration of a function with respect to another function.

Question

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.

Answer 1

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.

Answer 2

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, $x^2 + C$.
• $\int x^5\mbox{d}(x^5)$ 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: $\frac{1}{2}\dfrac{\mbox{d}(x^{10} + C)}{\mbox{d}(x^5)} = \frac{1}{2}\dfrac{\mbox{d}((x^5)^2)}{\mbox{d}(x^5)} + 0 = x^5$

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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 $\sin x$).

Answer 3

I found this relevant discussion on math overflow.

Visualisation of Riemann-Stieltjes Integral