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Math Help - Find the x−coordinate of the point A

  1. #1
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    Find the x−coordinate of the point A

    I'm given the function f(x)=-3x(x-2) and must work out the x coordinate on f(x), A, which a line dividing the area between the function and the x-axis from [0,2] into 2 parts of equal area passes through. The area total area between the function and x-axis from [0,2] is 4 square units.

    I've got it down to -A^3+3A^2-2=0

    I'm kinda stuck there. Is there another way to do it that doesn't require me to solve polynomial equations?
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  2. #2
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    Quote Originally Posted by blackdragon190 View Post
    I'm given the function f(x)=-3x(x-2) and must work out the x coordinate on f(x), A, which a line dividing the area between the function and the x-axis from [0,2] into 2 parts of equal area passes through. The area total area between the function and x-axis from [0,2] is 4 square units.

    I've got it down to -A^3+3A^2-2=0

    I'm kinda stuck there. Is there another way to do it that doesn't require me to solve polynomial equations?
    Assuming your calculations are correct, it's not difficult to see that A = 1 is a possible solution ....
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  3. #3
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    Quote Originally Posted by blackdragon190 View Post
    I'm given the function f(x)=-3x(x-2) and must work out the x coordinate on f(x), A, which a line dividing the area between the function and the x-axis from [0,2] into 2 parts of equal area passes through. The area total area between the function and x-axis from [0,2] is 4 square units.

    I've got it down to -A^3+3A^2-2=0

    I'm kinda stuck there. Is there another way to do it that doesn't require me to solve polynomial equations?
    Does the line have to pass through a certain point on the x-axis?
    Numerous lines will divide the area in two equal parts.
    The most obvious one is the line x=1, since the graph is symmetrical about x=1.
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    -a^3+3a^2-2=0=-(a-1) (a^2-2 a-2)
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  5. #5
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    I don't think I expressed the problem well enough...here's a picture of what I mean:



    Argh, I just realized that my initial calculation is totally wrong.
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  6. #6
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    Let (x_0,-3(x_0-1)^2+ 3) be the point where that line crosses the parabola. Then the line has equation y= \frac{-3(x_0-1)^2+ 3}{x_0}x and the area, above that line but below the parabola is given by
    \int_0^x_0 -3x^2+ 6+ 3\frac{(x_0-1)^2- 3}{x_0}x dx
    Since the area under the entire parabola is 4, find that integral, in terms of x_0, set it equal to 2, and solve for x_0.
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  7. #7
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    Quote Originally Posted by blackdragon190 View Post
    I don't think I expressed the problem well enough...here's a picture of what I mean:



    Argh, I just realized that my initial calculation is totally wrong.
    If you integrate to get the area under the parabola from x=0 to x=A,
    then subtract the area of the triangle, that resulting area is 2.

    The area of the triangle is \displaystyle\frac{Af(A)}{2}

    Alternatively, integrate from x=A to x=2 and add the triangle area.
    Those combined areas are also 2.

    f(x)=-3x^2+6x

    f(A)=-3A^2+6A

    \displaystyle\int_{0}^Af(x)}dx=2+\frac{Af(A)}{2}

    \displaystyle\int_{0}^A\left(-3x^2+6x\right)}dx=2+\frac{-3A^3+6A^2}{2}

    which leads to

    A^3=4
    Attached Thumbnails Attached Thumbnails Find the x−coordinate of the point A-parabola2.jpg  
    Last edited by Archie Meade; October 7th 2010 at 02:30 PM.
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