Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.
Problem 4
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 $\displaystyle{\int_{-1}^{1} P(x) dx = P(x_1) + P(x_2)}$, where $x_1$ and $x_2$ are independent of the polynomial $P$.
- Show that $x_1 = - x_2$.
- Find $x_1$ and $x_2$.
Hints
Hint #1 for part (a)
Since you need to show a linear relationship between $x_1$ and $x_2$, what kind of a function might you pick for $P(x)$?
Hint #2 for part (a)
Consider $P(x) = x$, which you can do since the statement given in the problem is true for all $P(x)$ up to degree $3$.
Hint #1 for part (b)
You may have seen that using a linear function gets you the linear relationship between $x_1$ and $x_2$, but using another linear function won’t get you much further. Consider a non-linear function $P(x)$ — what’s a very simple non-linear function you can consider that’s up to degree $3$?
Hint #2 for part (b)
Consider $P(x) = x^2$, the simplest non-linear function in $x$.
Solution
(a) Show that $x_1 = -x_2$
Open to see the solution; try solving it first or see the hints above.
Since the given statement is true for all polynomials up to degree 3, let’s take a very simple polynomial $P(x) = x$.
Then:
$$ \begin{align*} \int_{-1}^1 P(x) dx & = P(x_1) + P(x_2) & \quad \text{given} \\ \int_{-1}^1 x dx & = x_1 + x_2 & \quad \text{substitute $P(x) = x$} \\ \left( \frac{1}{2}x^2 + C \right) \Big|_{-1}^1 & = x_1 + x_2 & \quad \text{integrate} \\ \frac{1}{2} \cdot 1^2 - \frac{1}{2} \cdot (-1)^2 & = x_1 + x_2 & \quad \text{evaluate the integral} \\ 0 & = x_1 + x_2 & \text{simplify} \\ x_1 & = -x_2 \qquad \blacksquare \\ \end{align*} $$(b) Find $x_1$ and $x_2$
Open to see the solution; try solving it first or see the hints above.
Let’s consider another polynomial: $P(x) = x^2$. Then:
$$ \begin{align*} \int_{-1}^1 P(x) dx & = P(x_1) + P(x_2) & \quad \text{given} \\ \int_{-1}^1 x^2 dx & = x_1^2 + x_2^2 & \quad \text{substitute $P(x) = x^2$} \\ \left( \frac{1}{3} x^3 + C \right) \Big|_{-1}^1 & = x_2^2 + x_2^2 & \quad \text{using $x_1 = -x_2$} \\ \frac{1}{3} \cdot 1^3 - \frac{1}{3} \cdot (-1)^3 & = 2 x_2^2 & \quad \text{evaluate integral} \\ \frac{2}{3} & = 2 x_2^2 & \quad \text{simplify} \\ \frac{1}{3} & = x_2^2 & \quad \text{divide both sides by 2}\\ x_2 & = \pm \frac{1}{\sqrt{3}} & \quad \text{take square root of both sides} \\ \end{align*} $$Given that $x_1 = -x_2$, we have the following symmetric solutions:
$\boxed{x_1 = \frac{1}{\sqrt3}, x_2 = -\frac{1}{\sqrt{3}}} \quad$ or $\quad \boxed{x_1 = -\frac{1}{\sqrt3}, x_2 = \frac{1}{\sqrt{3}}} \qquad \blacksquare$