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A curve y has gradient Dy/dx=3x^2-6x+2a)if the curve passes through the origin find it equation b)Find the area of the finite region included between the curve in a)and the x-axis.?

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  1. dy/dx=3x^2 - 6x + 2
    y = x^3 - 3x^2 + 2x + c
    but it passes throught (0,0)
    0 = 0 - 0 + 0 + c -----> c = 0

    y = x^3 - 3x^2 + 2x

    b) y = x(x^2 - 3x + 2) = x(x - 1)(x - 2)
    so the x intercepts are 0, 1, 2
    giving you two enclosed areas between the curve and the x-axis

    You will need ∫ x^3 - 3x^2 + 2x dx from 0 to 1 + ∫ -x^3 + 3x^2 - 2x dx from 1 to 2

    You should have no difficulty with this straight-forward integral.

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  2. ∫ x^3 - 3x^2 + 2x dx from 0 to 1 + ∫ -x^3 + 3x^2 - 2x dx from 1 to 2

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