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.


4 Answers 4


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 $\dfrac{\mbox{d}F}{\mbox{d}x} = f$


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

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 $\sin x$).
  • $\begingroup$ Can we you this rule for definite integrals as well? Or do we have to modify the limits as we do it for integration by substitution? $\endgroup$
    – Rolandos
    Commented Dec 30, 2022 at 13:01

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.

  • 3
    $\begingroup$ I think if our function $g(x)$ is continuously differentiable, so $$d(g(x))=g'(x)dx$$ and so $$\int f(x)d(g(x))=\int f(x)g'(x)dx$$ $\endgroup$
    – Mikasa
    Commented Feb 5, 2013 at 12:40
  • $\begingroup$ @Glougloubarbaki, could you please elaborate what "parametrization of integral" means. $\endgroup$
    – user45099
    Commented Feb 5, 2013 at 12:49
  • 1
    $\begingroup$ as @BabakSorouh noted, your formula was incorrect. $\endgroup$
    – Albert
    Commented Feb 5, 2013 at 13:22

I found this relevant discussion on math overflow.

Visualisation of Riemann-Stieltjes Integral


This is an old question with a marked answer, but there is a better one.

This is how you find the area under the curve for a function with respect to another function given some hidden parameter. One such application is the receiver operating charateristic (ROC) Receiver Operating Characteristic

In the ROC, we are interested in how the dependent variables $f(t)$ and $g(t)$ vary with $t$. We have the probability of false detection on the $x$-axis and the probability of detection on the $y$-axis. So, the integral of $f(t)$ with respect to $g(t)$ is an excellent measure of how likely we are to correctly identify the presence of some signal. There are likely other similar interpretations, but this is one I'm familiar with.


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