Is $\int_{\mathbb{R}} |F(x)-G(x)|^2(f(x)-g(x))dx =0$ true, for $f,g$ probability densities and $F,G$ the corresponding distribution functions?

Integrating by parts gives \begin{align*} I&=\int_{\mathbb{R}} (F(x)-G(x))^2(f(x)-g(x))\,dx \\ &=(F(x)-G(x))^2 \bigg( \int (f(x)-g(x))\,dx \bigg) \bigg|_{-\infty}^\infty - \int_{\mathbb{R}} (F(x)-G(x)) \frac d{dx}\big( (F(x)-G(x))^2 \big) \,dx \\ &=(F(x)-G(x))^2 (F(x)-G(x)) \big|_{-\infty}^\infty - \int_{\mathbb{R}} (F(x)-G(x)) \cdot 2(F(x)-G(x))(f(x)-g(x) \big) \,dx \\ &=(0-0) - 2I \end{align*} and hence $I=0$. (The boundary term vanishes because $\lim_{x\to\infty} F(x) = \lim_{x\to\infty} G(x) = 1$ and $\lim_{x\to-\infty} F(x) = \lim_{x\to-\infty} G(x) = 0$.)