1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Average Value of a function

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    If fave [a,b] denotes the average value of f on the interval [a,b] and a<c<b, show that
    fave[a,b] = (c-a)/(b-a) fave[a,c] + (b-c)(b-a) fave[c,b]


    2. Relevant equations

    All i know is the mean value theorem for integrals is f(c) = fave = 1/(b-a) integral(f(x),x,b,a)

    3. The attempt at a solution

    Tried using the theorem, but had no idea how to get to that point.

    Thanks!
     
  2. jcsd
  3. Jul 24, 2011 #2
    Hi,

    Why don't you apply the definition of the fave to fave[a,c] and fave[c,b]?

    Once you do this, I think you'll see that your expression simplifies quite nicely.
     
  4. Jul 24, 2011 #3
    Hello, I tried it and i believe this is how it goes...

    fave[a,c] = 1/(c-a) [f(c)-f(a)]
    fave[c,b] = 1/(b-c) [f(b)-f(c)]

    then i add it together or what? Kind of confused on what to do because this gives something weird...

    I do know that u can split up the bounds from [a,c] and [c,b] to get [a,b].. does that correlate with what this gotta do?
     
  5. Jul 24, 2011 #4
    Perhaps I don't understand your notation, but shouldn't [f(c)-f(a)] be Int[f, a, c]?

    Try plugging in those expressions into the right-side of the equation that you're trying to prove.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Average Value of a function
Loading...