Question

Find the equation of a curve that passes through the point (1;4) and dy/dx=4x^3+4x

justaguide · Accepted Answer

We have dy/dx = 4x^3 + 4x.

dy/dx = 4x^3 + 4x

=> dy = (4x^3 + 4x) dx

Integrate both the sides

Int [ dy ] = Int [ (4x^3 + 4x) dx ]

=> y = 4x^4 / 4 + 4*x^2 / 2

=> y = x^4 + 2*x^2 + C

As the curve passes through (1 , 4)

4 = 1^4 + 2*1^2 + C

=> 4 = 1 + 2 + C

=> C = 1

This gives y = x^4 + 2*x^2 + 1

The required equation of the curve is y = x^4 + 2*x^2 + 1

giorgiana1976 · Answer

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

Int dy = Int (4x^3+4x)dx

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

Int (4x^3+4x)dx = Int 4x^3 dx + Int 4xdx

Int (4x^3+4x)dx = 4 Int x^3dx + 4Int xdx

Int (4x^3+4x)dx = 4*x^4/4 + 4*x^2/2 + C

We'll simplify and we'll get:

Int (4x^3+4x)dx = x^4 + 2x^2 + C

The expression represents a family of curves that depends on the values of the constant C.

We know, from enunciation that the point (1 , 4) is located on the curve. Therefore it's coordinates will verify the equation of the curve.

4 = (1)^4 +2*(1)^2 + C

4 = 1 + 2 + C

4 = 3 + C

C = 4 - 3

C = 1

The equation of the curve, whose derivative is dy/dx=4x^3+4x , is the complete square: y = x^4 + 2x^2 + 1 = (x^2 + 1)^2.