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An integrale function

  1. Nov 3, 2008 #1
    Let f:[-1,1]→ℝ an integrable function such that |f(x)|[tex]\leq2[/tex].
    Show that there exist x in [-1,1] such that:

    integral -1 to x f(t)dt-4x=0

    is unique?
  2. jcsd
  3. Nov 3, 2008 #2


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    First show that there exists a solution to that equation, did you do that? Once you've done that consider using the Mean Value Theorem to prove uniqueness.
  4. Nov 4, 2008 #3
    let [tex] g(x)=\int_{-1}^xf(t)dt-4x[/tex]
    Now let's test the sign of this function on at the points -1 and 1 respectively.

    [tex] g(-1)=\int_{-1}^{-1}f(t)dt-4(-1)=4>0[/tex]

    [tex] g(1)=\int_{-1}^1f(t)dt-4(1)\leq0[/tex] Since [tex] -2\leq f(x)\leq 2=>0=-2x|_{-1}^1=\int_{-1}^1(-2)dx\leq \int_{-1}^1f(x)dx\leq\int_{-1}^12dx=2x|_{-1}^1=4[/tex]

    Now, according to IVT there exists a number c in the interval (-1,1) such that

    [tex] g(c)=\int_{-1}^cf(t)dt-4c=0[/tex]

    So far we know that there is a solution. Now we need to prove its uniqueness.
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