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Math Help - Area of a region

  1. #1
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    Area of a region

    What is the area of the region between the graphs of y=x^2 and y = .x from x= 0 to x = 2?

    a. (2/3)
    b. (8/3)
    c. 4
    d. (14/3)
    e. (16/3)
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  2. #2
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    Quote Originally Posted by frozenflames
    What is the area of the region between the graphs of y=x^2 and y = .x from x= 0 to x = 2?

    a. (2/3)
    b. (8/3)
    c. 4
    d. (14/3)
    e. (16/3)
    Hello,

    as you may have noticed, the 2nd equation is damaged slightly. By the given borders (0 and 2) I guessed, that this equation should be linear and that the intercepting point between parabola and straight line should be (2,4). Therefore I guess that the 2nd equation should be y=2*x. If not, all following calculations are wrong!

    You calculate the enclosed area as integral of the difference of functions, from one intercepting point to the next one.

    The difference is here: d(x)=2x-x^2 and the intercepting points are (0,0) and (2,4).

    The area is \int^{2}_{0}{d(x)\ dx}=\int^{2}_{0}{\left(2x-x^2 \right) dx}=\left[x^2-\frac{1}{3}x^3 \right]^{2}_{0}=4-\frac{8}{3}=\frac{4}{3}

    This answer is missing in your list of possible answers, so my assumption about the 2nd equation must be wrong. Sorry that I couldn't help you.

    Greetings

    EB
    Last edited by earboth; March 19th 2006 at 12:27 AM.
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