# 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?

• Yes, continuous. Sorry, I forgot to mention it. – user759366 Mar 13 '20 at 19:58

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.

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.