Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    pasmith

    User Avatar
    Homework Helper

    [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
    [tex]
    \int_0^1 g(t)\,\mathrm{d}t = \int_0^1 (t,t^2)\,\mathrm{d}t = (\frac12, \frac13)
    [/tex]
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Generalization of Mean Value Theorem for Integrals Needed
Loading...