Question

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

Accepted Answer

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

Answer

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.

Calculus

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

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