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I was looking into why the property that $\int_a^b f(x) \ dx = -\int_b^a f(x) \ dx$ holds true. I found that 2 common answers were that

  1. It comes from $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$ for arbitrary $a \le b \le c$ (for example, in this answer).
  2. It comes from the fundamental theorem of calculus $\int_a^b f(x)\,dx = F(b) - F(a)$ (for example, in this answer).

From my understanding of these answers, the first one has a more lenient hypothesis, since to apply F.T.C. we need the function to have an antiderivative, which is not always the case.

Knowing this, I was wondering about the extension of this question into a line integral. Let's say that $C$ is a path that starts at point $p$ and ends at point $q$. If I define $C^*$ to be the same path but staring at $q$ and ending at $p$, is it generally true that

$$ \int_{C} \mathbf{F} \cdot d\mathbf{r} = - \int_{C^*} \mathbf{F} \cdot d\mathbf{r} \quad ? $$

where here $r:[t_0, t_f]\subset \mathbb{R} \to C$, with $r(t_0) = p$ and $r(t_f) = q$ being a biyective parametrization of our path.

I know that I can show this to be true if $\mathbf{F}$ happens to be a conservative field using the gradient theorem, in a similar manner as the 1D case can be shown by F.T.C., but since this is not always true I don't know if I can say that this holds in general.

Is there a way to show that this always holds? Or alternatively, is there a counterexample where this fails? Thank you!

Robert Lee
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1 Answers1

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You can deduce this property from the one dimensional case: Suppose we have a $C^1$ path $\gamma:[0,1]\to C$, where $C=\gamma([0,1])$.

Now we consider a $C^1$ 'reparametrization' $\varphi:[0,1]\to [0,1]$ with $\varphi(0)=1$, $\varphi(1)=0$ and $\varphi'(s)<0$ for all $s\in [0,1]$. Let's define a new path $\gamma^*:[0,1]\to C$ by $\gamma^*(s)=\gamma(\varphi(s))$. (Note that the new path 'runs through $C$ in the opposite direction')

Then by definition we have \begin{align} \int_\gamma F\cdot dr=\int_0^1 F(\gamma(t))\cdot \dot{\gamma}(t)dt \end{align} If we make the substitution $t=\varphi(s)$ then $dt=\varphi'(s)ds$ and \begin{align}\int_0^1F(\gamma(t)))\cdot \dot{\gamma} (t)dt&=\int_1^0F(\gamma(\varphi(s))\dot{\gamma}(\varphi(s))\varphi'(s)ds\\ &=\int_1^0F(\gamma^*(s))\dot{\gamma}^*(s)ds\\ &=-\int_0^1F(\gamma^*(s))\dot{\gamma}^*(s)ds\\ &=-\int_{\gamma^*}F\cdot dr\end{align}

(The second line follows from the chain rule and the third line by the identities for the 1-d case that you mentioned)

Hope this helps!

Jake28
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