What Is the Intermediate Value in the Mean Value Theorem for These Integrals?

[courtrigrad]

Let's say you want to find the intermediate value \xi of the mean value theorem of the integral calculus for:

(a)\int^b_a 1 \ dx
(b)\int^b_a x \ dx
(c)\int^b_a x^n \ dx
(d)\int^b_a \frac{dx}{x^2}

Using the Mean Value Theorem we know that a \leq \xi \leq b and \mu = f(\xi) following from:

\int^b_a f(x) \ dx = \mu(b-a) where m \leq \mu \leq M and m, M are the least and greatest values of f(x).

Hence \int^b_a x \ dx = (b-a) f(\xi) because dx = b-a. So f{(\xi) = \frac{(b-a)(b+a)}{2} \times \frac{1}{b-a} = \frac{a+b}{2} Is this right? I am not sure because I think this is for f(\xi) but I want \xi. Or are they the same thing?

Thanks for any help

 

[courtrigrad]

i think its right

 

[HallsofIvy]

In the case that f(x)= x they're the same!
You want to find ξ such that \int_a^b 1 dx= f(\xi)(b-a).

Okay, go ahead and do the integral \int_a^b 1 dx= b- a so f(ξ)= 1. Of course, that's true for any ξ.

You want to find ξ such that \int_a^b x dx= \frac{1}{2}(b^2- a^2)= \frac{b^2- a^2}{2}= \frac{b+a}{2}(b-a). Okay, you want f(\xi)= \frac{b+a}{2} and since f(x)= x, you want \xi= \frac{b+a}{2}, the midpoint of the interval (arithmetic mean of a and b).

You want to find ξ such that \int_a^b x^n dx= f(\xi)(b-a).
\int_a^b x^n dx= \frac{1}{n+1}(b^{n+1}- a{n+1})= \frac{1}{n+1}(b-a)(b^{n}+ ab^{n-1}+ ... + a^{n-1}b+ a^n) so you want f(\xi)= \xi^n= \frac{1}{n+1}(b^n+ ab^{n-1}+ ...+ a^{n-1}b+ a^n). Since that lies between
bⁿ and aⁿ, such a ξ exists.

You want to find ξ such that \int_a^b \frac{1}{x^2} dx= f(\xi)(b-a). \int_a^b x^{-2}dx= \frac{1}{a}- \frac{1}{b}.

You want \frac{1}{a}- \frac{1}{b}= \frac{b-a}{ab}= \frac{1}{ab}(b-a). You want f(\xi)=\frac{1}{x^2}= \frac{1}{ab}. That is, you want \xi^2= ab or \xi= \sqrt{ab}, the "geometric mean" of a and b.