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Average value

  1. Jun 11, 2008 #1
    For what value of c>0 is the average value of the graph on [-5,c] equal to zero and why?

    Graph is attached.

    I thought that that c should be 4 since at that point the areas cancel out. Not sure though. Could someone help me? Thanks!

    Attached Files:

  2. jcsd
  3. Jun 11, 2008 #2
    I don't see a graph since it's still pending approval, but use the Mean Value Theorem(for integrals) to solve for c.
  4. Jun 11, 2008 #3
  5. Jun 11, 2008 #4
    ive managed to get the graph
  6. Jun 11, 2008 #5
    i also looked at the mean value theorem. it deals with f'(x)=0 ?
  7. Jun 11, 2008 #6
    Ah, ok. Yes, c=4 since f is a continuous function on [-5,c], then the average value of f from -5 to c = [tex]\frac{1}{c-(-5)}\[ \int_{-5}^{c} f(x)\,dx.\] [/tex]. Since the average value = 0, then it's just the integral from -5 to 4(since it's a graph, yeah the areas cancel out)

    No, check out the MVT for integrals also known as the average value of a function.
    Last edited: Jun 11, 2008
  8. Jun 11, 2008 #7
    okay great! thanks so much =)
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