reducted

@above: There is a mistake in your proof. You can check with the function $f(x)=1-x.$

with Ramanujan, Galois, and Erdös $L=2\int_{0}^{1} xf(x)dx$ ok for f(x)=1-x or f(x)=cos(pi*x/2)

@above: What is the reference to 'Ramanujan, Galois and Erdös' supposed to mean?? Anyway, nice problem! (And also: How smart are grade-12 students in Romania???) The condition $f(1) = 0$ is somewhat strange, since the integral will not change if a constant is added to $f$. Heuristically you may want to switch integral and limit to obtain $\int_0^1 -x^2 f'(x) dx$ (assuming that $f$ is differentiable, which obviously is not given) and then proceed evaluating the integral via integration by parts to obtain the answer given above. Or heuristically, you may want to apply the definition of the derivative (assuming that the integral is differentiable w.r.t. $t$), so the answer is $- \frac d {dt} \int_0^1 x(f(tx) - f(x)) dx)|_{t = 0}$ and proceed by switching integration and differentiation (again assuming that $f$ is differentiable, ....) Actually the latter heuristics is not so bad: Since you only need to differentiate w.r.t. $t$, getting $t$ out of $f(tx)$ would help. This is a possible motivation for the following solution: Note that $\int_0^1 x f(tx) dx = \int_0^t \frac 1 {t^2} s f(s) ds$ (substituting $s = xt$). So let $G(x) := \int_0^x s f(s) ds$, then the given integral equals $\frac 1 {t^2} G(t) - G(1)$. Noting that $G$ is differentiable (by the FTC) the given limit exists and equals $$ - \frac d {dt} \frac 1 {t^2} G(t)|_{t = 1} = \frac 2 {t^3} G(t) - \frac 1 {t^2} G'(t)|_{t = 1} = 2G(1) - G'(1) = 2G(1) = 2 \int_0^1 x f(x) dx. $$since $G'(1) = x f(x)|_{x = 1}= 0$. This coincides with the answer given above.