# 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 $1-x$. May 8, 2016 at 14:19
• @Nikunj For ISI - Indian Statistics Institute. May 8, 2016 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 $f(x)=1-x.$ Jun 30, 2016 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 Jun 30, 2016 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. May 8, 2016 at 14:23
• It must be my lucky day today :) May 8, 2016 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. Jun 30, 2016 at 9:42
• @adityagupta $f$ is its own inverse. Jan 6, 2021 at 11:06
• @EgeErdil Nice solution. Sep 26, 2022 at 15:12

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 :)) May 8, 2016 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. May 8, 2016 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. May 8, 2016 at 15:54
• @Starfall Unfortunately this was not a multiple choice question. May 8, 2016 at 15:54

Fix a positive even natural $$\text{n}$$ (we care about the case in which $$\text{n}=2016$$, but the argument is general).

As $$f\left(x\right)$$ is continuous and injective, $$f\left(x\right)-x$$ must be monotonically decreasing with a unique root on $$\left[0,1\right]$$. It follows that $$0\ \leq\ \left(f\left(x\right)-x\right)^{\text{n}}\ \leq\ 1$$ for $$x\in\left[0,1\right]$$ and also that for any $$y\in\left[0,1\right]$$ there exists a (unique, continuous) choice of $$\alpha\left(y\right),\beta\left(y\right)\in\left[0,1\right]$$ such that $$\left(f\left(x\right)-x\right)^{\text{n}}\leq y\ \iff\ x\in\left[\alpha\left(y\right),\beta\left(y\right)\right]\text{.}$$ In particular, $$\alpha\left(y\right)$$ and $$\beta\left(y\right)$$ are the unique values satisfying $$f\left(\alpha\left(y\right)\right)-\alpha\left(y\right)\ =\ \sqrt[n]{y}$$ $$\beta\left(y\right)-f\left(\beta\left(y\right)\right)\ =\ \sqrt[n]{y}\text{,}$$ so that $$\beta\left(y\right)=f\left(\alpha\left(y\right)\right)\ \implies\ \beta\left(y\right)-\alpha\left(y\right)\ =\ \sqrt[n]{y}\text{.}$$ We conclude that \begin{align*} \int_{x=0}^{1} \left(f\left(x\right)-x\right)^{\text{n}}\ \text{d}x\ &=\ 1-\int_{y=0}^{1} \left(\beta\left(y\right)-\alpha\left(y\right)\right)\ \text{d}y\\ &=\ 1-\int_{y=0}^{1} \sqrt[n]{y}\ \text{d}y\\ &=\ \int_{x=0}^{1} x^{n}\ \text{d}x\\ &=\ \boxed{\tfrac{1}{n+1}}\text{.} \end{align*} (Note that we did not need the hypothesis that $$f$$ is differentiable—continuity suffices!)

• (Note also that essentially the above argument shows more generally that $\int_{x=0}^{1} g\left(f\left(x\right)-x\right)\text{d}x=\int_{x=0}^{1} g\left(x\right)\text{d}x$ for $g\colon\left[-1,1\right]\to\mathbb{R}$ an even continuous function satisfying $g\left(0\right)=0$ and with $g\colon\left[0,1\right]\to\mathbb{R}$ injective.)
– Rafi
Feb 9 at 2:36