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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.

  • 2
    You may be interested in [Young's inequality for increasing functions](http://en.wikipedia.org/wiki/Young%27s_inequality#Standard_version_for_increasing_functions).2012-02-07

5 Answers 5

28

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

enter image description here

  • 0
    This owuld be much better as an addition to his answer, rather than competing against it.2012-02-07
23

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 $u=f(x)$.

  • 0
    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=0.2012-04-07
12

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.

  • 0
    @math101 hey, thanks! :)2012-02-26
9

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).

4

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.)