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Integral equality paradox?

  1. Jul 10, 2007 #1
    if we have that:

    [tex] \int_{a}^{b}dxf(x) - \int_{a}^{b}dxg(x)= \int_{a}^{b}dx(f(x)-g(x)) [/tex]

    where the integral over (a,b) of f(x) and g(x) exist separately then my question is if

    [tex] \int_{a}^{b}dx(f(x)-g(x)) =0 [/tex] then

    does this imply necessarily that [tex] f(x)=g(x)+h'(x) [/tex]

    where h(a)=h(b)=0 and its derivative is 0 almost everywhere on the interval (a,b)
  2. jcsd
  3. Jul 10, 2007 #2

    Gib Z

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    We have that [tex]\int^b_a f(x) dx = \int^b_a g(x) dx[/tex] and from that condition you ask if that implies [itex]f(x) = g(x) \in [b,a][/itex](Neglecting small value of h'(x) you proposed, i know don't why you did, though it doesn't end up making a difference). As a simple counter example, f(x) and g(x) may be very different Odd functions, and the integral of these two between b and -b will be 0.
  4. Jul 10, 2007 #3


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    No, this is not correct, you just need that h(a) = h(b), since:

    [tex]\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = \int_a ^ b \left( f(x) - g(x) \right) dx = \int_a ^ b \left( g(x) + h'(x) - g(x) \right) dx[/tex]

    [tex]= \int_a ^ b h'(x) dx = \left. h(x) \right|_a ^ b = h(b) - h(a)[/tex]

    Now, if h(a) = h(b), then we'll have: [tex]\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = 0[/tex]

    No, why should it's derivative is 0? =.="
  5. Jul 10, 2007 #4

    Gib Z

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    Are you sure that worked out right VietDao29 >.<? He asks if [tex] \int^b_a f(x) dx = \int^b_a g(x) dx[/tex] then does that imply f(x) = g(x) + h'(x), where h(a)=h(b)=0. It seems from your post you assumed f(x)=g(x) + h'(x), to get that all we need is h(a)=h(b), but that is somewhat the converse of what he is asking I think >.<
  6. Jul 11, 2007 #5
    He also said h'=o a.e.
  7. Jul 11, 2007 #6


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    I was pointing out that his claim that h(a) = h(b) = 0 is wrong, and it's even wronger, when he said that h' = 0.

    You can find h'(x) simply by subtracting f(x) from g(x).

    Say, we have:
    [tex]\int_0 ^ \frac{\pi}{2} \sin x = \int_0 ^ \frac{\pi}{2} \cos x = 1[/tex]

    We also have:
    [tex]\sin x = \cos x + (\sin x - \cos x)[/tex]

    So h'(x) = sin(x) - cos(x)
    Integrate it with respect to x, we have:
    h(x) = -cos(x) - sin(x) + C

    [tex]h(0) = -1 + C[/tex]

    [tex]h \left( \frac{\pi}{2} \right) = -1 + C[/tex]

    [tex]\Rightarrow h(0) = h \left( \frac{\pi}{2} \right)[/tex] (but not equal to 0, unless C = 1)

    However, h(0) = h(pi/ 2) does not imply that h'(x) = 0 on the interval (0; pi/2).

    So, in conclusion, both of his assertions (?, can I use plural here) are wrong. =.="
    Last edited: Jul 11, 2007
  8. Jul 11, 2007 #7

    Gib Z

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    Ok we'll good we agree on that :) Pretty pointless though, "Klaus" never comes back to his threads when he's proven wrong, and matt grime reckons its some other guy named Jose whose not good >.<
  9. Jul 11, 2007 #8
    ok,.. thank you i fooled myself once again.
  10. Jul 11, 2007 #9


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    Well, he did come back. :rolleyes: o:) ^.^
  11. Jul 11, 2007 #10
    If he is Jose he's calmed down a lot. Which is annoying actually, because jose was crazy enough to be funny, where as this person is simply obnoxious.
  12. Jul 11, 2007 #11


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    Well, it must be some of my missing days here, at PF, but who's the well-known Jose that you guys are talking about? I don't think I know anyone named Jose. :cry: And, btw, what has he done to make him this famous?
  13. Jul 11, 2007 #12
    Well his username was eljose. You can search for posts by him. Let's just say he was a genius on the level of gauss, and solved multiple millennium prize problems but the mathematical community kept holding him down because they didn't appreciate proper mathematics.
  14. Jul 12, 2007 #13

    matt grime

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    The snobbish mathematical community, I think you'll find, who only let famous people publish in journals.
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