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Two limited integration questions

  1. Dec 13, 2004 #1

    I want to know the conditions that must be satisfied by a function
    [tex] f(x) [/tex] for any of the following two cases to be true (each case independent from the other);

    1- [tex] \int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx [/tex]

    2- [tex] \int^a_0 f(x) dx = \int^0_a f(x) dx [/tex]

    They gave me confusion when I was solving problems related to electric field and electric potential.
  2. jcsd
  3. Dec 13, 2004 #2


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    I don't know if this condition is sufficient but a condition would be, if we write the integrals in terms of their primitives,

    [tex] \int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx \Leftrightarrow \mathcal{F}(a) - \mathcal{F}(-a) = 2\mathcal{F}(a) \Leftrightarrow \mathcal{F}(-a) = -\mathcal{F}(a)[/tex]

    The condition is that it is true iff the primitive of f is a function F such that F(-a) = -F(a)
  4. Dec 13, 2004 #3
    First case...

    [tex] \int^a_{-a} f(x) dx = \int^0_{-a} f(x) dx + \int^a_0 f(x) dx[/tex]

    Then use the fact that :

    [tex] \int^0_{-a} f(x) dx = - \int^{-a}_0 f(x) dx [/tex]

    and replace x by -x...the limit -a will then change to a because of this substitution. and dx will become -dx. Now f(x) becomes f(-x) and there are two possibilities. Either f(-x) = -f(x) or f(-x) = f(x)....you know what you will need to achieve so which one of the two is it...

    Question 2 :

    Just put the integral in right hand side to the left hand side and use the above property to get rid of the minus-sign...what do you get ???

  5. Dec 13, 2004 #4
    Can you put like , first case is true if f(x) is even and if such and such....
    I shall study your answer. But from my first reading I need more.
    I need a condition to apply to f(x) so I can use each of the above properties.

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