Let $P(x)$ be any polynomial of degree at most $3$.$\int_{-1}^{1}P(x)dx = P(x_1)+P(x_2)$, where $x_1$ and $x_2$ are independent of the polynomial P. Let $P(x)$ be any polynomial of degree at most $3$. It can be shown that there
are numbers $x_1$ and $x_2$ such that $$\int_{-1}^{1}P(x)\,\operatorname{d}x = P(x_1)+P(x_2),$$ where $x_1$ and $x_2$ are independent of the polynomial $P$.


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*Show that $x_1=−x_2$.

*Find $x_1$ and $x_2$.



I let $P(x) := ax^3+bx^2+cx+d$
According to the question,
$$\int_{-1}^{1}\left(ax^3+bx^2+cx+d\right)\,\operatorname{d}x=ax_1^3+bx_1^2+cx_1+d+ax_2^3+bx_2^2+cx_2+d$$
$$\frac{2b}{3}+2d=ax_1^3+bx_1^2+cx_1+d+ax_2^3+bx_2^2+cx_2+d$$
I am stuck here. I dont know how to solve further. Please help me.

By expanding and simplifying the equation, we get
$$0a + \frac{2}{3}b + 0c + 2d = a(x_1^3 + x_2^3) + b(x_1^2 + x_2^2) + c(x_1 + x_2) + 2d$$
Since this equation holds true for all values of $a, b, c, d$, we sub $(a, b, c, d) = (0 , 0, 1, 0)$ to get
$$0 = x_1 + x_2$$
$$x_1 =- x_2$$
as required.
Now to find the values of $x_1, x_2$, sub $(a, b, c, d) = (0, 1, 0, 0)$ to get
$$\frac{2}{3} = x_1^2 + x_2^2 = 2x_1^2$$
$$x_1^2 = \frac{1}{3}$$
$$x_1 = \pm \frac{1}{\sqrt{3}}$$
so that
$$(x_1, x_2) = \pm\frac{1}{\sqrt{3}}(1, -1)$$
