# (Edited) Confusing on proving reverse direction of Line Integral

Now i'm taking complex varibles course and learning about complex integration. But this is still beginning and i need to improve my knowledge about line integral. (I forgot a little about line integral and its parametrization on vector analysis, so i'm sorry if my question might be too easy for you guys.) And i have this problem.

To make it easy, I will numbering each equation so that you can call the number of the equation.

I wanna ask how to prove this equality, (Actually, in here, i'm using real line integral) :

$$\oint_C M(x,y)\,\mathbb ds=-\oint_C M(x,y)\,\mathbb ds\quad\tag{1}$$

It said, whenever we have a closed path no matter what path you choose, then the value of line integral is unchanged.

Even if it might not be relevant with complex line integral, i have a link that talks about that.

Line Integral

Properties of Line Integral

I'm starting to confuse with the second paragraph (proving the properties).

Cz it has another properties like :

$$\int_C M(x,y)\, \mathbb dx=-\int_C M(x,y)\,\mathbb dx\quad\tag 2$$

I don't know if i misunderstood or something. Is that mean on that form (general form and could be not a closed path) no matter the line integral is, we always have the same value in different direction?

$$\int_C M(x,y)\, \mathbb dx=\int_{-C} M(x,y)\,\mathbb dx\quad\tag 3$$

and

$$\int_{-C} M(x,y)\, \mathbb dx=-\int_C M(x,y)\,\mathbb dx\quad\tag 4$$

And in the second page of link i found :

$$\int_{-C} M(x,y)\, \mathbb dx=-\int_{-C} M(x,y)\,\mathbb dx\quad\tag 5$$

All of those just make me more confused. I know, i need to learn more, but i can't find the best reference anymore for answer my question and prove the first equation. I've read journals, books and watching some videos on youtube, but i got nothing.

Edit :

What i'm trying to say is, How to prove :

$$\oint_{-C} M(x,y) \,\mathbb ds=\oint_{C} M(x,y) \,\mathbb ds$$

This is similar to $$\int _a^b f(x) dx = - \int _b ^a f(x)dx$$
Now, if instead of $$dx$$ you have $$ds$$ then the orientation does not matter because $$ds$$ is always positive.
That is $$\oint_C M(x,y)\,\mathbb ds=\oint_{-C} M(x,y)\,\mathbb ds\quad\tag{1}$$
• @user516076 the difference is with $ds$ verses $dx$. Note that with $ds$ the direction does not matter because $ds$ is positive with both orientations. – Mohammad Riazi-Kermani Oct 12 '19 at 1:19