Find the average value of a function

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Homework Statement


Find the average value of a function of f(x)= |x+1| sgnx on the interval [-2,2]


Homework Equations



The average value formula.

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?
 
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Answers and Replies

  • #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].
 
  • #3
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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].
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)
 
  • #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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