Show that $\int_{\left(-6,2\right)}^{\left(-4,10\right)}fdx+gdy = \int_{\left(-6,2\right)}^{\left(-4,10\right)}40x^{3}y^{2}dx+20yx^{4}dy$ is independent of path:
$\frac {\partial f}{\partial y}$ =
$\frac {\partial g}{\partial x}$ =
$\int_{\left(-6,2\right)}^{\left(-4,10\right)}40x^{3}y^{2}dx+20yx^{4}dy$ =