# Finding the value of $3(\alpha-\beta)^2$ if $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all $f$

Let $f$ be a polynomial of degree $n$ at most $3$ such that there exists some $\alpha,\beta$ satisfying $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all such $f$. Find the value of $3(\alpha-\beta)^2$

Attempt: We need to find the value of $3(\alpha-\beta)^2$. Now, this expression reminds of the function $f(x) = (x-\beta)^3$. Differentiating this function at $x=\alpha$ will give us the desired result.

Hence, we just need to know the derivative of $k(x)=(x-\beta)^3$ at every point.

However, I am unable to move forward with this intuition. Could anyone please help me to move ahead in this problem.

Thank you very much for your help in this regard.

#### Solutions Collecting From Web of "Finding the value of $3(\alpha-\beta)^2$ if $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all $f$"

Since

$$\frac{8}{3} = \int_0^2 x^2dx = \alpha^2+\beta^2$$
$$2 = \int_0^2 x \ dx = \alpha + \beta$$

You get

$$(\alpha-\beta)^2 = 2(\alpha^2+\beta^2) -(\alpha+\beta)^2 = 2 \cdot \frac{8}{3} – 4 = \frac{4}{3}$$
so $3(\alpha-\beta)^2 = 4$

However, note that from this one is able to compute $\{\alpha, \beta\} = \{ 1\pm\frac{1}{\sqrt{3}}\}$. And now
$$\int_0^2 x^3 \ dx = 4 \neq \alpha^3 + \beta ^3$$

so, the formula can be valid only for polynomials of degree $\leq 2$.

$$f(x)=ax^3+bx^2+cx+d\\ \int_0^2f(x)dx=4a+8b/3+2c+2d\\ =a\alpha^3+b\alpha^2+c\alpha+d+a\beta^3+b\beta^2+c\beta+d\text{ for all a,b,c,d}\\ 4=\alpha^3+\beta^3,8/3=\alpha^2+\beta^2,2=\alpha+\beta,2=1+1$$
Try to calculate $3(\alpha-\beta)^2$ from those equations.