Are limits of integration defined with respect to the pronumeral or the differential? Consider the indefinite integral $$\int x^2 \,d(x^2).$$ It evaluates fairly easily to $\tfrac{x^4}{2} + C$. My question is about what happens when we start evaluating definite integrals with respect to these functions. In a specific example, how would $\displaystyle \int_0^2 x^2 \,d(x^2)$ be evaluated?
Would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot\left(x^2\right)^2\right]_0^2 =\frac{4}{2}-0=2 $$ with the limits applying to $x^2$, or would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot x^4\right]_0^2 =\frac{16}{2}-0=8 $$ with the limits applying to $x$.
Hope someone can clarify, thanks in advance!
 A: Confusion
If this is a Riemann integral I think I have to disappoint you. I don't think there is something like
$$\int_a^b f(x)\,\mathrm{d}g(x) = \int_a^b f(x)\,\frac{g(x)}{x}\,\mathrm{d}x.$$
It just does not make sense to define something like that. What is the problem with the integral on the right hand side? Why do we need a new notation? If there is, I take everything back. But it is not widely known, you should have defined it.
A similar integral
In fact there is a well-defined integral that looks like yours. It's called the Stieltjes integral, a generic term for the Riemann-Stieltjes integral and the Lebesgue-Stieltjes integral.
$$\int_a^bf(x)\,\mathrm{d}g(x).$$
You may look up it's definition here. I will not go into detail but it has a nice property if $g(x)$ is absolutely continous
$$\int_a^b f(x)\,\mathrm{d}g(x) = \int_a^b f(x)g'(x)\,\mathrm{d}x.$$
Stieltjes integral of your problem
In your example take $f(x)=g(x)=x^2$. Since $g(x) = x^2$ is absolutely continous we can write
$$\int_0^2 x^2\,\mathrm{d}x^2 = 2\int_0^2 x^3\,\mathrm{d}x=\left[\frac{x^4}{2}\right]_{0}^{2} = 8.$$
A: There are two ways to interpret $I= \int_0^2 x^2 \,d(x^2):$
$$I=\int_0^2 x^2 2x\,dx={{2x^4}\over 4}\big|_0^2=8$$
and, substituting $u=x^2,$
$$I=\int_0^4 u\,du={u^2\over 2}\big|_0^4=8.$$
