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 $P$.


*

*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.
 A: 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
$$(x_1, x_2) = \pm\frac{1}{\sqrt{3}}(1, -1)$$
A: You already have two fine answers, under the assumption that such $x_1, x_2$ exist.  Your own argument can be completed to show that they do indeed exist.  You have already reached the equivalent statement:
$$a(x_1^3+x_2^3) + b(x_1^2+x_2^2-\frac23) + c(x_1 + x_2)=0$$
which must hold for all $a, b, c$.  Clearly $x_1+x_2=0$ will ensure it holds for any $a, c$, and then you need $x_1^2+x_2^2=\frac23$ if it has to hold for any $b$.  
Solving these two conditions, we do get two real solutions - i.e $x_1 = -x_2 = \pm \frac1{\sqrt3}$, so in fact we have also shown the existence.
A: (a) Choose $P(x)=x$. $\int_{-1}^1xdx=0=x_1+x_2$.
(b) Choose $P(x)=x^2$. $\int_{-1}^1x^2dx=\dfrac23=x_1^2+x_2^2=2x_1^2$.
$$x_1=\frac{1}{\sqrt{3}},x_2=-\frac{1}{\sqrt{3}}$$
