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Average Value Question

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]


    3. The attempt at a solution

    Hope you can see what I've written on the top of the page, but basically I want to find the antider. of f(x) from 1 to 4 (from the graph I suppose). So that should be the negative area under the graph plus the little triangle just above the x axis?
     
  2. jcsd
  3. May 6, 2007 #2

    Gib Z

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    Break it up into 2 integrals,
    [tex]\int^{3.5}_1 f(x) dx + \int_{3.5}^4 f(x) dx[/tex]

    We can see the 1st integral is just the negative of a trapezium, and it is -7/4.
    The 2nd integral is just a triangle, area 1/4. So the integral evaluates to -1.5.

    A better way to have done it would have been to see the area from 3 to 3.5 and 3.5 to 4 cancel out making the integral [tex]\int_1^3 f(x) dx[/tex], which is -3/2 as expected. Turns out you don't see an anti derivative, but if you were really really desperate, you could have written a piece wise definition of f(x) where f(x) = -x+1 for x between 0 and 2, f(x)=-1 for x between 2 and 3, so on, so forth..and done it the long way.

    Anyway, we have the integral now, you can do the rest.
     
  4. May 6, 2007 #3
    So basically average value = (1/3)*(-3/2)
    = -1/2
     
  5. May 6, 2007 #4

    Gib Z

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    Exactly. Good Work.
     
  6. May 6, 2007 #5
    *bows down to maths god*
     
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