Let $f : [0, 1] \to \mathbb{R}$, and continuous. The goal is to prove that there exists a $c$ in $[0, 1]$ such that
$$\int_{0}^{1}x^{4}f(x)\, dx = \dfrac{1}{5}f(c)$$
So far however, what I have is this. By MVT we have that $ \int_{0}^{1}x^{4}f(x)\,dx = (1 - 0) \cdot c^{4}f(c)$ for $c$ in $(0, 1)$, which is strictly less than $ f(c)$ for some $c$ in $[0, 1]$. This is just an inequality however and not the equality wanted, which is stronger. Are there any good hints for getting an actual equality?
Edit: as commenters here pointed out, this directly follows from a more general version of the mean value theorem. But I am also interested in more elementary ways to prove it, possibly using other techniques like the intermediate value theorem.
Assume $f(x)\not\equiv 0$. Consider $g(x)=\int_0^1 y^4\left[f(y)-f(x)\right]dy$. Since $\{f(x):x\in[0,1]\}$ is compact, its maximum and minimum are attained at two (not necessarily unique) points $x_\text{max}$ and $x_\text{min}$, respectively. The function $g(x)$ is continuous and we have $$ { g(x_\text{min})<0\quad,\quad g(x_\text{max})>0. } $$Therefore, there exists a $c\in [x_\text{min},x_\text{max}]$ such that $g(c)=0$. This proves what we wanted.