equation of curve?

Question

the gradient of the tangent of a point on a curve is given by X^2+X-2=0 now how will we find the equation of the curve if it pass through (2,1)

Answer 1

The slope of a curve is found by taking the derivative. So taking the integral of the slope should give you the curve plus a constant. Taking the integral we have:

∫(x² + x - 2)dx = x³/3 + x²/2 - 2x + C

y = x³/3 + x²/2 - 2x + C

Plugging in the point (2,1) we have

1 = 8/3 + 4/2 - 4 + C = 2/3 + C
1/3 = C

The equation of the curve is

y = x³/3 + x²/2 - 2x + 1/3