An interesting relation of integrals

Question

Proof that there does not exist an $F: \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ such that

$\int_{0}^{1} f(x)g(x)dx =F(\int_{0}^{1}f(x)dx, \int_{0}^{1}g(x)dx).$

Answer 1

Hint: First consider $f(x)=g(x) = x-\frac{1}{2}$. Then consider $f(x)=g(x)=\frac{1}{2}(x-\frac{1}{2})$.

Hint2: Consider $f(x) = x-\frac{1}{2}+a$ and $g(x) =x-\frac{1}{2}+b$. Then $f(x)=a$ and $g(x) = b$.

Answer 2

Let $a,b \in \mathbb R$, $f=2a$ for $x<1/2$, 0 elsewhere, $g=2b$ on $(1/2,1)$, 0 elsewhere. We get $F(a,b)=0$. Obviously a contradiction.