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Math Help - Integration:area between curve

  1. #1
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    Integration:area between curve

    area between curves f(x)= x^2 and g(x)=-x^2 +1

    is the answer (2/3)x^3 + x + C please?
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  2. #2
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     x^2 = -x^2 + 1
     x = \pm \sqrt{\frac{1}{2}}
     \int_{\sqrt{\frac{-1}{2}}}^{\sqrt{\frac{1}{2}}} [(-x^2 + 1) - (x^2)] dx =
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  3. #3
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    YES!(WELL THIS IS AN EDIT AND ACTUALLY NO, ITS OFF BY A NEGATIVE SIGN...SEE SECOND POST) This is the answer except you might want to include the limits. Because this is a closed region you can find the area.

    You set everything up right just like this: you might want to include limits depending on if you questions wants the actual area or the expresssion for the area.
    integral of (-x^2+1)-(x^2) dx - Wolfram|Alpha

    Here is a picture of their graphs. You want the area inbetween them.
    Plot x^2, -x^2+1 - Wolfram|Alpha

    -x^2+1 has the graph on top of the other for the region we are looking for. So the expression that we find the integral over is would be:

    (-x^{2}+1)-(x^{2}) or
    -2x^{2}

    The limits of integral are from where the two graphs intersect. Set the two equations equal to eachother to find what x values they have in common....The intersection points are -Sqrt(2)/2 and +Sqrt(2)/2.

    This means your integral should look like:
    \int_{-Sqrt(2)/2}^{+Sqrt(2)/2} -2x^{2} dx

    You should get (2/3)Sqrt(2) as the area between the curves. (or just substitute in the limits into your expression that you already have, these are the same things!!).
    integral of (-x^2+1)-(x^2) dx from -Sqrt(2)/2 to Sqrt(2)/2 - Wolfram|Alpha
    Last edited by snaes; April 6th 2010 at 02:45 PM.
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  4. #4
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    How come the limits is sqrt 1/2
    and the other time its sqrt2/2?

    I am getting limits of sqrt(1/2)
    and a final answer of 1.65 area?
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  5. #5
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    Oh, I see, this confused me for a mintue, but check this out:

    Sqrt(.5)=.7071037812
    Sqrt(2)/2=.7071067812

    They actually turn out the be the same thing.

    We are both right .


    Okay, so we want (-x^{2}+1)-(x^2) because we want the top curve minus the bottom curve.
    -2x^{2}+1

    Lets integrate this without the limits first...
    \int-2x^{2}+1
    \dfrac{-2}{3}x^{3}+x+c
    ***Please not that this has a \dfrac{-2}{3} NOT \dfrac{+2}{3}. In my first post, i overlooked that you missed the negative sign.
    Now put in the limits of integration: \sqrt{.5} and -\sqrt{.5}....

    I get .9428....or the exact answer is: \dfrac{2}{3}\sqrt{2}

    Hopefully, I got everything this time?
    Last edited by snaes; April 6th 2010 at 02:47 PM. Reason: typos
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