Calculus Challenge

We note that $\begin{aligned} f(x) & = \; -\int_x^1 f'(t)\,dt \\ \int_0^1 x^2f(x)\,dx & = \; -\int_0^1 x^2 \int_x^1 f'(t)\,dt\,dx \; = \; -\int_0^1 f'(t) \int_0^t x^2\,dx\,dt \; = \; -\tfrac13\int_0^1 t^3 f'(t)\,dt \end{aligned}$ so that $\int_0^1 t^3 f'(t)\,dt \; = \; -1$ But this means that $\left(\int_0^1 t^3 f'(t)\,dt\right)^2 \; = \; \int_0^1 t^6\,dt \times \int_0^1 f'(t)^2\,dt$ and hence, by the integral version of the Cauchy-Schwarz Inequality, we deduce that $f'(t)$ is a constant multiple of $t^3$ . It is now easy to show that $f'(t) = -7t^3$ , and hence $f(t) = \tfrac74(1-t^4)$ . This makes the desired integral equal to $\boxed{\tfrac75}$ .

@Jon Haussmann

I found that $f(x) = -\frac{7}{4} x^4 + \frac{7}{4}$ works, but I have no idea if there are any other functions that do. @MS HT , can you give any insight?