$\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}}$
Definition 1. If $L$ is an oriented curve, parametrized $(x(t),y(t))$ with $t$ running from $t_0$ to $t_1$, so $(x(t_0),y(t_0))$ is it's start-point and $(x(t_1),y(t_1))$ is it's end-point, then \begin{multline} \int_L \bigl(M(x,y)\,dx + N(x,y)\,dy\bigr) :=\\ \int_{t_0}^{t_1} \bigl(M(x(t),y(t))x'(t) + N(x(t),y(t))y'(t)\bigr)\,dt \label{eqn-1} \end{multline} is called line integral. Here either $t_0\le t_1$ or $t_0\ge t_1$.
Line integral has the usual properties of the integral (integral of the sum equals the sum of integrals, one can pull a constant factor from the integral) and \begin{gather} \int_{L_1+L_2} \bigl(Mdx + Ndy\bigr)= \int_{L_1} \bigl(Mdx + Ndy\bigr)+ \int_{L_2} \bigl(Mdx + Ndy\bigr), \label{eqn-2}\\ \int_{-L} \bigl(Mdx + Ndy\bigr)= -\int_{L} \bigl(Mdx + Ndy\bigr), \label{eqn-3} \end{gather} where if $L_1=\{ (x(t),y(t)) \}$, $L_2=\{ (x(t),y(t))\}$ with $t$ running from $t_0$ to $t_1$ and from $t_1$ to $t_2$ correspondingly, then
Remark 1. Line integral is different from the length integral \begin{equation} \int_L f(x,y)\,ds := \int_{t_0}^{t_1} f(x(t),y(t)) \underbracket{\sqrt{(x'(t))^2+(y'(t))^2}\,dt} \label{eqn-4} \end{equation} where $ds=\sqrt{(x'(t))^2+(y'(t))^2}\,dt$ is a length element. We will (almost) never use the length integral in our course.
Definition 2. If the start-point and end-point coincide, then the curve $L$ is called closed and integral (\ref{eqn-1}) is often written as \begin{equation} \oint_{L} \bigl(M\,dx + N\,dy\bigr). \label{eqn-5} \end{equation}
Remark 2. If $L$ is a closed curve, then the line integral does not depend on the choice of a start-and-end-point.
Exercise. Prove it.
Theorem 2 (Green's theorem).
\begin{equation} \oint_{L} \bigl(M\,dx + N\,dy\bigr)= \iint_{D} \bigl(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\bigr)\,dxdy, \label{eqn-6} \end{equation} where $D$ is a bounded domain in $\mathbb{R}^2$ and $L=\partial D$ is it's border, properly oriented, $M$ and $N$ are smooth in $\bar{D}=D\cup L$.
Remark 3. What is properly oriented? If $D$ is simply connected (with no holes, then it means counter-clock-wise.
If $D$ is not a simply connected (with holes inside) then $L=L_0 + L_1+\ldots + L_N$ where $L_0$ is the external boundary, counter-clockwise oriented and $L_1,\ldots,L_N$ are boundaries of the holes, clock-wise oriented.
Why? Let us make an infinitely thin cut (if $N>1$ we need $N$ cuts) so that domain $D^*$ after cuts will be connected and simply-connected. Its boundary $L^*=L_0+\ell+\ell'+L_1$ is connected and the counter-clockwise orientation of $L_0$ implies clockwise orientation of $L_1$ (and $L_2,\ldots,L_N$) and the proper orientations of $\ell'$ and $\ell$.
Applying to $D^*$, $L^*$ Green's formula we see that $\iint_{D^*}=\iint_{D}$, and $\oint_{L^*}=\int_{L_0}+\int_{\ell}+\int_{\ell'}+\int_{L_0}=\oint_{L}$ because $\ell'=-\ell$, so corresponding integrals cancel one another.
Sure, if functions were not good at the points of the cut, these equalities could fail.