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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 $9x^2$.

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    $\begingroup$ $f(x) = 18x$ doesn't work since the LHS is $9x^2$ but the RHS is $324x^2$. However, $f(x) = \tfrac{1}{2}x$ does work. $\endgroup$
    – JimmyK4542
    Sep 24, 2020 at 6:38
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    $\begingroup$ 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. $\endgroup$ Sep 24, 2020 at 11:41
  • $\begingroup$ @JimmyK4542 If you substitute $x = 1$ into the LHS you get 0, so $f(x)= \frac{x-1}{2}$. $\endgroup$
    – 1123581321
    Sep 28, 2020 at 21:39
  • $\begingroup$ When I posted that comment, the lower bound in the integral on the LHS was $0$ not $1$. The question has since been edited. $\endgroup$
    – JimmyK4542
    Sep 28, 2020 at 21:45

1 Answer 1

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

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