74th Romanian Mathematical Olympiad

Consider $g:[0,1]\to \mathbb{R}$ defined by $g(x)=xf(x)$, for any $x\in [0,1]$. Being continuous, $g$ has an antiderivative $G:[0,1]\to \mathbb{R}$ with $G(0)=0$. Then $g(1)=f(1)=0$ and $${\int }_0^{1}xf(x)dx={\int }_0^{1}g(x)dx=G(1),$$ and $${\int }_0^{1}xf(tx)dx=\frac{1}{t^2}\cdot {\int }_0^{1}tg(tx)dx=\frac{1}{t^2}\cdot G(t),$$ for $t\in (0,1]$. Then $$\frac{1}{1-t}\cdot {\int }_0^{1}x(f(tx)-f(x))dx=\frac{1}{1-t}\cdot \left(\frac{1}{t^2}\cdot G(t)-G(1)\right)=\frac{1}{t^2}\cdot \frac{G(t)-t^{2}G(1)}{1-t}.$$ Let $u,v:[0,1]\to \mathbb{R}$ defined by $u(t)=G(t)-t^{2}G(1)$, and $v(t)=1-t$, respectively, for $t\in [0,1]$. Both are differentiable and $u^{\prime }(t)=g(t)-2tG(1)$, $v^{\prime }(t)=-1$, for $t\in (0,1)$, and: $$\underset{t\to 1^{+}}{\lim }u(t)=\underset{t\to 1^{+}}{\lim }v(t)=0\text{and}\underset{t\to 1^{+}}{\lim }\frac{u^{\prime }(t)}{v^{\prime }(t)}=2G(1)-g(1)=2G(1).$$ By the l'Hospital rule, $$\underset{t\to 1^{+}}{\lim }\left(\frac{1}{1-t}\cdot {\int }_0^{1}x(f(tx)-f(x))dx\right)=\underset{t\to 1^{+}}{\lim }\frac{1}{t^2}\cdot \frac{u(t)}{v(t)}=\underset{t\to 1^{+}}{\lim }\frac{u(t)}{v(t)}$$ so $$ \lim_{t \to 1^+} \frac{u'(t)}{v'(t)} = 2G(1) = 2 \int_{0}^{1} x f(x) \, dx.