Math 61 part 2

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Find the centriod of the boundEd by the curvez y=x^2 and y=2x+3

  • Math 61 part 2 -

    The area of the region is

    A = ∫[-1,3] (2x+3)-x^2 dx
    = -1/3 x^3 + x^2 + 3x [-1,3]
    = (-9+9+9)-(1/3 + 1 - 3)
    = 32/3

    For xbar, find
    X = ∫[-1,3] x((2x+3)-x^2) dx
    = -1/4 x^4 + 2/3 x^3 + 3/2 x^2 [-1,3]
    = (-81/4 + 18 + 27/2)-(-1/4 + 2/3 + 3/2)
    = 32/3

    For ybar, find
    Y = ∫[-1,3] 1/2 ((2x+3)^2 - (x^2)^2) dx
    = -1/5 x^5 + 4/3 x^3 + 6x^2 + 9x [-1,3]
    = 544/15

    xbar = X/A = 1
    ybar = Y/A = 17/5

  • Math 61 part 2 -

    To solve this category of problems where you need the area/centroid of a region bounded by two curves, you need to first find TWO intersection points of the two curves by equating y1(x)=x^2 and y2(x)=2*x+3.
    The intersection points are at x=-1 and x=3, with y2(x) above y1(x).
    Then you need to find the area by integrating
    A=∫(y2(x)-y1(x))dx between the limits x=-1 and x=3.
    =32/3

    To find the centroid(xBar,yBar), you need to find the first moments,
    xBar=∫x(y2(x)-y1(x))dx / A
    and
    yBar=∫(1/2)(y2(x)+y1(x))(y2(x)-y1(x))dx /A
    from which I get
    (xBar,yBar)=(1,17/5)

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