Is $\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dx$? 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.
 A: 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).
A: No.
Suppose that $f(x)$ is a continuous, strictly increasing function on $[a,b]$. Then $\int_a^b f(x)dx$ gives the area between the curve and that segment $[a,b]$ on the $x$-axis, while $\int_{f(a)}^{f(b)}f^{-1}(x)dx$ gives the area between the curve and the segment $[f(a),f(b)]$ on the $y$-axis, and it’s easy to see that these two areas need not be the same. (There are easy concrete examples $-$ I see now that Leandro has given one $-$ but even a few pictures should convince you.)
A: For what it's worth, here's a diagram to  accompany   Brian M. Scott's and Leandro's answers:

A: No, but there is a relationship between those two quantities if $f$ is continuously differentiable and strictly increasing (you can relax the last assumption).
By parts, we have $\int_a^b f(x)dx = x f(x)|_a^b - \int_a^b x f'(x)dx = x f(x)|_a^b - \int_a^b f^{-1}(f(x)) f'(x)dx$ = $x f(x)|_a^b - \int_{f(a)}^{f(b)} f^{-1}(u) du $ where in the last step we do the substitution $u=f(x)$.
A: No. Take $f(x) = x^2$, $a = 0$, $b = 1$. Then in this interval $f$ is integrable and invertible, with $f^{-1}(x) = \sqrt{x}$, but the integrals are easily seen to be different.
