A curve has dy/dx=3x^2-2x. The curve passes through the point (2;5). What is the equation of the curve?

To determine the equation of the curve, we'll have to
determine the antiderivative of the given expression.

Int
dy = Int (3x^2-2x)dx

We'll use the property of integrals to
be additive:

Int (3x^2-2x)dx = Int 3x^2 dx - Int
2xdx

Int (3x^2-2x)dx = 3 Int x^2dx - 2Int
xdx

Int (3x^2-2x)dx = 3*x^3/3 - 2*x^2/2 +
C

We'll simplify and we'll
get:

Int (3x^2-2x)dx = x^3 - x^2 +
C

What we've get is not a curve, but a family of curves
that depends on the values of the constant C.

We know, from
enunciation that the point  (2 , 5) is located on the curve. Therefore it's coordinates
will verify the equation of the curve.

5 = (2)^3 - (2)^2 +
C

5 = 8 - 4 + C

5 = 4 +
C

C = 5 - 4

C =
1

The equation of the curve, whose derivative
is dy/dx=3x^2-2x, is: y =  x^3 - x^2 + 1.