I got the following question in multivariable calculus class:
Let $f:[a,b]\to \mathbb{R}$ be a continuous function. Prove that $$2\int_a^b\int_x^bf(x)f(y)dydx=\left(\int_a^bf(x)dx\right)^2.$$
I was able to prove this using the fundamental theorem of calculus:
Define $F:[a,b]\to\mathbb R$ by $F(t)=\int_a^tf$. Then, $F$ is differentiable and $F'=f$. Therefore, we can write
$$2\int_a^b\int_x^bf(x)f(y)dydx=2\int_a^bf(x)\ \left( \int_x^bf(y)dy\right)dx=2\int_a^bf(x)[F(b)-F(x)]dx.$$
We can now split this integral into the sum of two integrals:
$$2\int_a^bF(b)f(x)dx +\int_a^bF(x)f(x)dx,$$
which are both immediate integrals. It can be easily shown that the sum above yields $$(F(b)-F(a))^2,$$
and that is exactly $\left(\int_a^bf(x)dx\right)^2$.
Nonetheless, in class we have almost always worked with diffeomorfisms, and I suspect there is a way to prove this result without using FTC, but using some sort of change of variables instead.
When graphing the integration region we cleary see it is the triangle with vertices $(a,a),(a,b)$ and $(b,b)$, so intuition tells me that if we could transform this triangle into the square with vertices $(a,a),(a,b),(b,a)$ and $(b,b)$ via a diffeomorfism, perhaps things could cancel out and the desired result would show up. However, I am unable to find such a diffeomorfism.
If you find any other solution to this problem that involves diffeomorfisms, please share it too! Thank you.
