I am supposed to evaluate $F(x,y)=(4x^3y^2-2xy^3)\,\mathbf{i} + (2x^4y-3x^2y^2+4y^3)\,\mathbf{j}$ along the curve $r(t)=(t+\sin(t\pi))\,\mathbf{i}+(2t+\cos(t\pi)\,\mathbf{j}$, $0\le t\le 1$.
I could put try to do the $\int_0^1 f(r,t) r'(t)\,dt$, but that seems really messy. Is there a simpler way?
Note that $\vec{F}$ is conservative, as it is the gradient of $$ f(x,y)=x^4y^2-x^2y^3+y^4. $$
It follows by the fundamental theorem of calculus that
$$ \int_C \vec{F}\cdot d\vec{r}=\int_C \nabla f\cdot d\vec{r}=f(\vec{r}(1))-f(\vec{r}(0))=f(1,1)-f(0,0)=1. $$