21Let t = f(z) Then \int_{x}^{y}{}f^{-1}(t)dt =\int_{f^{-1}(x)}^{f^{-1}(y)}{}zf^{^{,}}(z)dz = \left(zf(z) \right)|_{f^{-1}(x)}^{f^{-1}(y)} - \int_{f^{-1}(x)}^{f^{-1}(y)}{}f(z)dz (integrating by parts)
So, we get \int_{x}^{y}{}f^{-1}(t)dt = yf^{-1}(y)- xf^{-1}(x) - \int_{f^{-1}(x)}^{f^{-1}(y)}{}f(t)dt
Substituting x= f(a) , y = f(b) we get \boxed{\, \, \int_{a}^{b}f(x)dx+\int_{f(a)}^{f(b)}f^{-1}(y)dy=bf(b)-af(a)}