1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Definite integrals of absolute values
Loading...