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Generalization of Mean Value Theorem for Integrals Needed

  1. Aug 3, 2013 #1
    Hi all,

    I'm having trouble finding a certain generalization of the mean value theorem for integrals. I think my conjecture is true, but I haven't been able to prove it - so maybe it isn't.

    Is the following true?

    If [itex] F: U \subset \mathbb{R}^{n+1} \rightarrow W \subset \mathbb{R}^{n} [/itex] is a continuous function

    and [itex]x: I \subset \mathbb{R} \rightarrow V \subset \mathbb{R}^{n}[/itex] is a continuous function

    then [itex]\exists t^* \in [t_1,t_2][/itex] such that

    [itex]\int_{t_1}^{t_2} F(x(t),t)\,dt = F(x(t^*),t^*)(t_2-t_1)[/itex]

    I can see that it holds for each of the component functions of ##F##, but I'm not sure about the whole thing.
  2. jcsd
  3. Aug 3, 2013 #2
    I'm actually fairly certain now that my conjecture is false...
  4. Aug 4, 2013 #3


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    [itex]g(t) = F(x(t),t)[/itex] is a function from [itex][t_1,t_2] \subset \mathbb{R}[/itex] to [itex]\mathbb{R}^n[/itex], so by definition one integrates it component by component with respect to the standard basis.

    For each component [itex]g_k[/itex], there exists [itex]t^*_k \in [t_1,t_2][/itex] such that [itex](t_2 - t_1) g_k(t^*_k) = \int_{t_1}^{t_2} g_k(t)\,\mathrm{d}t[/itex] by the mean value theorem applied to [itex]G_k(t) = \int_{t_1}^t g_k(s)\,\mathrm{d}s[/itex].

    The value of [itex]t[/itex] where [itex]g_k[/itex] attains its average value is not necessarily unique, so for [itex]g[/itex] to attain its average there must be at least one [itex]t[/itex] where every component attains its average, which is not necessarily the case.

    For example, consider [itex]g : [0,1] \to \mathbb{R}^2 : t \mapsto (t,t)[/itex]. Then
    \int_0^1 g(t)\,\mathrm{d}t = \int_0^1 (t,t^2)\,\mathrm{d}t = (\frac12, \frac13)
    Since each component is strictly increasing on [itex][0,1][/itex], that component attains its average at exactly one point, and we have [itex]t_1^{*} = \frac 12[/itex] and [itex]t_2^{*} = \frac{1}{\sqrt{3}}[/itex]. These are not equal, so [itex]g[/itex] does not attain its average on [itex][0,1][/itex].

    One can obtain [itex]g[/itex] by taking [itex]F(x,y,z) = (x,y^2)[/itex] and [itex](x(t),y(t)) = (t,t)[/itex].
  5. Aug 8, 2013 #4
    That makes sense. Thanks a lot for the help.
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