Problem 1. Determine the continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ with the property that

$$
\left(a^{2}+a b+b^{2}\right) \int_{a}^{b} f(x) \mathrm{d} x=3 \int_{a}^{b} x^{2} f(x) \mathrm{d} x
$$

for any $a, b \in \mathbb{R}$.
Solution. We choose $a=0$ and $b=t>0$. Then

$$
t^{2} F(t)=3 \int_{0}^{t} x^{2} f(x) \mathrm{d} x
$$

where $F(t)=\int_{0}^{t} f(x) \mathrm{d} x$.

1 point

By differentiation, we get $2 t F(t)+t^{2} f(t)=3 t^{2} f(t)$, hence $2 t(t f(t)-F(t))=0$, which means $\left(\frac{F(t)}{t}\right)^{\prime}=0$, for any $t \in(0, \infty)$. 3 points

It follows that the function $g(t)=\frac{F(t)}{t}$ is constant on the interval $(0, \infty)$, so $F(t)=k t$, that is,

![](https://cdn.mathpix.com/cropped/2024_06_07_5b0be9c0b0d77c339529g-3.jpg?height=49&width=1556&top_left_y=1537&top_left_x=282)

Proceeding similarly on the interval $(-\infty, 0)$, we find that there exist constants $k_{1}, k_{2}, k$ such that

$$
f(t)= \begin{cases}k_{1}, & t>0 \\ k, & t=0 \\ k_{2}, & t<0\end{cases}
$$

## 1 point

Since $f$ is continuous, it follows that $k_{1}=k_{2}=k$, so $f(t)=k$, for any $t \in \mathbb{R}$. It is immediately verified that constant functions on $\mathbb{R}$ satisfy the relation in the statement.