In $\mathbb{R}^n$ for $n>1$ the substitution rule for an injective, differentiable function $\phi$ is given by

$$\int_{\phi(U)} f(\mathbf{v})\, d \mathbf{v} = \int_U f(\phi(\mathbf{u})) \left|\det(\operatorname{D}\phi)(\mathbf{u})\right| \,d \mathbf{u}.$$

In the one-dimensional case is the absolute value missing and results as

$$\int_{\phi([a,b])} f(\mathbf{v})\, d \mathbf{v} = \int_a^b f(\phi(\mathbf{u})) \phi'(u) \,d \mathbf{u}.$$

Why is the right hand side not $$\int_a^b f(\phi(\mathbf{u}))\, |\phi'(u)| \, \,d \mathbf{u} \quad?$$

  • $\begingroup$ It is that, as long as you adopt the convention that when you integrate over $\phi([a,b])$, you always integrate from the lower bound to the higher one, rather than from $\phi(a)$ to $\phi(b)$, which sometimes results in a "backwards" integral. A reversal occurs when $\phi$ is decreasing. $\endgroup$ – user208259 Jan 21 '15 at 14:29

If $\varphi:[a,b]\to\mathbb R$ is differentiable in $(a,b)$, continuous in $[a,b]$ and injective, then it is monotone. If $\varphi$ is increasing, then $\varphi'\geq0$, $\varphi(a)\leq\varphi(b)$, and $|\det \varphi'|=\varphi'$, so the one-dimensional case follows from the higher-dimensional case.

If $\varphi$ is decreasing, then $\varphi'\leq0$, $\varphi(a)\geq\varphi(b)$ and $|\det\varphi'|=-\varphi'$. In this case we have \begin{align*} \int_{\varphi(a)}^{\varphi(b)} f(v)dv=-\int_{\varphi([a,b])}f(v)dv=-\int_a^bf(\varphi(u))(-\varphi'(u))du=\int_a^bf(\varphi(u))\varphi'(u)du. \end{align*} So in this case, we have again, that the one-dimensional case is a special case of the higher-dimensional formula.


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