Let $A=(0,0)$ and $B=(0,1)$.
Using $r_1:[0,1]\longrightarrow\mathbb{R}^2$. $r_1(t)=(0,1-t)$
$$\displaystyle\int_B^A 1=\int_0^11\,dt=1.$$
On the other hand,
Using $r_2:[0,1]\longrightarrow\mathbb{R}^2$. $r_2(t)=(0,t)$
$$\displaystyle\int_A^B 1=\int_0^11\,dt=1.$$
So,
$$\boxed{\displaystyle\int_A^B 1=\int_B^A 1}$$
Is it right? So, in line integrals, not depend on the orientation?
But I thought that
$$\boxed{\displaystyle\int_A^B 1={\color{red}-}\int_B^A 1}$$
Indeed, we have that $$\int_A^B=-\int_B^A$$ However, for two parameterized paths $r_1:[a,b]\to\Bbb R^2$ and $r_2:[a,b]\to\Bbb R^2$ with $r_1(a)=A,r_1(b)=B$ and $r_2(a)=B,r_2(b)=A$ (i.e. $r_1$ and $r_2$ have opposing direction), we have $$\int_{r_1}=\int_A^B=-\int_B^A=-\int_{r_2}$$ since $\int_{r_1}=\int_A^B$ and $\int_{r_2}=\int_B^A$.
So when you use two paths with different directions, and simultaneously switch the limits of integration, you produce two negatives which cancel each other out.