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Find the average value of a function

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the average value of a function of f(x)= |x+1| sgnx on the interval [-2,2]


    2. Relevant equations

    The average value formula.

    3. The attempt at a solution

    I know I can divide |x+1| into two integrals from [-2,0] and [0,2] (I think?? can anyone confirm this is the proper split?) and add and solve. But what can I do about sgnx?
     
    Last edited: Dec 6, 2009
  2. jcsd
  3. Dec 6, 2009 #2

    Dick

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    I would split |x+1| into two intervals. x<(-1) and x>=(-1). And sgn(x) I would split at x=0. You might want to think about splitting |x+1|*sgn(x) into three intervals [-2,-1], [-1,0] and [0,2].
     
  4. Dec 7, 2009 #3
    Ohh I see why you would split into three pieces. But...if I do split sgn(x) at x=0, which would mean [-2,0] and [0,2] and evaluate with |x| as my integral, wouldn't I get 0? (I know I'm suppose to get 4)
     
  5. Dec 7, 2009 #4

    HallsofIvy

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    No, [-2,0] and [0,2} is not three pieces! Dick said to use [-2,-1], [-1,0], and [0,2].
    On [-2,-1], x+1< 0 so |x+1|= -(x+1). x< 0 so sgn(x)= -1. |x+1|sgn(x)= -(x+1)(-1)= x+1. Integrate that from -2 to -1.

    On [-1,0], x+1> 0 so |x+1|= x+1. x< 0 so sgn(x)= -1. |x+1|sgn(x)= (x+1)(-1)= -(x+1). Integrate that from -1 to 0.

    On [0,1], x+1> 0 so |x+1|= x+1. x>0 so sgn(x)= 1. |x+1|sgn(x)= (x+1)(1)= x+ 1. Integrate that from 0 to 2.
     
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