You actually have that:
$\int_a^b f(x) dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1} (x) dx $
Here's a graph:

The rectangle $ObCB$ has area $b\cdot f(b)$. The rectangle $OaDA$ has area $a \cdot f(a)$. The curved trapezium $ADCB$ has area $\int_{f(a)}^{f(b)} f^{-1} (x) dx $ so it is expected that:
$\int_a^b f(x) dx = \mathcal{A}(ObCB) - \mathcal{A}(OaDA) - \mathcal{A}(ADCB)$ $\int_a^b f(x) dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1} (x) dx $
which is actually true.
Remeber that the starting function has to be one-to-one and onto for the inverse to be defined.
You can check a full proof in Michael Spivak's Calculus (though he want's you to do it, he provides the steps necessary to do so).