# function f that is squared of integral

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 $$9x^2$$.

• $f(x) = 18x$ doesn't work since the LHS is $9x^2$ but the RHS is $324x^2$. However, $f(x) = \tfrac{1}{2}x$ does work. Sep 24, 2020 at 6:38
• Please don't delete your questions right after an answer was posted. That's rather rude towards the person who spent the effort to write an answer. Sep 24, 2020 at 11:41
• @JimmyK4542 If you substitute $x = 1$ into the LHS you get 0, so $f(x)= \frac{x-1}{2}$. Sep 28, 2020 at 21:39
• When I posted that comment, the lower bound in the integral on the LHS was $0$ not $1$. The question has since been edited. Sep 28, 2020 at 21:45

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 $$f(x)=\dfrac x2-\dfrac12$$.