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Solving integrals with absolute values

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data
    solve the integral [abs(x+1)(3+abs(x))]/(x+1) between -3 and 1


    2. Relevant equations



    3. The attempt at a solution
    when x<-1 then [abs(x+1)(3+abs(x))]/(x+1) = [-(x+1)(3-x)]/(x+1) = -(3-x)
    when -1<x<0 then [abs(x+1)(3+abs(x))]/(x+1) = (x+1)(3-x)/(x+1) = 3-x
    when x>0 then [abs(x+1)(3+abs(x))]/(x+1) = (x+1)(3+x)/(x+1) = 3+x

    so now i have:
    (x-3)dx(between -3 and -1)+(3-x)dx(between -1 and 0)+(3+x)dx(between 0 and 1)
    = ((1/2)x^2-3x)(between -3 and -1)+(3x-(1/2)x^2)(between -1 and 0)+(3x+(1/2)x^2)(between 0 and 1)
    = (1/2)+3-((9/2)+9)+(0)-(-3-(9/2))+(3+(1/2))-0
    = 1/2+3-9+3+3+1/2 = 1
    i should have got -3... help??
     
  2. jcsd
  3. Aug 12, 2011 #2

    I like Serena

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    Hi again Shannabel! :smile:

    Can you recalculate (3x-(1/2)x^2)(between -1 and 0)?
     
  4. Aug 12, 2011 #3
    thankyou!! :)
     
  5. Aug 12, 2011 #4

    I like Serena

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    I take it you did? And that you found the proper solution?

    Then you're welcome! :smile:
     
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