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Let $U\subset \mathbb{R}^2$ be open, and $F:U\to \mathbb{R}^2$ a $C^1$-vector field. Assume that:

$$\frac{\partial{F_1}}{\partial{x_2}}(x)=\frac{\partial{F_2}}{\partial{x_1}}(x)\quad\forall x\in U$$

Let $\triangle \subset U$ be a closed right-angled triangle.

Show that $\int_\gamma F\cdot dx=0$, where $\gamma$ is any rectifiable path( a $C^1$ function $\gamma: [a, b] \to \mathbb{R}^2$ with $a,b\in\mathbb{R}$)describing $\partial \triangle$(the boundary of $\triangle$). Deduce from this, that $\int_\gamma F\cdot dx=0$, where $\gamma$ is the boundary of any closed triangle contained in $U$.

I do not see that the equality of partial derivatives helps. Here F is unknown, that makes this question confusing. I am stuck in the beginning. What are relations between the partial derivative and the integral of this vector field?

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