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Math Help - Area between curves

  1. #1
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    Area between curves

    1) y = |x| and y = x^2-6

    Using the definition of absolute value of x as a piecewise function, the intersects are -3 and 3.

    \int^3_{-3} (|x| - x^2-6)dx = \int^0_{-3}(-x-x^2-6)dx + \int^3_0 (x-x^2-6)dx

    \frac{-x^2}{2}-\frac{x^3}{3}-6x |^0_{-3} + \frac{x^2}{2} -\frac{x^3}{3}-6x|^3_0

    0 - (\frac{9}{2} - \frac{27}{3}-18)+(\frac{9}{2}-\frac{27}{3}-18)-0

    0 -(\frac{-45}{2})+(\frac{-45}{2})
    But that can't be right....

    2) y = 7x^2 and y = \frac{1}{7}x

    This one I think i could do if I could get started. But I keep end up having to find the square root of 1/7, which I can't seem to figure out how to express as a fraction. Maybe just a may problem, but I don't know.
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  2. #2
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    Quote Originally Posted by Open that Hampster! View Post
    1) y = |x| and y = x^2-6

    Using the definition of absolute value of x as a piecewise function, the intersects are -3 and 3.

    \int^3_{-3} (|x| - x^2-6)dx = \int^0_{-3}(-x-x^2-6)dx + \int^3_0 (x-x^2-6)dx

    \frac{-x^2}{2}-\frac{x^3}{3}-6x |^0_{-3} + \frac{x^2}{2} -\frac{x^3}{3}-6x|^3_0

    0 - (\frac{9}{2} - \frac{27}{3}-18)+(\frac{9}{2}-\frac{27}{3}-18)-0

    0 -(\frac{-45}{2})+(\frac{-45}{2})
    But that can't be right....

    2) y = 7x^2 and y = \frac{1}{7}x

    This one I think i could do if I could get started. But I keep end up having to find the square root of 1/7, which I can't seem to figure out how to express as a fraction. Maybe just a may problem, but I don't know.


    1. take advantage of the symmetry of the two graphs ...

    2 \int_0^3 x - (x^2-6) \, dx = 27    <br />
... fixed error


    2. find the points of intersection ...

    7x^2 = \frac{x}{7}

    49x^2 = x

    49x^2 - x = 0

    x(49x-1) = 0

    x = 0 , x = \frac{1}{49}

    \int_0^{\frac{1}{49}} \frac{x}{7} - 7x^2 \, dx
    Last edited by skeeter; February 18th 2010 at 05:19 PM. Reason: too many doubles ...
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  3. #3
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    Quote Originally Posted by skeeter View Post
    1. take advantage of the symmetry of the two graphs ...

    2 \int_0^3 x - (x^2-6) \, dx = 54<br />


    2. find the points of intersection ...

    7x^2 = \frac{x}{7}

    49x^2 = x

    49x^2 - x = 0

    x(49x-1) = 0

    x = 0 , x = \frac{1}{49}

    \int_0^{\frac{1}{49}} \frac{x}{7} - 7x^2 \, dx

    as usual u nailed it Skeetor, but the first one evaluates to 27 not 54.
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