Let $f:[0, 1] \rightarrow \mathbb{R} $ a differentiable function such that $$ \int_{0}^{1} f(x) \, dx = 0 \, . $$ Prove that there exists a $c \in (0, 1)$ such that $$ f'(c) +9 \int_{0}^{1/3} f(x) \, dx = 0 \, . $$
I think that the solution of this problem can be obtained by applying the mean value theorem for integrals. Since the integral of $f$ on $[0, 1]$ is $0$, we obtain that $$ 0 = \int_{0}^{1} f(x) \, dx=f(a) (1-0) = f(a) $$ for some $a\in (0, 1)$. Next $$ 0 = \int_{0}^{1} f(x) \, dx=\int_{0}^{1/3} f(x) \, dx + \int_{1/3}^{1}f(x) \, dx \, , $$ so we have to prove that there exists a point $c \in (0, 1) $ such that $$ f'(c) = 9 \int_{1/3}^{1}f(x) \, dx \, . $$ From now on I don't have any idea. I just verified that for the function $f(x) =2x-1$, the assertion is true.
The function $$ F(x) = \int_0^x f(t) \, dt $$ satisfies $F(0) = F(1) = 0 $, so that $$ g(x) = F(x) - \frac 9 2 x(1-x) F(\frac 13) $$ satisfies $g(0) = g(1/3) = g(1) = 0$. (The idea is to subtract from $F$ the quadratic polynomial which interpolates $F$ at $x = 0$, $x= 1/3$, and $x= 1$.)
Repeated application of Rolle's theorem gives $g''(c) = 0$ for some $c \in (0, 1)$, and that is the desired conclusion because $$ g''(x) = F''(x) + 9 F(\frac 1 3) = f'(x) + 9 \int_0^{1/3} f(x)\, dx \, . $$
If we replace $1/3$ by an arbitrary $a \in (0, 1)$ then the same method gives the existence of an $c\in (0, 1)$ such that $$ f'(c) + \frac{2}{a(1-a)} \int_0^a f(x) \, dx = 0 \, . $$