1.This function is NOT one to one on the interval $[0,2]$, because $f(1-x)=f(1+x)$, therefore there is no inverse on $[0,2]$ , consider $f(.5),f(1.5) and f(.9),f(1.1)$ but there are inverses for $[0,1]$ and $[1,2]$ separately.
2.Draw the funtcion on $[0,2]$, then rotate it 90 degrees, you can see the inverse of the function, and what the integral of the function and it's inverse add up to ( hint adding them up should give you area of a rectangle).
3.Try a simpler function instead, for example $f(x)=\frac{1}{4x}$, then try to move up to a function that looks more similar to this one e.g. $f(x)=\frac{1}{4\sqrt{x}}$