# Show that a line integral is path independent?

I have to show that a line integral is path independent between two points, and while I know how to check if one is, I have no idea where to begin proving that one is. The equation looks very simple to integrate, but I don't even know where to start. The line integral is (3x-5y)dx+(7y-5x)dy, and it is over the line segment from (1,3) to (5,2). The only thing I have figured out is that the path is (-t,4t) with t ranging from 0 to one, but after that I'm lost.

• No, the path cannot be $(-t,4t)$. That connects $(0,0)$ with $(-1,4)$. Do you know the definition of line integral? Use it., – Pedro Tamaroff Dec 4 '14 at 20:37
• Hint: use (Green's theorem)[en.wikipedia.org/wiki/Green%27s_theorem] – sid Dec 4 '14 at 20:45

Hints:

To show path independence, suppose you have a closed curve $C$ through any two points. Then, applying Green's theorem, we have:

$$\int_{C} Ldx + Mdy = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right)dxdy$$

Where $D$ is the region enclosed by $C$ and $L = 3x - 5y$ and $M = 7y-5x$. So what happens to the integrand on the RHS above, and what can we conclude?

Based on this work, what is a sufficient condition for showing that a general vector field is path-independent?

To compute the specific path integral for the line segment connecting $(1, 3)$ and $(5, 2)$, we first need to get a correct parametrization. Pedro is correct in the comments. You've gone astray since at $t = 0$, your parametrization is putting you at the origin.

All lines can be parametrized in the form $\alpha(t) = \mathbf{p_0} + \mathbf{c}t$. Given that we want to be at $(1, 3)$ at time $t = 0$, then what is a good choice for $\mathbf{p_0}$? What, then, must $\mathbf{c}$ be if we wish to be at $(5, 2)$ when $t = 1$?

Of course, that is the most difficult part. Once you have a good parametrization, the rest is just plug-and-chug.

Find a potential function for the vector field $<3x-5y,7y-5x>$ (you can succeed because the mixed partial condition holds). The value of the line integral equals the change in potential between the endpoints, which of course only depends on the endpoints; that's the definition of path independence.