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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.

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You may be interested in Young's inequality for increasing functions. – Jonas Meyer Feb 7 '12 at 3:20

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

enter image description here

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I am curious: what program did you use for drawing this illustration? – user2468 Feb 7 '12 at 4:58
@J.D. – David Mitra Feb 7 '12 at 13:01
This owuld be much better as an addition to his answer, rather than competing against it. – Pureferret Feb 7 '12 at 15:45

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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I think this is the best answer. No need to bring geometry in. Upvoted. – Pedro Tamaroff Feb 7 '12 at 4:52
Furthermore, we can get $\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(u=f(x))=x f(x)|_a^b - \int_{f(a)}^{f(b)} f^{-1}(x) dx $, so your guess is close to the fact that $\int_a^b f(x) dx = bf(b)-af(a) - \int_{f(a)}^{f(b)}f^{-1}(x)dx$ – Huang Feb 7 '12 at 6:51
@Huang You just wrote the same as him. (?) – Pedro Tamaroff Feb 13 '12 at 4:01
@Peter Because the op guessed $\int_a^b f(x) dx = (f(b)-f(a))b - \int_{f(a)}^{f(b)}f^{-1}(x)dx$ , I just wan to show him his guess is close to the fact. – Huang Feb 13 '12 at 5:47
Also when the function f and its inverse are analytic in a domain about the origin including b, the results are related to the Legendre transformation with f(0)=0 and a=0. – Tom Copeland Apr 7 '12 at 3:26

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.

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My vote here helped you jump from 998 to 1008! – Herband Feb 26 '12 at 20:07
@math101 hey, thanks! :) – student Feb 26 '12 at 22:18

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:

enter image description here

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).

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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.)

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