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Homework Help: Definite integrals of absolute values

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\int|x^{2}+x-2|dx[/tex] from -2 to 2


    2. Relevant equations
    The integral of f(x) from a to b = F(b) - F(a)
    |x| = { x if x >= 0; -x if x < 0

    ---

    Ok, I don't know how to do the definite integrals of absolute values.. was never shown an example of it in class, but I kind of have an idea...

    we know that
    |x| = { x if x >= 0; -x if x < 0,

    and then im lost from here....

    btw, x= 1, -2
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2

    Dick

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    If you want to apply your definition of abs, you'll first have to figure out where x^2+x-2 is positive and negative.
     
  4. Dec 9, 2007 #3

    rock.freak667

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    quick sketches of the function will help you...
     
  5. Dec 9, 2007 #4
    ok so from -2 to 1, it is negative and from 1 to 2 it is positive...

    do i just split up the integral from here?
    like from -2 to 1, its negative, so the integral would be negative, and from 1 to 2, it is postive so the integral would be postive...

    and just from inspection, would the integral be equal to 0 because its symmetric and one is a negative the other is postive...
     
  6. Dec 9, 2007 #5

    Dick

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    The integral of x^2+x-2 from -2 to 1 IS negative. The integral of |x^2+x-2| is not. Do you see why?
     
  7. Dec 9, 2007 #6
    nvm. i got 19/3...
     
    Last edited: Dec 9, 2007
  8. Dec 9, 2007 #7
    I dunno if I do..

    is it because the absolute value of any function is always positive?
     
  9. Dec 9, 2007 #8

    rock.freak667

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    Well from 2 to one it is +ve so you want to get
    [tex]\int_1 ^{2} (x^2+x-2)dx[/tex]

    and from 1 to -2 it is -ve so you want to find

    [tex]\int_{-2} ^{1} -(x^2+x-2)dx[/tex]

    and sum them up
     
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