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Absolute Value Integration

  1. Nov 3, 2008 #1
    1. Evaluate

    [tex]\int_{-1}^{3} \left|x^2 -4\right| dx[/tex]




    3. The attempt at a solution

    This is the first time I'm trying this type of question & I think I need to use the following theorem for such questions;

    f is integrable on a closed interval a to b.

    [tex]\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx[/tex]

    (c; any point in between, no matter how the points are ordered)

    Therefore in my problem [tex]\int_{-1}^{3} \left|x^2 -4\right| dx[/tex], I choose c = 2

    Now since the integrand is absolute value we have two cases:

    1) x2-4
    2) 4-x2

    I'm not which one of the following is correct;

    [tex]\int_{-1}^{2} (4-x^2)dx + \int_{2}^{3}(x^2 -4)dx[/tex]

    OR

    [tex]\int_{-1}^{2} (x^2-4)dx + \int_{2}^{3}(4-x^2)dx[/tex]

    Which one is correct and why?

    Thanks!

     
  2. jcsd
  3. Nov 3, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    |x^2-4| is nonnegative everywhere. So the pair of integrals which have a nonnegative integrand over the region of integration is the correct one. Which is it? The first or the second?
     
  4. Nov 3, 2008 #3
    The first one is non-negative!
    I think I understand the idea now, thank you very much!!!
     
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