TITLE: What does it mean for an integral to be independent of a path?
QUESTION [7 upvotes]: What does it mean for an integral to be independent of a path? Take the path $x^4-6xy^3=4y^2$ for
$$
\int_{(0, 0)}^{(2,1)} \left(10x^4-2xy^3\right)\,{dx}-3x^2y^2\, dy
$$
My textbook says the integral is independent of the path since
$$
\frac{\partial}{\partial y}\left(10x^4-2xy^3\right)
 = \frac{\partial}{\partial x}\left(3x^2y^2\right)\,{dy}.
$$
But I've no idea what these partial derivatives (from Green's theorem) being equal has to do with it.

REPLY [4 votes]: Integrating a Gradient
Since
$$
\nabla\!\left(2x^5-x^2y^3\right)=\left(10x^4-2xy^3,-3x^2y^2\right)\tag{1}
$$
we get
$$
\begin{align}
\int_{(x_0,y_0)}^{(x_1,y_1)}\nabla\!\left(2x^5-x^2y^3\right)\cdot(\mathrm{d}x,\mathrm{d}y)
&=\int_{(x_0,y_0)}^{(x_1,y_1)}\mathrm{d}\!\left(2x^5-x^2y^3\right)\\
&=\left.2x^5-x^2y^3\vphantom{\int}\right|_{(x_0,y_0)}^{(x_1,y_1)}\tag{2}
\end{align}
$$

Exact Differential
This is also saying, using the chain rule, that
$$
\begin{align}
\mathrm{d}\!\left(2x^5-x^2y^3\right)
&=\frac{\partial}{\partial x}\left(2x^5-x^2y^3\right)\mathrm{d}x+\frac{\partial}{\partial y}\left(2x^5-x^2y^3\right)\mathrm{d}y\\
&=\left(10x^4-2xy^3\right)\mathrm{d}x+\left(-3x^2y^2\right)\mathrm{d}y\tag{3}
\end{align}
$$
which is to say that $\left(10x^4-2xy^3\right)\mathrm{d}x+\left(-3x^2y^2\right)\mathrm{d}y$ is an exact differential.

Equality of Mixed Partials
We can check that $(3)$ is an exact differential using the Equality of Mixed Partials. That is, since
$$
\frac{\partial^2}{\partial x\,\partial y}\left(2x^5-x^2y^3\right)=\frac{\partial^2}{\partial y\,\partial x}\left(2x^5-x^2y^3\right)\tag{4}
$$
we have
$$
\frac\partial{\partial y}\overbrace{\left(10x^4-2xy^3\right)}^{\frac\partial{\partial x}\left(2x^5-x^2y^3\right)}=\frac\partial{\partial x}\overbrace{\left(-3x^2y^2\right)}^{\frac\partial{\partial y}\left(2x^5-x^2y^3\right)}\tag{5}
$$
Thus, the fact that $(3)$ is an exact differential implies $(5)$. However, to show that $(5)$ implies that $(3)$ is an exact differential, we need to use Green's Theorem.