Generalization of Mean Value Theorem for Integrals Needed

1. Aug 3, 2013

Only a Mirage

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 $F: U \subset \mathbb{R}^{n+1} \rightarrow W \subset \mathbb{R}^{n}$ is a continuous function

and $x: I \subset \mathbb{R} \rightarrow V \subset \mathbb{R}^{n}$ is a continuous function

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

$\int_{t_1}^{t_2} F(x(t),t)\,dt = F(x(t^*),t^*)(t_2-t_1)$

I can see that it holds for each of the component functions of $F$, but I'm not sure about the whole thing.

2. Aug 3, 2013

Only a Mirage

I'm actually fairly certain now that my conjecture is false...

3. Aug 4, 2013

pasmith

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

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

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

For example, consider $g : [0,1] \to \mathbb{R}^2 : t \mapsto (t,t)$. 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 $[0,1]$, that component attains its average at exactly one point, and we have $t_1^{*} = \frac 12$ and $t_2^{*} = \frac{1}{\sqrt{3}}$. These are not equal, so $g$ does not attain its average on $[0,1]$.

One can obtain $g$ by taking $F(x,y,z) = (x,y^2)$ and $(x(t),y(t)) = (t,t)$.

4. Aug 8, 2013

Only a Mirage

That makes sense. Thanks a lot for the help.

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