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Find the area of the figure bounded by the given pair of curves y = x2 - x + 3 a...



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  • Area bounded by
    y = x2 - x + 3 and y = 3
    x2 - x + 3 = 3
    x2 - x = 0
    x(x -1) = 0
    x= 0 and x = 1
    \(\int^{1}_{0}\)(x2 - x + 3)dx = [1/3x3 - 1/2x2 + 3x]\(^1 _0\)
    (1/3(1)3 - 1/2(1)2 + 3(1) - (1/3(0)3 - 1/2(0)2 + 3(0))
    = (1/3 - 1/2 + 3)- 0
    = \(\frac{2-3+18}{6}\)
    = 17/6

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