Is $\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dx$?

calculus integration

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.


For what it's worth, here's a diagram to accompany Brian M. Scott's and Leandro's answers:

enter image description here


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 $u=f(x)$.

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