Generalization of Mean Value Theorem for Integrals Needed

[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.

 

[Only a Mirage]

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

 

[pasmith]

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.

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
<br /> \int_0^1 g(t)\,\mathrm{d}t = \int_0^1 (t,t^2)\,\mathrm{d}t = (\frac12, \frac13)<br />
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).

 

[Only a Mirage]

That makes sense. Thanks a lot for the help.