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.?

Answer 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.

Answer 2

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