What is the Convention in Arc Length Parametrization? My text gave a statement but not a very thorough explanation.  For simplicity, let $\ f:  \mathbb R^2 \rightarrow \mathbb R$ be continuous in a domain containing the smooth curve $C$.  The text says that the line integral with respect to the arc-length parameter is independent of the orientation.  Thus, if $s$ is the arc-length parameter, $\int_C\ f(x, y)\ ds = \int_{-C}\ f(x, y)\ ds$.  The reason is that, in the definition of the line integral, $\Delta s_k$ is always positive.
$$\int_C\ f(x, y)\ ds = \lim_{n \rightarrow \infty} \sum_{k = 1}^n\ f(x(s_k^*), y(s_k^*)) \Delta s_k$$
However, since $\ f$ is continuous, I can rewrite the line integral as follows:
$$\int_C\ f(x, y)\ ds = \int_{a}^{b}\ (\ f \circ \sigma)(t)| \sigma'(t)|\ dt$$
where $\sigma:  [a, b] \rightarrow C$ is the smooth parametrization of $C$.  Now, if I go from point $b$ to $a$, then I'll definitely get
$$\int_{-C}\ f(x, y)\ ds = \int_{b}^{a}\ (\ f(\sigma(t))|\sigma'(t)|\ dt = - \int_C\ f(x, y)\ ds$$
Also, $\Delta s_k = |\sigma'(t_k^*)|\ \Delta t_k$ is no longer positive since each $\Delta t_k$ is negative.  Unless I re-parametrize $C$ so that $t\ \underline{\text{increases}}$ as it goes from $b$ to $a$, the statement is simply not true.  Can you kindly confirm on my understanding?
 A: Scalar line integrals (or line integrals over a scalar field)
Let $C$ be a curve with endpoints $P$ and $Q$ and $f$ defined on $C$. Let $a<b$. Let $\mathbf{c}:[a,b]\to \mathbb R^2$ be a parametrization of $C$ with $\mathbf{c}(a)=P$ and $\mathbf{c}(b)=Q$. On the other hand, let $\mathbf{d}:[a,b]\to\mathbb R^2$ be the parametrization of $C$ given by $\mathbf{d}(t)=\mathbf{c}(a+b-t)$, so that $\mathbf{d}(a)=Q$ and $\mathbf{d}(b)=P$. That is, $\mathbf d$ is simply the reverse of the parametrization $\mathbf{c}$. Then
\begin{align}
\int_C f(x,y)\,ds&=\color{blue}{\int_a^b f(\mathbf{c}(t))\|\mathbf{c}'(t)\|\,dt}\\
&=-\int_b^a f(\mathbf{c}(a+b-\tau))\|\mathbf{c}'(a+b-\tau)\|\,d\tau\\
&=\color{blue}{\int_a^b f(\mathbf{d}(\tau))\|\mathbf{d}'(\tau)\|\,d\tau}\\
&=\int_{-C} f(x,y)\,ds.
\end{align}
In other words, the scalar line integral is independent of the orientation of the curve $C$.
For example, if $f>0$ and $a<b$, then both blue integrals above will be positive, so reversing the orientation of $C$ does not negate the value of the integral.

Line integrals over a vector field
In contrast, the orientation of $C$ does (in general) matter when line integrating over a vector field $\mathbf{F}$. Using the same notation as above, note $\mathbf{d}'(t)=[\mathbf{c}(a+b-t)]'=-\mathbf{c}'(a+b-t)$ so that
$$
\int_C \mathbf{F}\cdot d\mathbf{s}=\int_a^b \mathbf{F}(\mathbf{c}(t))\cdot\mathbf{c}'(t)\,dt=-\int_a^b \mathbf{F}(\mathbf{d}(t))\cdot\mathbf{d}'(t)\,dt=-\int_{-C}\mathbf{F}\cdot d\mathbf{s},
$$
that is,
$$
-\int_C \mathbf{F}\cdot d\mathbf{s}=\int_{-C}\mathbf{F}\cdot d\mathbf{s},
$$
In other words, reversing the orientation of $C$ negates the value of the vector line integral.
