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Math Help - Area bounded by curves(check answer)

  1. #1
    Member javax's Avatar
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    Area bounded by curves(check answer)

    Hello.
    Calculate the area bounded by these curves:
    y=-x^2+4x-3
    y=4x-3
    x=2

    My answer is \frac{77}{12} . I went like this:
    For the area above x
    S_1 = \int\limits_{\frac{3}{4}}\limits^{2}{(4x-3)}dx-\int\limits_{1}\limits^{2}{(-x^2+4x-3)dx}
    and for the area under x
     <br />
S_2 = -\Bigg(\int\limits_{0}\limits^{1}{(-x^2+4x-3)dx}-\int\limits_{0}\limits^{\frac{3}{4}}{(4x-3)dx}\Bigg)<br />
    So S = S_1+S_2 Someone tell me if the answer is correct cause I don't have the result.

    Thanks
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  2. #2
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    skeeter's Avatar
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    the parabola and line intersect at x = 0, and , except at x = 0,
    4x-3 > -x^2 + 4x - 3

    A = \int_0^2 (4x - 3) - (-x^2 + 4x - 3) \, dx = \frac{8}{3}
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  3. #3
    Member javax's Avatar
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    I'm confused. Why you didn't need to separate the area, area above x-axis and area under x-axis(cause under x it will have a -)

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  4. #4
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    it does not make a difference.

    if f(x) > g(x), then the area between them is A = \int_a^b f(x) - g(x) \, dx no matter where f and g reside relative to the x-axis.

    here is a simple analogy ... consider two points (1, 5) and (1, 2) ... the distance between the two points is 5 - 2, correct?

    now, what about (1,5) and (1,-2) ... isn't the distance 5 - (-2) ?

    it's the same idea with the functions.
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