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Average Value Function

  1. Nov 30, 2004 #1
    Find the average value of

    [tex]
    \int_{x}^{1} cos(t^2)dt
    [/tex]
    on [0,1]

    I have no idea where to even begin..
     
  2. jcsd
  3. Nov 30, 2004 #2

    shmoe

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    You understand that your expression is just a function of x? namely you can write:

    [tex]f(x)=\int_{x}^{1} cos(t^2)dt[/tex]

    Now what's the general expression for the average value of a function f(x) on the interval [0,1]? Substitute the above in and see what happens.
     
  4. Nov 30, 2004 #3
    I just get
    1/1 * (sin 1 - sin(x^2))

    but that seems too simple? am I doing something wrong?
     
  5. Nov 30, 2004 #4

    shmoe

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    How did you get that? Please provide some details..

    The end asnwer should be a number, and will have no x's or other variables in it.
     
  6. Nov 30, 2004 #5
    I used 1/b-a *[tex]
    \int_{x}^{1} cos(t^2)dt
    [/tex]

    and I tried to evaluate the integral
     
  7. Nov 30, 2004 #6

    shmoe

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    sin(t^2) is not an antiderivative of cos(t^2), but that's beside the point here.

    You want the average value of the function f(x), the thing you were given. Ignore for a moment that it's defined by an integral and just treat it like any old function. The average value is given by [tex]\frac{1}{1-0}\int_{0}^{1}f(x)dx[/tex]

    Now substitute your integral equation for f(x):

    [tex]\int_{0}^{1}\left(\int_{x}^{1} cos(t^2)dt\right)dx[/tex]
     
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