A calculus problem by Sutirtha Datta

Inverse of f exists hence its onto, now f(0)=1 & f(1)=0

Again f''(x). f'(x) =1

So its strictly decreasing one

Hence f(x) =1-x

$\int_{0} ^1(2x-1)^{2017} dx =0$

EDIT: ANY OTHER DIRECT SOLUTION THAN MINE WILL BE HIGHLY APPRECIATED.

We will first prove that $f(x)$ is an one-to-one function.
For every $x_{1},x_{2}$ that belong in $[0,1]$ with $f(x_{1})=f(x_{2})\Rightarrow f(f(x_{1}))=f(f(x_{2}))\Rightarrow x_{1}=x_{2}$ .
So, $f(x)$ is an one-to-one function, which suggets that it's inverse exists.
$f(f(x))=x\Rightarrow f(x)=f^{-1}(x)$ .
Now we will prove that $f(x)$ is strictly decreasing.
$f(f(x))=x\Rightarrow f'(f(x))f'(x)=1$ $(1)$
Let's assume that there exists a number $x_{0}$ such that $f'(x_{0})=0$ . If we substitute $x=x_{0}$ in $(1)$ we get:
$f'(f(x_{0}))f'(x_{0})=1\Rightarrow 0=1$ .
That is a contradiction, which means that $f'(x)$ is never zero. So, the function can either be strictly increasing or scrictly decreasing.
We know that: $f(0)=1$ .
In $(1)$ for $x=0$ we get:
$f(f(0))=0\Rightarrow f(1)=0$ .
So, for $0<1$ , $1=f(0)>f(1)=0$ .
Therefor, $f(x)$ must be strictly decreasing.
But, if $f(x)$ is strictly decreasing and it's inverse function is the function itself, $f(x)$ can either be:
$f(x)=x$ (which is not acceptable since $f(x)$ strictly decreases) or $f(x)=c-x$ for some constant c.
$f(0)=1\Rightarrow c=1\Rightarrow f(x)=1-x$ .
So, we want the integral:
$\int_{0}^{1} (2x-1)^{2017}dx$
which, by substituting and using the power rule, works out to be $\boxed{0}$