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Quick question about total area (involving integrals)

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the total area betwen the curve y=x^2-4 and the interval [0,3].

    2. Relevant equations



    3. The attempt at a solution

    Okay, so I already know the answer, but I'm unsure about one thing.

    It's ture that you have to break the equation such that it says [tex]\int_{2}^{3} x^2 dx[/tex] - [tex]\int_{0}^{2} 4 dx[/tex], but how would you go about finding which upper and lower limits to set for the two integration problems like that?

    I know how to take care of the rest once I find those, but I'm not sure how to get the upper and lower limits once I break the equation into two.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 26, 2007 #2

    Dick

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    The integral will give you the area above the x-axis minus the area below the x-axis. If you want the sum of the two instead of the difference then you need to figure out where the curve crosses the x-axis.
     
  4. Feb 27, 2007 #3

    Gib Z

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    And you can't break up the integral like you did. You can either do
    [tex]\int^3_0 x^2 - 4 dx= \int^3_0 x^2 dx - \int^3_0 4 dx[/tex] or
    [tex]\int^3_0 x^2-4 dx = \int^3_2 x^2 - 4 dx + \int^2_0 x^2 - 4 dx[/tex]
     
  5. Feb 27, 2007 #4

    HallsofIvy

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    Do you mean "find the area between the curve y= x2- 4 and the x-axis for x in [0,3]?

    No. y= x2- 4 crosses the x-axis at x= 2. For x> 2, y is negative and the integral is the negative of the area. The area between the curve and the x-axis for x between 2 and 3 is
    [tex]-\int_2^3 (x^2- 4)dx[/tex]
    so the "total area" between the curve and the x-axis is
    [tex]\int_0^2 (x^2- 4)dx- \int_2^3 (x^2- 4)dx[/tex]



     
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