function f that is squared of integral

Question

I’m trying to find some function $f$ that satisfies the following

$$\int_{1}^{x} f(u)\mathrm{d}u = f(x)^2.$$

I was thinking that maybe $f(x)=18x$ works, because the antiderivative of $18x$ is $9x^2$.

Answer 1

If$$(\forall x\in\Bbb R):\int_1^xf(u)\,\mathrm du=f^2(x),$$then $(\forall x\in\Bbb R):f(x)=2f(x)f'(x)$. So, take $f$ such that $f(1)=0$ and that $2f'(x)=1$. In other words, take $f(x)=\dfrac x2-\dfrac12$.