Area between $y=x^{6} - x$ and $x = y^{6} - y$?

Question

I realize that they are the same equations but with $x$ and $y$ switched, but I'm stuck from there on how to approach the problem.

Here's one possible approach: Find where the two curves intersect, and call that point $P_1$. Find the area between one of the curves and the line, $L$, going through the origin and $P_1$. Simply double that area, and that's your answer. — 高田航, Jul 04 '17 at 23:01
https://www.desmos.com/calculator/vrbqgbiyfd Here is the graph ... — Donald Splutterwit, Jul 04 '17 at 23:05

Simply Beautiful Art · Accepted Answer · 2017-07-05T01:02:32.607

3

Inspired by 泥九奈加's comment.

Note these functions are symmetric across the line $y=x$ (by definition, since they are inverses). Likewise, solving for $f(x)=x$, we get

$$x^6-x=x\implies x^6=2x\implies\begin{cases}x=0\\x^5=2\implies x=\sqrt[5]2\end{cases}$$

Thus, the given integral is given symmetrically by

$$I=2\int_0^{\sqrt[5]2}x-(x^6-x)~\mathrm dx$$

And I imagine you can take it from here?

edited Jul 05 '17 at 01:02

answered Jul 04 '17 at 23:34

Simply Beautiful Art

74,685

@MichaelRozenberg Which statement? That a function and its inverse are symmetric across $y=x$? As far as I can tell, $y=-x$ and $x=-y$ are symmetric across $y=x$. – Simply Beautiful Art Jul 05 '17 at 00:59
Simply Beautiful Art $(-5,5)$ is also common point. I think your solution is not full. – Michael Rozenberg Jul 05 '17 at 01:03
@MichaelRozenberg What do you mean? I'm fairly certain that $f(x)\equiv x^6-x=x$ has no other real solutions and $f(f(x))=x$ has no other solutions either. To what does $(-5,5)$ have anything to do here? – Simply Beautiful Art Jul 05 '17 at 01:20
OK. I'll post a right solution. – Michael Rozenberg Jul 05 '17 at 01:21
1

it looks good to me @MichaelRozenberg – Saketh Malyala Jul 05 '17 at 01:22
@MichaelRozenberg Well, I'll check your solution in the morning. Good night! – Simply Beautiful Art Jul 05 '17 at 01:33
@Simply Beautiful Art Try $y=x^2-2x$ and $x=y^2-2y$. – Michael Rozenberg Jul 05 '17 at 01:36

Michael Rozenberg · Answer 2 · 2017-07-05T01:43:55.317

Firstly we'll find common points of our curves.

$x^6-x=y$ and $y^6-y=x$ gives $x^6=x+y$ and $y^6=x+y$, which gives $x=y$ or $x=-y$.

In the first case we have $x=0$ or $x=\sqrt[5]2$.

The second case gives an unique solution $x=0$, which says that our curves intersect in two points only and since our curves are symmetric with respect to the line $y=x$, we obtain that our area is $$2\int\limits_{0}^{\sqrt[5]2}(x-(x^6-x))dx=...$$

Area between $y=x^{6} - x$ and $x = y^{6} - y$?

2 Answers2