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Math Help - Finding Area Enclosed by Three Curves

  1. #1
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    Question Finding Area Enclosed by Three Curves

    Find the area enclosed by
    y=3x
    y=6x
    y=(2/x)+1

    In the first quadrant of the Cartesian plane.

    I have already point the points on intersection to be (0,0) (2/3,4) and (1,3)

    I have posted a picture of my attempted solution, if anyone could tell me where I'm going wrong I'd really appreciate it.
    Thanks!
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  2. #2
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    Re: Finding Area Enclosed by Three Curves

    Quote Originally Posted by shennara123 View Post
    Find the area enclosed by
    y=3x
    y=6x
    y=(2/x)+1
    In the first quadrant of the Cartesian plane.
    Did you draw a graph?

    It should be \int_0^{2/3} {\left[ {6x - 3x} \right]dx}  + \int_{2/3}^1 {\left[ {\frac{2}{x} + 1 - 3x} \right]dx}

    Look at the graph
    Attached Thumbnails Attached Thumbnails Finding Area Enclosed by Three Curves-untitled.gif  
    Thanks from MarkFL
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