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Math Help - Definite integrals

  1. #1
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    Definite integrals

    I need to express the area of the region as a definite integral.
    f(x)=-x^(2)-2x+3
    Thanks for your help.
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  2. #2
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    Quote Originally Posted by nystudent2729 View Post
    I need to express the area of the region as a definite integral.
    f(x)=-x^(2)-2x+3
    Thanks for your help.


    \int_{a}^{b}{f(x)}dx = F(b) - F(a)

    where F is the antiderivative of f(x)

    You need to give us the a and b so that we can tell you the area of the reagion.

    For example if you want the area of a region of f(x)=-x^(2)-2x+3 from x=0 to x=6

    the answer should be

    \int_{0}^{6}{(-x^{2}-2x+3)} dx
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  3. #3
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    Definite Integral

    I need to express the area of the region from -3 to 1 as a definite integral:
    f(x)=-x^(2)-2x+3
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  4. #4
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    Quote Originally Posted by nystudent2729 View Post
    I need to express the area of the region from -3 to 1 as a definite integral:
    f(x)=-x^(2)-2x+3
    \int_{-3}^{1}{(-x^{2}-2x+3) } dx = area of region
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  5. #5
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    Dear nystudent2729,

    you can easy check out that -3 and 1 is the roots of equation f(x) = 0. And f(x) > 0 in the given interval, therefore the area is the integral from -3 to 1.

    This is elemtary integral, what is the problem?
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  6. #6
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    Quote Originally Posted by fonso_gfx View Post
    \int_{a}^{b}{f(x)}dx = F(b) - F(a)

    where F is the antiderivative of f(x)

    You need to give us the a and b so that we can tell you the area of the reagion.

    For example if you want the area of a region of f(x)=-x^(2)-2x+3 from x=0 to x=6

    the answer should be

    \int_{0}^{6}{(-x^{2}-2x+3)} dx
    This doesn't give the area of the region. It gives the signed area of the region. There's a big difference.
    Last edited by mr fantastic; December 18th 2008 at 10:38 PM.
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