Show: $ f(a) = a,\ f(b) = b \implies \int_a^b \left[ f(x) + f^{-1}(x) \right] \, \mathrm{d}x = b^2 - a^2 $ 
If $a,b$ are fixed points of $f$, then $$ \int_a^b \left[ f(x) + f^{-1}(x) \right] \, \mathrm{d}x = b^2 - a^2 $$

In the words of 2014 MIT Integration Bee Champion (Carl Lian), the above property was responsible for the champion's victory in the 2013 MIT Integration Bee. How does one go about proving this property? 


The area of the two whole thing is $b^2-a^2$ (sum of areas of two rectangles)
$$b(b-a)+a(b-a)=b^2-a^2$$
