If F(x,y)=⟨3x2+4xy,2x2+1⟩\mathbf{F}(x,y) = \langle 3x^2 + 4xy, 2x^2 + 1 \rangleF(x,y)=⟨3x2+4xy,2x2+1⟩ is conservative, find the potential function f(x,y)f(x,y)f(x,y) such that ∇f=F\nabla f = \mathbf{F}∇f=F.
f(x,y)=x3+2x2y+y+Cf(x,y) = x^3 + 2x^2y + y + Cf(x,y)=x3+2x2y+y+C
f(x,y)=x3+x2y+y+Cf(x,y) = x^3 + x^2y + y + Cf(x,y)=x3+x2y+y+C
f(x,y)=3x2+4xy+2x2+Cf(x,y) = 3x^2 + 4xy + 2x^2 + Cf(x,y)=3x2+4xy+2x2+C
f(x,y)=x3+2xy2+y+Cf(x,y) = x^3 + 2xy^2 + y + Cf(x,y)=x3+2xy2+y+C