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Integration of inverse function

  1. May 19, 2008 #1

    Integration is the 'area under the curve' for a one dimensional function. Say we have a function [tex]f(x)=x^2[/tex], if we compute the integral of this function between two limits we get the area under it between the two limits on the x axis.

    What if we wanted to find the area between two limits on the 'f(x) axis'?

    The usual way (i.e. the way I've been taught) would be define a new function, say g(x) which is the inverse of the function f(x) and calculate the integral between the desired limits for g(x).

    I was thinking of 'swapping' x and f(x) in the integrand. We just need to make sure the limits are right and hey presto. I've tried it so far for a couple of functions and it seems to work out.

    It looks like this:
    [tex]\int_a^b f(x) dx[/tex]
    We now want the integral on the y axis for the same limits.
    So swap x with f(x).
    [tex]\int_{f(a)}^{f(b)} x df(x)[/tex] = [tex]\int_{f(a)}^{f(b)} x \frac{d}{dx}f(x) dx[/tex]

    Is this right? The whole reason for doing it this way is you don't have to find an inverse function, even though its not that difficult to do in this case.

    I imagine that not all inverses are defined for an arbitrary region of interest, so the whole process of finding an inverse is more general than this method.

    A quick counterexample that I can think of would be trying to find the integral this way of f(x)= sin(x) from 0 to [tex]4\pi[/tex].
  2. jcsd
  3. May 20, 2008 #2
    lol 65 views and not one comment :tongue:
  4. May 20, 2008 #3
    I don't think this formula is a true outside of a few coincedences, since by integration by parts we have:

    \int_{f(a)}^{f(b)} x \frac{d}{dx}f(x) dx = x f(x) |_{f(a)} ^{f(b)} - \int_{f(a)}^{f(b)} f(x) dx

    This is not a disproof, it just makes your result look even more unlikely to me.

    What functions did you try it on that it worked?


    Here is what I believe is an accurate formula (assuming the inverse is well-defined, differentiable, etc) for the integral of the inverse function in terms of a double integral of the reciprocal of the derivative of the original function:

    \int_{f(a)}^{f(b)} \int_{0}^y \frac{1}{\frac{d}{dx}f(x)} dx dy

    I based this on the surprisingly little taught result that:

    [tex] \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} [/tex]

    as long as the inverse exists and is differentiable.
    Last edited: May 20, 2008
  5. May 20, 2008 #4
    Haha, I tried it for a number of integer powers of x i.e. x^2, x^3, x^4 and it worked great. Except that I did it over a unit interval from 0 to 1 :tongue: If you do that, since 1 to the power 'anything' is still one, it works. Very foolish indeed!

    Heres an example:

    [tex]\int_0^1 x 3x^2dx = \frac{3}{4}x^4 [/tex]

    Juxtaposed with:
    [tex]\int_0^1 x^{1/3}dx = \frac{3}{4}x^{4/3} [/tex]

    Which both yield the same answer between the limits 0 and 1. I didn't write it in this form though, I skipped steps since I was in a rush :blushing:

    Thanks for pointing out my error, I'll try using your approach and tell of results. :smile:
  6. May 20, 2008 #5


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    This is a well known result. The problem using it is one must verify some conditions. It is usefull for seveal common integrals.
  7. May 20, 2008 #6


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    of course that result is little taught most people only need to be taught it once. How many times would you like to be taught it?
    It also is a trivail consequence of the chain rule.

    Darn coincedences
    I am convinced
    (x y)'=y+x y'
    is a coincedence
    one day I will find a conter example
    y=x no
    y=sin(x) no
    y=exp(x) no
    y=7+pi*x no
    one day!!!
  8. May 20, 2008 #7
    OP: You might need to require that f be a monotonically increasing function in x.
  9. May 20, 2008 #8


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    with a few simple restrictions like monotone, this formula is an old way to view integration by parts, as i recall. as always look in courant for a picture.

    the point is simply that the area under the inverse plus that under the original function is the area of a rectangle, and thats the product expression uv, on the RHS of the integration by parts formula.

    im just babbling without thinking here, could be wrong, but you sounded desirous of some commentary.
  10. May 21, 2008 #9
    Aha thanks all, makes a great deal more sense now :biggrin:

    Also in the OP the limits are wrong, they should remain a and b. Then it works.
    I'll fix that now.

    Thanks to David and Mathwonk for your explanations, helped alot :smile:

    Edit: can't seem to fix the error, probably meant to prevent backtracking during a debate :P

    What it should be is:
    [tex] \int_{a}^{b} x df(x )[/tex] = [tex]\int_{a}^{b} x \frac{d}{dx}f(x) dx [/tex]

    For good measure here is the example I gave later on:
    Integrations using method above between limits 0 and 2
    \int_0^2 x 3x^2dx = \frac{3}{4}x^4
    Which equals

    Integrations using the inverse between the limits 0 and 8.
    \int_0^8 x^{1/3}dx = \frac{3}{4}x^{4/3}

    Which equals

    Last edited: May 21, 2008
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