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Does absolute makes a difference in area

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data

    y = 2 − x^2, y = |x|
    sketch and find the area of the region determined
    by the intersections of the curves

    2. Relevant equations

    regular f - g of integral

    3. The attempt at a solution

    The absolute value puzzles me. How would it work then?

    I mean we know how each equation looks like.
    So I would do
    integral of (2-x^2) - |x| from a to b

    Now how would I get the intersection point and evaluate this integral with |x|?

  2. jcsd
  3. Mar 5, 2010 #2


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    Split into the cases x<0 and x>=0. What's a simpler expression for |x| in each case? What's the intersection point in each case?
  4. Mar 5, 2010 #3
    2-x-x^2 = 0
    this gives me -2, 1

    and how do i integrate it?
    treat it as a regular x but keep || when i do the calculation?
  5. Mar 5, 2010 #4


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    The point here is that for x>=0 |x|=x. For x<0 |x|=(-x). You have to split the integration into two parts. And you also have to find the intersection separately for each part. x=(-2) is NOT an intersection.
  6. Mar 6, 2010 #5
    hi thanks for the help.
    i still can't figure out the x <0
    but after examining the graph, i see they are symmetry, so instead, i take 2 * integral of the positive, will produce the answer

    yet, i still want to know how to find the intersection point for x <0
  7. Mar 6, 2010 #6


    Staff: Mentor

    Your values of x above are the x-coordinates at the points of intersection of y = 2 - x^2 and y = x. The only one that is valid is x = 1, because |x| = x only if x >= 0.

    The other point of intersection is found in almost the same way that you found the first one. The only difference is that you are looking for the point of intersection of y = 2 - x^2 and y = -x. The latter equation is due to the fact that |x| = -x for x <=0.

    So, solve 2 - x^2 = - x for x. The domain this time is x <= 0, so discard any positive x values. You have already commented on the symmetry of the two graphs, so your result should not be surprising.
  8. Mar 6, 2010 #7
    Hi, thanks for your help. Now I get it. ^^
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