Are there two distinct intervals having the same length where the area under the curve of a polynomial function is the same? This is a follow-up for my previous question,

Is there always a point $(h,k)$ in a polynomial function $P(x)$ for which $P$ becomes odd or even by translating $(h,k)$ to $(0,0)$?


Before anything else, define $P$ as the basis polynomial where $P$ is odd or even, and $f$ for polynomials $P$ where it is translated from $(0,0)$ to $(h,k)$. If $P$ does not exist, let $P = f$.
The linked question is the same as asking if $P$ exists for all $f$ and we got the answer that $P$ does not always exist.
Now, we know that for a closed interval $[a,b]$, the area bounded by $x = a$, $x = b$, $y = 0$, and $y = P(x)$ is $\int_a^b |P(x)|\,dx$. Now, are there two distinct intervals having the same length such that the area will be the same?
All $f$ of degree zero to three can be translated to the form $P(x) = x^n$. For higher degrees, this only occurs when the roots of $P$ are evenly spaced. In this question, let's remove the case where the degree of $f$ is from zero to three, or if $f$ can be translated in such a way where the roots of $P$ are evenly spaced.

I think that there are distinct intervals, but I don't know how to prove it yet. Any help would be appreciated. By the way, this is just my mere curiosity as a result of experimenting with polynomials on Desmos.
 A: Yes. Fixing an interval length $c$, define
$$g(x)=\int_x^{x+c}|P(x)|dx.$$
Note that $g(x)$ is continuous and tends towards $+\infty$ as $x$ goes to $\pm\infty$. As a result, it has some minimum value $m$, reached at some $x_0$ and any $M>m$ is reached by $g(x)$ in each of $(-\infty,x_0)$ and $(x_0,\infty)$ by the intermediate value theorem.
A: How about the following, I will assume for now that the polynomial has odd degree. Fix $\alpha>0$, this will be our target area, and I will show there are two intervals of equal width giving this area. For an interval $[a,b]$, I will denote the area enclosed by this integral $A(a,b)$.
Since the polynomial has odd degree, it crosses the $x$-axis at some point, say $x_0$. Choose $\varepsilon>0$ such that $A(x_0-\varepsilon,x_0+\varepsilon)<\alpha$. This is possible since taking the limit as $\varepsilon\to 0$ this area clearly vanishes. Now translate this interval a long way to the left and right of $x_0$. Then two observations

*

*The area varies continuously as a function of the distance we translate the interval since polynomials are continuous, and

*The area is unbounded as we translate the interval to $\pm\infty$ since the polynomial is unbounded.

Therefore there will be a point as we translate the interval in either direction when the area will equal our target value (apply the intermediate value theorem). This will give the two intervals you want. If in addition you want the intervals to be disjoint, it seems intuitively clear that taking $\varepsilon$ sufficiently small to begin with allow you to achieve this, but I'll leave that to you to try and prove.
No suppose that the degree of the polynomial is even. It may be that the the polynomial meets the $x$-axis, in which case proceed as above. Otherwise, let $x_0$ be a local extreme point of the polynomial, and proceed in the same way.
Remark: in fact, I think my construction will work for any choice of $x_0$, but I think the intuitive picture of what's going on is clearest if you make the choice as above.
