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Math Help - Area

  1. #1
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    Area

    a) Find the area bounded by the curve y - 3x + 3 = 0 and the line x = 4.

    b) Find the area bounded by the curve y = 4x - x and the line x = 0 and y = 4.
    Last edited by cazimi; February 2nd 2008 at 09:48 AM.
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  2. #2
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    Quote Originally Posted by cazimi View Post
    a) Find the area bounded by the curve y - 3x + 3 = 0 and the line x = 4.

    ...
    1. Transcribe this equation of the relation into equations of functions:

    f(x) = \sqrt{3} \cdot \sqrt{x-1} or f(x) = -\sqrt{3} \cdot \sqrt{x-1}

    2. Calculate the zero(s) of the function (or relation): y = 0

    -3x +3 = 0~\iff~ x = 1

    3. Therefore the area is:

    a=2\cdot \sqrt{3} \cdot \int_1^4\left(\sqrt{x-1}\right)dx = 2\cdot \sqrt{3} \cdot \left[\frac23 \cdot (x-1)^{\frac32}\right]_1^4 = 2\cdot \sqrt{3} \cdot \frac23 \cdot 3^{\frac32} = 12
    Attached Thumbnails Attached Thumbnails Area-area_parab_gerade.gif  
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  3. #3
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    Quote Originally Posted by cazimi View Post
    ...

    b) Find the area bounded by the curve y = 4x - x and the line x = 0 and y = 4.
    1. Calculate the difference of functions:

    g(x)=4-(-x^2+4x)

    2. Calculate the zeros of g:

    g(x) = x^2-4x+4 = (x-2)^2 thus x = 2 is the only zero.

    3. Calculate the area:

    a = \int_0^2\left( (x-2)^2\right)dx=\left[\frac13 (x-2)^3\right]_0^2 = \frac83
    Attached Thumbnails Attached Thumbnails Area-area_parab_gerade2.gif  
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