## Problem 1

Find all continuous functions $f:[1,8] \rightarrow R$, such that

$$
\int_{1}^{2} f^{2}\left(t^{3}\right) d t+2 \int_{1}^{2} f\left(t^{3}\right) d t=\frac{2}{3} \int_{1}^{8} f(t) d t-\int_{1}^{2}\left(t^{2}-1\right)^{2} d t
$$
Solution. Using the substitution $t=u^{3}$ we get

$$
\frac{2}{3} \int_{1}^{8} f(t) d t=2 \int_{1}^{2} u^{2} f\left(u^{3}\right) d u=2 \int_{1}^{2} t^{2} f\left(t^{3}\right) d u
$$

Hence, by the assumptions,

$$
\int_{1}^{2}\left(f^{2}\left(t^{3}\right)+\left(t^{2}-1\right)^{2}+2 f\left(t^{3}\right)-2 t^{2} f\left(t^{3}\right)\right) d t=0
$$

Since $f^{2}\left(t^{3}\right)+\left(t^{2}-1\right)^{2}+2 f\left(t^{3}\right)-2 t^{2} f\left(t^{3}\right)=\left(f\left(t^{3}\right)\right)^{2}+\left(1-t^{2}\right)^{2}+2\left(1-t^{2}\right) f\left(t^{3}\right)=\left(f\left(t^{3}\right)+1-t^{2}\right)^{2} \geq$ 0 , we get

$$
\int_{1}^{2}\left(f\left(t^{3}\right)+1-t^{2}\right)^{2} d t=0
$$

The continuity of $f$ implies that $f\left(t^{3}\right)=t^{2}-1,1 \leq t \leq 2$, thus, $f(x)=x^{2 / 3}-1,1 \leq x \leq 8$.

Remark. If the continuity assumption for $f$ is replaced by Riemann integrability then infinitely many $f$ 's would satisfy the given equality. For example if $C$ is any closed nowhere dense and of measure zero subset of $[1,8]$ (for example a finite set or an appropriate Cantor type set) then any function $f$ such that $f(x)=x^{2 / 3}-1$ for every $x \in[1,8] \backslash C$ satisfies the conditions.