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Average value

  • Thread starter chukie
  • Start date
  • #1
80
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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!
 

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Answers and Replies

  • #2
238
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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
80
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ive managed to get the graph
 
  • #5
80
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i also looked at the mean value theorem. it deals with f'(x)=0 ?
 
  • #6
238
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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)

i also looked at the mean value theorem. it deals with f'(x)=0 ?
No, check out the MVT for integrals also known as the average value of a function.
 
Last edited:
  • #7
80
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okay great! thanks so much =)
 

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