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

Expert Answers
justaguide eNotes educator| Certified Educator

To find the curve we integrate the given dy/dx = 3x^2 - 2x.

Int [ 3x^2 - 2x dx ]

=> 3*x^3 / 3 - 2x^2 / 2 + C

=> x^3 -  x^2 + C

As the curve passes through (2 , 5)

5 = 2^3 - 2^2 + C

=> 5 = 8 - 4 + C

=> C = 1

The required curve is y = x^3 - x^2 + 1

giorgiana1976 | Student

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.

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question