The gradient of a curve is defined as Dy/dx = 3x^2 + 3x and it passes through the point (0,0), what is the equation of the curve
Integrate this = (3x^3)/3 + (3x^2)/2 + c So y = x^3 + (3x^2)/2 + c Using point (0,0), 0 = 0 + 0 + c so c = 0. Full equation of the curve is therefore x^3 + (3x^2)/2
LT
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