Prove there is $c\in (0,1)$ such that $c^3f(c)+cf(c)-1=0$ Let $f:[0,1] \rightarrow \mathbb{R}$ continuous such that $\int_0^1 xf(x)\,dx=\frac{\pi}{4}$. Prove that there is $c\in (0,1)$ such that $c^3f(c)+cf(c)-1=0$.
Here is what I think, MVT for integrals gives that there is some $c\in(0,1)$ such that $cf(c)=\frac{\pi}{4}$, but I doubt this is relatable with the condition to prove. Maybe someone can give me a hint?
 A: Since $$\displaystyle\frac{\pi}{4}=\int_0^1\frac{1}{1+x^2}\, dx$$ 
we can write the condition as:
$$\int_0^1\left(xf(x)-\frac{1}{1+x^2}\right)\,dx=0$$
and from the mean value theorem, there exists some $c\in (0,1)$ such that:
$$cf(c)-\frac{1}{1+c^2}=0$$
This is equivalent with the equality to prove.
A: You can get this somewhat easily by examining cases: since $f$ is continuous, so is the function taking $c$ to $c^3f(c)+cf(c)-1$, meaning that if this quantity is never zero, it must either always be negative or positive. So there are two cases: If this quantity is always negative we get
$$c^3f(c)+cf(c)-1<0$$
which, for $c\in (0,1]$ rearranges to
$$cf(c)<\frac{1}{c^2+1}.$$
The other case would rearrange to
$$cf(c)>\frac{1}{c^2+1}.$$
However, if you integrate both sides of either inequality over $[0,1]$, you get that the integral on the right is $\int_{0}^1\frac{1}{c^2+1}\,dc=\frac{\pi}4$ and that the integral on the left is either greater than or less than this - but cannot be equal because the inequality of functions is strict everywhere. This is basically an application of the intermediate value theorem.
