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Can't figure out integral question

  1. Mar 19, 2006 #1
    what is the proper method for finding the average of an integral? For example, the question I'm trying to answer is this:

    "Find the average value of the function f(x) = 4 - x² over the interval [-2, 2]."

    Now the answer is suppost to be 8/3. But I keep getting an answer of -11/3 no matter what I do. WTF!?!?
  2. jcsd
  3. Mar 19, 2006 #2


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    f is nonnegative on the interval [-2,2], so how can the average possibly be negative?
  4. Mar 19, 2006 #3
    Well, what did you do to get -11/3?

    The function you are given here is [tex]\geq 0[/tex] in [-2,2]. This makes things quite a lot easier.
    By integration you can find out the area under the curve. When you say you'd like to know the average value, it means this: replace the function by a constant one that has the same area under its curve (this time it must be a rectangle).
    I hope it's clear now.
    Best regards

  5. Mar 19, 2006 #4
    How did you get -11/3? Are you sure you integrated the function correctly?
  6. Mar 19, 2006 #5
    Greetings Flatland:

    For any continuous function, f(x), on the interval [a,b], the average value of f on that interval is given by:

    f_ave = 1/(b-a) * Integral f(x) dx ; [a,b]. So, in the present case,

    f_ave = 1/(2+2) * Integral (4 - x^2) dx; [-2,2]
    = (1/4)(4x - (1/3) x^3) = (x/12)(12 - x^2); [-2,2]
    = (1/6)(12-4) - (-1/6)(12-4)
    = (1/6)(8+8)
    = 8/3

    I hope this helps.


    Rich B.
  7. Mar 19, 2006 #6
    nvm I figured out what I did wrong thx
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