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Which function suits this description?

  1. Apr 25, 2007 #1
    Which function's definite integral from a to b will be equal to its definite integral from b to b + (b-a)/2?

    [tex] \int_{a}^{b} f(x) dx = \int_{b}^{b+\frac{b-a}{2}} f(x) dx [/tex]
    [tex] \int_{a}^{b} f(x) dx = \int_{b}^{\frac{3b-a}{2}} f(x) dx [/tex]

    I know that the exponential function e^x will have a definite integral for the second area that is twice as large as the first, and that the next one ((3b-a)/2 to (9b-3a)/2) will be twice as large as the second... so would this function be the exponential function divided by some function like 2^(n-1), where n is the number of the term in the sequence of definite integrals? How would I represent this function without the use of n?

    Thanks for any help!
     
  2. jcsd
  3. Apr 26, 2007 #2
    Does it have to be a continuous function? If not, the following piecewise function will work:

    Define [tex] f(x) = c [/tex] for [tex] a \leq x < b [/tex] and

    [tex] f(x) = 2c [/tex] for [tex] b \leq x \leq b + \frac {b-a}{2} [/tex]

    Where c is any constant. If you want a continuous function, I'll have to do more thinking (as will you...)

    (of course, if c = 0, then you've got your answer, albeit a trivial one)
     
    Last edited: Apr 26, 2007
  4. Apr 26, 2007 #3
    Well, a continuous function would be preferable, if possible, as this function should have the same behaviour over any other interval (e.g. from one point to any other point, the area under the first half of that domain will be equal to the area between the halfway point and the 3/4 point, etc).

    Thanks for any further suggestions.
     
  5. Apr 26, 2007 #4

    D H

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    This is not correct in general.

    If [itex]f(x) = \exp x[/itex] then
    [tex] \int_{a}^{b} f(x) dx
    = e^b-e^a
    = e^a\left(e^{(b-a)}-1\right)[/tex]
    [tex]\int_{b}^{\frac{3b-a}{2}} f(x) dx
    = e^{(b+\frac{b-a}2)}-e^b
    =e^b\left(e^{\left(\frac{b-a}2\right)}-1\right)[/tex]

    The ratio of the two integrals is thus

    [tex]\frac{e^b}{e^a}\,\frac{e^{\left(\frac{b-a}2\right)}-1}{e^{(b-a)}-1} = \frac{e^{(b-a)}}{e^{\left(\frac{b-a}2\right)}+1}[/tex]

    Is there some relation between a and b that you haven't told us about?
     
    Last edited: Apr 26, 2007
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