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Mean value ordinate

  1. Jan 31, 2007 #1
    The graph of the function f(x)=[tex]\sqrt{a^2-x^2}[/tex] (-a<= x <=a) is a semicircle, centre the origin and bounding diameter the portion of the x-axis between the points A(-a,0) and B(a,0). Show that the mean value with respect to x of the ordinates of the semicircle is [tex]\pi[/tex]a/4. ( This question can be done without using calculus.)
    Does ordinates mean the x-coordinates only of the points?
    My attempt, mean value =[tex]\frac{1}{b-a}\int_a^{b}f(x)dx = \frac{1}{2a}\int_{-a}^{a}(\sqrt{a^2-x^2})dx = \frac{1}{4a}\int_{-a}^{a}\ln({a^2-x^2})dx[/tex].
    I'm interested to know how it can be done without using calculus? Thanks for the help.
     
    Last edited: Jan 31, 2007
  2. jcsd
  3. Jan 31, 2007 #2

    Dick

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    In fact, it's the average value of y. So your first two integrals are just fine. Can't say the same about the third. Where did the wacky ln come from???? Hmm. How to do this without calculus?? Don't know.
     
  4. Feb 2, 2007 #3

    HallsofIvy

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    The "average value" of a function, over some interval, is the integral of the function over that interval divided by the length. As Dick said, the "ordinal" is the y component, not the x.

    As for doing it without calculus, isn't the integral of a function the area under the curve? What is the area here?
     
  5. Feb 2, 2007 #4
    To solve the problem try [tex] x=asin\theta[/tex], then [tex]\sqrt{a^2-x^2} =\sqrt{a^2-a^2sin^2\theta} = \sqrt{a^2(1-sin^2\theta)} = acos\theta[/tex]. Therefore [tex]\\ \frac{1}{2a}\int_{x=-a}^{x=a}\sqrt{a^2-x^2}dx =
    \frac{1}{2a} \int_{x=-a}^{x=a}\sqrt{a^2-x^2}acos\theta d\theta \\[/tex]
    [tex]= \frac{1}{2a} \int_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}} \sqrt{a^2(1-sin^2\theta)}acos\theta d\theta \\ =
    \frac{a}{2} \int{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(1+cos\theta) d\theta \\ = a\frac{\pi}{4}[/tex]. So I learned how to solve the problem using calculus.
     
    Last edited by a moderator: Feb 2, 2007
  6. Feb 2, 2007 #5

    HallsofIvy

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    Great! Now how would you do it without calculus? This is in "pre-calculus mathematics"
     
    Last edited: Feb 2, 2007
  7. Feb 5, 2007 #6
    The area in question is 1/2 the area of a circle = [tex] \frac{1}{2}\pi \times r^2 [/tex] multiplied by [tex] \frac{1}{2r} \mbox{ to get the mean value, which =} a\pi/4 \mbox{ where a=r.}[/tex]
     
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