Find the increasing function $f:\mathbb{R} \to \mathbb{R}$ such that $\int_{0}^{x}f(t)\,dt=x^2$ for all $x\in \mathbb{R}$

By Lebesgue differentiation theorem, we have $F' = f$ almost everywhere. So we have $f(x) = 2x$ almost everywhere. In particular, for at least two different points $x_1 < x_2$, we have $f(x_1) = 2 x_1$ and $f(x_2) = 2 x_2$, so $f$ can't be non-increasing.

Now, if $f(x_0) \neq 2 x_0$ for some $x_0$, then either $f(x_0) < 2 x_0 - \varepsilon$ or $f(x_0) > 2 x_0 + \varepsilon$. In the former case, there is some $y \in [x_0 - \varepsilon / 2, x_0)$ s.t. $f(y) = 2 y$, and thus $f(y) > f(x_0)$ while $y < x_0$ - so $f$ can't be non-decreasing. Similarly, if $f(x_0) > 2 x_0$, for some $y$ we have $y > x_0$ but $f(x_0) < f(y)$.

So, if $f(x) \neq 2x$ for some $x$, then $f$ can't be non-decreasing. $f$ also can't be non-increasing, so if $f$ is monotone, it should be equal to $2x$.

More generally, if $F(0) = 0$, $F$ is everywhere differentiable and $F'$ is continuous and monotone, then there is only one monotone function - $F'$ s.t. $F(x) = \int_0^x f(t)\,dt$.

If $F$ is just absolute continuous, then there can be several solutions - for example, if $F(x) = |x|$, then for any $f$ s.t. $f(x) = \operatorname{sgn}(x)$ if $x \neq 0$ and $f(0) \in [-1, 1]$ the equality will hold, and $f$ is monotone.

$f(x)=2x$ whenever $f$ is continuous at $x$. Since $f$ is increasing it is continuous except at a countable number of points Hence it is continuous on a dense set. Given any $x$ choose an increasing sequence $(x_n)$ and a decreasing sequence $(y_n)$ both tending to $x$ such that $f$ is continuous at each $x_n$ and at each $y_n$. Then $2x_n=f(x_n)\leq f(x) \leq f(y_n)=2y_n$. Letting $n \to \infty$ you get $f(x)=2x$.