Respuesta :

The integral is path-independent if there is a scalar function [tex]f[/tex] whose gradient is

[tex]\nabla f=(2e^x\cos2y,-\sin2y)[/tex]

(at least, that's what it looks like the given integrand is)

Then

[tex]\dfrac{\partial f}{\partial x}=2e^x\cos 2y\implies f(x,y)=2e^x\cos2y+g(y)[/tex]

Differentiating both sides with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=-4e^x\sin 2y\neq-\sin2y[/tex]

so the line integral *is* dependent on the path. (again, assuming what I've written above actually reflects what the question is asking)

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