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Math Help - finding area

  1. #1
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    Exclamation finding area

    sketch and find the area between y=-x,& x=y-y^2

    im having trouble drawing the (x=y-y^2) any help pls
    thnx
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  2. #2
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    Quote Originally Posted by alex83 View Post
    sketch and find the area between y=-x,& x=y-y^2

    im having trouble drawing the (x=y-y^2) any help pls
    thnx
    Completing the square:

    -x = (y^2-y+\frac{1}{4}) - \frac{1}{4}

    -x + \frac{1}{4} = (y-\frac{1}{2})^2

    y-\frac{1}{2} = \pm \sqrt{-x + \frac{1}{4}}

    y = \frac{1}{2} \pm \sqrt{-x + \frac{1}{4}}

    You should be able to sketch that and find the area.
    Remember that the positive and negative square roots are different functions: you may need to divide the area up into several integrals.
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  3. #3
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    Another thought: the area between two functions f(x) and g(x) is the same as the area between the inverse function f^{-1}(x) and the original function g(x). It will be easier to calculate the area this way, as it requires only one integral.
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  4. #4
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    Quote Originally Posted by calum View Post
    Another thought: the area between two functions f(x) and g(x) is the same as the area between the inverse function f^{-1}(x) and the original function g(x). It will be easier to calculate the area this way, as it requires only one integral.
    This simply is not true. You may be thinking of the area between [tex]x= f^{-1}(y)[/itex] and x= g^{-1}(y)

    alex83, to find the graph, I would write x= y- y^2= y(1- y) so the graph crosses the y-axis at (0, 0) and (0, 1). Since the coefficient of y^2 is negative, it is a parabola opening to the left going through those points.

    More important is the fact that y= -x is the same as x= -y so the two graphs intersect when x= -y= y- y^2 or, adding y to both sides, 2y- y^2= y(2-y)= 0 so y= 0 or y= 2. The two graph intersect at (0,0) and (-2, 2).

    Yes, you could do this integrating in terms of x, one integral from -2 to 0, the other from 0 to 1/4.

    But it might be simpler to do the integral in terms of y (perhaps that was what calum was thinking of in his second post). Drawing a horizontal line from from x= -y to x= y- y^2, we see that the height of a rectangle (in the Riemann sum) is (y- y^2)- (-y)= 2y- y^2 and that the "width" is dy. The area of any such rectangle is (2y- y^2)dy and so the area is their "sum": \int_0^2 (2y- y^2)dy.

    That should be very easy.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    This simply is not true. You may be thinking of the area between [tex]x= f^{-1}(y)[/itex] and x= g^{-1}(y)
    Hmmm...that is what I meant, I got confused because when g(x) = -x,    g^{-1}(x) = -x ie the same, so for this particular problem this statement works, but not in general.

    What I mean to say is if you have two functions f(x) and g(x), the area between the functions is equal to the area between the two inverse functions f^{-1}(x) and g^{-1}(x)
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