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Thread: Area between curves

  1. #1
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    Area between curves

    1) $\displaystyle y = |x|$ and $\displaystyle y = x^2-6$

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

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

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

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

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

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

    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) $\displaystyle y = |x|$ and $\displaystyle y = x^2-6$

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

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

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

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

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

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

    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 ...

    $\displaystyle 2 \int_0^3 x - (x^2-6) \, dx = 27
    $ ... fixed error


    2. find the points of intersection ...

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

    $\displaystyle 49x^2 = x$

    $\displaystyle 49x^2 - x = 0$

    $\displaystyle x(49x-1) = 0$

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

    $\displaystyle \int_0^{\frac{1}{49}} \frac{x}{7} - 7x^2 \, dx$
    Last edited by skeeter; Feb 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 ...

    $\displaystyle 2 \int_0^3 x - (x^2-6) \, dx = 54
    $


    2. find the points of intersection ...

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

    $\displaystyle 49x^2 = x$

    $\displaystyle 49x^2 - x = 0$

    $\displaystyle x(49x-1) = 0$

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

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

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