Prove that $g(x)=\sqrt{x}$ is not an integral function on $[0,1]$, i.e. $\nexists f \in R[0,1]$ so that $$\sqrt{x}=\int_0^x f(t) \mathrm{d}t$$
I see that $f$ cannot be continuous, because if it were continuous, we would have that $f(x)=(\sqrt{x})'=\frac{1}{2\sqrt{x}} \forall x \in (0, 1)$ (because the FTC1 and $x \mapsto \sqrt{x}$ is continuous on $[0,1]$ and differentiable on $(0,1)$), but $f(x)=\frac{1}{2\sqrt{x}}$ is not Riemann-integrable on $[0, 1]$ (with any kind of definition of $f(0)$). But how could I prove that there isn't any non-continuous Riemann-integrable function wih that property?
Recall that a Riemann integrable function is bounded. If $\sqrt{x}$ were equal to the integral above, then
$$\sqrt{1/n}-\sqrt{1/2n}=\int_{1/2n}^{1/n}f(t)\,dt$$ The l.h.s is equal to $(1/2n)\frac{1}{2\sqrt{c_n}}$ for some $1/2n<c_n<1/n$ and the r.h.s is bounded above by some $C/n$ for some constant. This is a contradiction, because the order of magnitudes do not match when $n\to\infty$. More precisely, we would get: $$\frac{1}{4}\frac{1}{\sqrt{n}}<(1/2n)\frac{1}{2\sqrt{c_n}}\leq \frac{C}{n}$$ and for sufficiently large $n$ this is of course false.