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Tricky definite integral

  1. May 4, 2014 #1
    Am trying to get a solution to the definite integral below. Looking for some direction.

    I = 01 xf(x)dx where

    01 f(x)dx = A, is known.

    Also, its is know that when x =1, f(x) =0 and when x =0, f(x) =1.

    Can we get a solution of I in terms of A?

    I have tried going the integration by-parts route which did not lead to any success. Any help is much appreciated.
     
  2. jcsd
  3. May 4, 2014 #2

    mfb

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    Without more details about f(x), this is impossible. It is easy to find some examples where the constraints are met, A is the same but I differs.
     
  4. May 4, 2014 #3
    Unfortunately, f (x) is a transcendental function

    f(x) = a - be[px+qf(x)] -c[px+qf(x)]

    where, a, b, c, p and q are all constants. I don't think this helps?
     
  5. May 4, 2014 #4

    mfb

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    I don't know if that helps, but it is at least some hope to make it possible (I just don't know how). Without knowing anything about f(x) it would be completely impossible.
     
  6. May 4, 2014 #5

    micromass

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    It's going to be annoying, but if ##f## has an expansion as a power series, then you might be able to find an approximation to ##I##. I really doubt a analytic solution exists.
     
  7. May 4, 2014 #6

    micromass

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  8. May 4, 2014 #7
    wolframalpha gives the result in terms of Lambert's W function which I have already looked at. Looking for another route...
     
  9. May 15, 2014 #8
    I am wondering if I can use rotation of coordinates to solve this integral, like here, Example D.9? Looking for some direction.....
     
  10. May 16, 2014 #9
    try this way, set u=x, dv=f(x)dx, you get du=dx and v=∫f(x)dx
    then I=Ax-∫∫f(x)dxdx. Can you solve this?
     
  11. May 19, 2014 #10
    Nope, not possible with what I know.
     
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