Finding the Intermediate Value

[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.