[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec s=\int_0^1(t^2+t^4,4t^3)\cdot(1,2t)\,\mathrm dt=\int_0^1(t^2+9t^4)\,\mathrm dt=\boxed{\frac{32}{15}}[/tex]
[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1(2t^2,4t^2)\cdot(1,1)\,\mathrm dt=\int_0^16t^2\,\mathrm dt=\boxed{2}[/tex]
The value of the line integral depends on the path, so [tex]\vec F[/tex] is not a gradient vector field.
