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  "body": "\n## Proof\n\nWe work directly with the sums in the definitions. Reversal maps each partition of $[a, b]$ to another partition of the same mesh.\n\n1. Let $\\rho(t) = a + b - t$. It is a decreasing bijection of $[a, b]$ onto itself, and $\\gamma^- = \\gamma \\circ \\rho$.\n2. Since $\\gamma$ is piecewise $C^1$, there is a [[partition|069P20KYQT]] $a = p_0 \u003c \\dots \u003c p_m = b$ such that $\\gamma$ is a [[$C^1$ curve|0JF1XXGWB4]] on each $[p_{l-1}, p_l]$. By the [[extreme value theorem|0WS1A8E2P9]] there is $M \\ge 0$ with $\\norm{\\gamma'} \\le M$ on each piece, using one-sided derivatives at the $p_l$.\n3. Let $s \u003c t$ lie in one piece. By the [[fundamental theorem of calculus|0NHGY0N0JB]] in each component, $\\gamma(t) - \\gamma(s) = \\int_s^t \\gamma'(u) \\, du$. With $w = \\gamma(t) - \\gamma(s)$, the [[Cauchy–Schwarz inequality|0RFWW56Y22]] gives $\\norm{w}^2 = \\int_s^t w \\cdot \\gamma'(u) \\, du \\le \\norm{w} M (t - s)$, so $\\norm{w} \\le M (t - s)$.\n4. For any $s \u003c t$ in $[a, b]$, insert the points $p_l$ between them. The [[triangle inequality|03HF5PMY5T]] and step 3 give $\\norm{\\gamma(t) - \\gamma(s)} \\le M (t - s)$.\n5. Hence each inscribed polygon of $\\gamma$ has length at most $M (b - a)$. So $\\gamma$ is [[rectifiable|0C28RT5BG7]], and the line integrals over $\\gamma$ exist.\n6. Let $[c, d] \\subseteq [a, b]$, and let $P$ be a partition $c = t_0 \u003c \\dots \u003c t_k = d$. Put $u_j = \\rho(t_{k-j})$. Then $P^* = (u_0, \\dots, u_k)$ is a partition of $[\\rho(d), \\rho(c)]$, and $P \\mapsto P^*$ is a bijection between the partitions of the two intervals.\n7. The terms $\\norm{\\gamma^-(t_i) - \\gamma^-(t_{i-1})} = \\norm{\\gamma(u_{k-i+1}) - \\gamma(u_{k-i})}$ are the terms of the polygon length of $\\gamma$ for $P^*$, in reverse order. So the two polygon lengths are equal.\n8. Taking suprema, the length of $\\gamma^-$ on $[c, d]$ equals the length of $\\gamma$ on $[\\rho(d), \\rho(c)]$. In particular $\\gamma^-$ is rectifiable.\n9. Now let $a = t_0 \u003c \\dots \u003c t_k = b$ with points $\\tau_i \\in [t_{i-1}, t_i]$. Let $u_j = \\rho(t_{k-j})$ and, for $j = k + 1 - i$, let $\\sigma_j = \\rho(\\tau_i)$. Then $\\sigma_j \\in [u_{j-1}, u_j]$, and the two partitions have the same mesh.\n10. For $j = k + 1 - i$ we have $\\gamma^-(t_i) = \\gamma(u_{j-1})$, $\\gamma^-(t_{i-1}) = \\gamma(u_j)$ and $\\gamma^-(\\tau_i) = \\gamma(\\sigma_j)$.\n11. So $\\sum_i F(\\gamma^-(\\tau_i)) \\cdot \\big( \\gamma^-(t_i) - \\gamma^-(t_{i-1}) \\big) = -\\sum_j F(\\gamma(\\sigma_j)) \\cdot \\big( \\gamma(u_j) - \\gamma(u_{j-1}) \\big)$.\n12. By step 8, the length of $\\gamma^-$ on $[t_{i-1}, t_i]$ equals the length $\\Delta s_j$ of $\\gamma$ on $[u_{j-1}, u_j]$. So $\\sum_i f(\\gamma^-(\\tau_i)) \\Delta s^-_i = \\sum_j f(\\gamma(\\sigma_j)) \\Delta s_j$, where $\\Delta s^-_i$ is the length of $\\gamma^-$ on $[t_{i-1}, t_i]$.\n13. Let the mesh tend to $0$. The right sides of steps 11 and 12 tend to $-\\int_\\gamma F \\cdot d\\mathbf{r}$ and $\\int_\\gamma f \\, ds$. This holds by the definitions of the [[vector line integral|0YS8RX87F4]] and the [[scalar line integral|0M50E4WYWJ]].\n14. Hence the left sides converge to the same values, so $\\int_{\\gamma^-} F \\cdot d\\mathbf{r} = -\\int_\\gamma F \\cdot d\\mathbf{r}$ and $\\int_{\\gamma^-} f \\, ds = \\int_\\gamma f \\, ds$.\n\n## Idea\n\nA partition of the reversed curve is a partition of the original curve read backwards. Each chord $\\gamma(t_i) - \\gamma(t_{i-1})$ turns into its negative, so the vector sums change sign. Lengths of pieces do not change, so the scalar sums stay the same.\n",
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