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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

$$\boxed{\displaystyle\int_A^B 1={\color{red}-}\int_B^A 1}$$

Michael Hardy
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yemino
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    The paths $r_1$ and $r_2$ are in opposite directions. So when you switch $A$ and $B$ in the limits of integration, this cancels out the fact that you're also switching the path direction. – Dave Jul 25 '17 at 03:16
  • I don't understand where the error is, both integrals are equals to $\displaystyle\int_0^1 1=1$, so are equals, ¿where is the error? – yemino Jul 25 '17 at 03:21
  • But the paths $r_1$ and $r_2$ are in different directions. So to go from $A$ to $B$ on $r_1$ you go from $t=1$ to $t=0$, whereas with $r_2$ you would go from $t=0$ to $t=1$. Since you simultaneously switch the path direction and the order of the limits of integration, these negatives cancel each other out. – Dave Jul 25 '17 at 03:23
  • really, I can't see the error. Is $\displaystyle\int_A^B1=\int_B^A 1=\int_0^1 1=1$? Where ir the error? Really, I can't see it. – yemino Jul 25 '17 at 03:27
  • If you used the same path, say $r_1$, you would have $\int_B^A=\int_0^1dt=1$ and then $\int_A^B=\int_1^0dt=-1$. Because you used a new path, $r_2$ when computing $\int_A^B$, the effect of the negative coming from switching the limits of integration is canceled out, because the path $r_2$ has the opposite direction to $r_1$. – Dave Jul 25 '17 at 03:31
  • In other words, you swapped order twice. Once in the parameterization, and once in the limits. – A. Thomas Yerger Jul 25 '17 at 03:32

1 Answers1

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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.

Dave
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