# Calculate $\int_{0}^{1} (x-f(x))^{2016} dx$, given $f(f(x))=x$.

This question was asked in an entrance test for an undergraduate mathematics program today, held all over India.

Question: $f$ is a differentiable function in $[0,1]$ such that $f(f(x))=x$ and $f(0)=1$.

Find the value of $\int_{0}^{1} (x-f(x))^{2016} dx$.

I tried to solve it by substituting $f(f(x))$ in place of $x$, but could not proceed much further. Any hints or solutions will be appreciated.

• Hint: the obvious function satisfying the conditions is $1-x$. – almagest May 8 '16 at 14:19
• @Nikunj For ISI - Indian Statistics Institute. – Gummy bears May 8 '16 at 15:52
• @almagest Is this operation valid for reaching that conclusion? If we apply Lagrange's MVT on $f$ which is differentiable in $(0,1)$ & continuous on $[0,1]$, $\exists$ at least one $c$ $\in (0,1)$ such that $f'(c)=\frac{f(1)-f(0)}{1-0}=-1$ [since $f(1)=0$ ]. So one possible function is $f(x)=-x+p.$ Now $p=1$ for $x=0,$ hence $f(x)=1-x.$ – StubbornAtom Jun 30 '16 at 9:45
• @StubbornAtom we can further assume that $c=1/2$, because of the symmetry. Infinite such functions are possible, so you can't really say anything based on that alone – Aditya De Saha Jun 30 '16 at 11:15

Let

$$I = \int_0^1 (x - f(x))^{2016}\, dx$$

Substitute $x = f(u)$ and note that $f(0) = 1$, $f(1) = 0$ to obtain

$$I = \int_1^0 (f(u) - u)^{2016} f'(u)\, du = - \int_0^1 (x - f(x))^{2016} f'(x)\, dx$$

Then,

$$2I = I + I = \int_{0}^{1} (x - f(x))^{2016} (1 - f'(x))\, dx$$

and we may substitute $w = x - f(x)$, $dw = (1 - f'(x))\, dx$ (noting that $1 - f(1) = 1$ and $0 - f(0) = -1$) to obtain

$$2I = \int_{-1}^{1} w^{2016}\, dw = \frac{2}{2017}$$

and

$$I = \int_0^1 (x - f(x))^{2016}\, dx = \frac{1}{2017}$$

• Why are you so quick at typing? I had been typing this. Well, doesn't matter. +1. – S.C.B. May 8 '16 at 14:23
• It must be my lucky day today :) – Ege Erdil May 8 '16 at 14:27
• So we need not know what $f(x)$ is in this solution. But I was wondering if for every real number $k>0$, $f(x) = \displaystyle\left(1-x^{\frac{1}{k}}\right)^k$ satisfies all conditions in the problem. – StubbornAtom Jun 30 '16 at 9:42

An alternative solution is to note that $f(f(x))=x$ means that $f(x)$ is its own inverse. Geometrically, this means that the function will be perpendicular to the line $y=x$ at the point of intersection and symmetric on each side of $y=x$. Since it is differentiable over $[0,1]$ and we are given that $f(0)=1$, one such function that comes to mind is $f(x)=1-x$.

Now that a possible function is known, the calculation of the integral is easy:

$$\int_0^1 (x-f(x))^{2016}\,dx=\int_0^1(2x-1)^{2016}\,dx=\frac{1}{2017}$$

• Yes, but you have to show that the value of the integral is the same for all such functions, which makes this solution invalid. (Unless you are in a multiple choice test, then it is clearly the superior solution :)) – Ege Erdil May 8 '16 at 14:36
• That is correct, my solution does not show that $\frac{1}{2017}$ is the value of the integral in all such cases. It is purely here to give an explicit example for the OP if he is interested in such an example. – Ethan Hunt May 8 '16 at 14:40
• I was so close to solving this question.... I kept on trying with $x-1$ and thinking it seems right... but won't satisfy the condition. facepalm from my side. – Gummy bears May 8 '16 at 15:54
• @Starfall Unfortunately this was not a multiple choice question. – Gummy bears May 8 '16 at 15:54