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Homework Help: Finding the Intermediate Value

  1. Jan 26, 2005 #1
    Let's say you want to find the intermediate value [tex] \xi [/tex] of the mean value theorem of the integral calculus for:

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

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

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

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

    Thanks for any help

    Last edited: Jan 26, 2005
  2. jcsd
  3. Jan 26, 2005 #2
    i think its right
  4. Jan 26, 2005 #3


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

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

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

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

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

    You want [itex]\frac{1}{a}- \frac{1}{b}= \frac{b-a}{ab}= \frac{1}{ab}(b-a)[/itex]. You want [itex]f(\xi)=\frac{1}{x^2}= \frac{1}{ab}[/itex]. That is, you want [itex]\xi^2= ab [/itex] or [itex]\xi= \sqrt{ab}[/itex], the "geometric mean" of a and b.
    Last edited by a moderator: Jan 26, 2005
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