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    Find the area of the region enclosed by the curves with equations y=x^2-16 and y=4x-x^2?

    I know you find the points of intersection first but then i'm stuck could anyone pleaseeeee help me ?
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    Might help you to draw a diagram, and then break it down into areas which you know how to calculate.
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    The area between two curves is the same as the area between the difference of the two curves and the x-axis. [If it's negative, get rid of the minus sign.]

    So for example the area between \sin x and \cos x between 0 and \dfrac{\pi}{4} is \displaystyle \int_0^{\frac{\pi}{4}} (\cos x - \sin x)\, dx.
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    Make them equal each other so you have one equation in terms of x.

    Find x (2 values).

    Integrate the first minus the second and use the x values you just found for the limits.
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    i'm still stuck on this
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    how do you find the points of intersection
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    find intersection points (-2,4)
    plot on graph make it visible where R is. Integration of upper curve minus lower curve integral. Using the intersection points as limits
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    y=x^2-16 and y=4x-x^2

    x^2-16=4x-x^2

    solve for x
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    ive tried that i keep getting the wrong answer
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    You may have to use the quadratic forumla, and you will get 2 intersection points
    which are roughly 6.472 and -2.472

    Those are the x values, so you use the definite intergral of (x^2-16)-(4x-x^2) between 6.472 and -2.472 to get the area
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    (Original post by moesizlak)
    ive tried that i keep getting the wrong answer
    show what you have done and someone will be able to point out your error
 
 
 
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Updated: March 28, 2011
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