How to prove an integral equality.

Question

I was reading this post, i would like to know how to prove the following inequality: $$\frac{1}{2}\int_a^b\int_a^b (f(x)g(y) - g(x)f(y))^2\,dx\,dy \\= \int_a^b f^2(x)\,dx\int_a^b g^2(x)\,dx - \left(\int_a^b f(x)g(x)\,dx\right)^2$$ the person who gave that equality said that the integrand $\displaystyle (f(x)g(y) - g(x)f(y))^2 \ge 0$, hence the equality follows. But I never been doing something like that. Any tips.

Answer 1

Left hand side: $(1/2)\int_a^b\int_a^b(f(x)g(y)-f(y)g(x))^2dxdy={(1/2)\int_a^bf(x)^2dx\int_a^bg(y)^2dy}+{(1/2)\int_a^bf(y)^2dy\int_a^bg(x)^2dx}{-\int_a^bf(x)g(x)dx\int_a^bf(y)g(y)dy}$

Since $y$ is a dummy, it can be replaced by $x$ to get right hand side..