TITLE: Is $\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dx$?
QUESTION [12 upvotes]: Is it true that
$$\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dx$$
Just making sure. 
If not, how about:
$$\int_a^b f(x) dx = (f(b)-f(a))b - \int_{f(a)}^{f(b)}f^{-1}(x)dx$$ 
I'm having a hard time concentrating right now, and I'm trying to figure out how to get the area under a curve when the function is inverted.

REPLY [9 votes]: 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).