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Math Help - Area between two functions

  1. #1
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    Area between two functions

    How do I find the area of the region bounded by the graphs y = root x, y= x-2 and y= 1

    I did the integral of root x minus 1 from 1 to 3 plus the integral of root x minus x plus 1 from 3 to 4. Apparently this is wrong, how should it be set up as?

    I apologize for not entering it in a proper math format, I have no idea how to do so.
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  2. #2
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    You MUST have some familiarity with basic functions.

    Find the intersections:

    (3,1), (1,1), (4,2)

    Decide on applicable regions.

    \sqrt{x} > 1\;on\;(1,3)

    The it switches to

    \sqrt{x} > x-2\;on\;(3,4)

    And you are almost done.

    \int_{1}^{3} \sqrt{x} - 1\;dx + \int_{3}^{4} \sqrt{x} - (x-2)\;dx

    Alternatively, and I'll let you explore where this comes from, it's a little easier on the y-axis:

    \int_{1}^{2} (y+2) - y^{2}\;dy
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  3. #3
    Member integral's Avatar
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    Find the area bounded by the equations bellow.

    y=\sqrt{x}
    y=x-2
    y=1


    \int^{1}_{0}\sqrt{x}dx=\frac{1^{1.5}}{1.5}=\frac{2  }{3}=.\overline{66}

    \int^{3}_{1}1d=3(1)-1(1)=2
    \int^{3}_{2}x-2dx=\int^{1}_{0}xdx=\frac{1}{2}

    2.\overline{66}-\frac{1}{2}=2.1\overline{66}

    edit: sorry this is wrong... forgive my stupidity
    Last edited by integral; February 20th 2010 at 08:21 PM.
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  4. #4
    MHF Contributor
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    Please rethink that, Integral. Just exactly what region is bounded by those equations? The bounds do not specify either the x-axis or the y-axis.
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