What is the equation of the curve?

Dy/dx = 3x^2 +4, find the equation of the curve?

• The gradient of a curve at the point (x, y) on the curve is given by dy/dx = 3x^2 +4 . Given that the point (1,7) lies on the curve find the equation of the curve. --------------------------- dy/dx = 3x^2 +4 I integrated 3x^2 +4: dy/dx = 3x^2 +4 f(x) = x^3 +4x + c (1, 7) 7 = (1)^3 + 4(1) + c 7 = 1 + 4 + c 7 = 5 + c c = 7 - 5 c = 2 Equation of the line: x^3 +4x + 2 Is this correct?

• Answer:

  Yup. f(1) = 1^3 + 4*1 + 2 f(1) = 7 and f(x) = x^3 + 4x + 2 f '(x) = 3x² + 4