Let $u\in C^{\infty}_{c}(\Omega)$ for some open $\Omega \subset \mathbb{R}^n$. We make the assumptions that $u$ is extended to all of $\mathbb{R}^n$ with it equaling $0$ outside of $\Omega$ and that with $x=(x_1, x_2, \ldots, x_n)$ we have $a <x_1<b$ for all $x\in \Omega$. For simplicity we denote $x=(x_1 ,x')$. Can someone please help explain to me how we get the last equality in the following computation?
$||u||^{2}_{L^{2}}=\int_{\mathbb{R}^n}u^2(x)dx = \int_{\mathbb{R}^{n-1}} \int_{a}^{b} (\int_{a}^{x_1} 2uu_{x_1}(t,x')dt)dx_1 dx = \int_{\mathbb{R}^{n-1}} \int_{a}^{b} (b-x_1) 2uu_{x_1}(x_1 , x') dx_1 dx$
Actually, it is a formula upon interchange the order of integration. $$ \int_0^x\int_0^t f(s)dsdt=\int_0^x\int_s^x f(s) dtds=\int_0^x(x-s)f(s)ds $$ Also, we can use integration by parts to get the same result: $$ \int_0^x(x-t)f(t)dt=\int_0^x(x-t)(\int_0^t f(s)ds)'dt=\int_0^x\int_0^t f(s)dsdt $$