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  "body": "\n## Proof\n\nWe compare the sums in the definitions with Riemann sums of the right sides.\n\n1. Let $\\gamma = (\\gamma_1, \\dots, \\gamma_n) \\colon [a, b] \\to \\R^n$ be a [[$C^1$ curve|0JF1XXGWB4]], with [[tangent vector|0D98771SM0]] $\\gamma'$. For a continuous $v \\colon [c, d] \\to \\R^n$, let $\\int_c^d v(t) \\, dt$ be the vector of the integrals of its components.\n2. For $a \\le c \u003c d \\le b$, the [[fundamental theorem of calculus|0NHGY0N0JB]] applied to each $\\gamma_k$ gives $\\gamma(d) - \\gamma(c) = \\int_c^d \\gamma'(t) \\, dt$.\n3. For continuous $v$, let $w = \\int_c^d v(t) \\, dt$. Then $\\norm{w}^2 = \\int_c^d w \\cdot v(t) \\, dt \\le \\norm{w} \\int_c^d \\norm{v(t)} \\, dt$, by [[linearity of the integral|080RWVVSB6]] and the [[Cauchy–Schwarz inequality|0RFWW56Y22]]. Hence $\\norm{w} \\le \\int_c^d \\norm{v(t)} \\, dt$.\n4. Let $P$ be a partition $c = u_0 \u003c \\dots \u003c u_k = d$ of $[c, d]$. By steps 2 and 3, $\\sum_j \\norm{\\gamma(u_j) - \\gamma(u_{j-1})} \\le \\sum_j \\int_{u_{j-1}}^{u_j} \\norm{\\gamma'} \\, dt = \\int_c^d \\norm{\\gamma'} \\, dt$.\n5. So the restriction of $\\gamma$ to $[c, d]$ is [[rectifiable|0C28RT5BG7]], and its length is at most $\\int_c^d \\norm{\\gamma'(t)} \\, dt$. In particular the line integrals over $\\gamma$ are defined.\n6. The interval $[a, b]$ is compact by the [[Heine–Borel theorem|07PFYV65MB]]. The maps $\\gamma'$, $f \\circ \\gamma$ and $F \\circ \\gamma$ are continuous on $[a, b]$. By the [[Heine–Cantor theorem|06B8BNJCD6]] they are uniformly continuous.\n7. By the [[extreme value theorem|0WS1A8E2P9]] there are $M, N \\ge 0$ with $\\norm{\\gamma'(t)} \\le M$ and $\\abs{f(\\gamma(t))} \\le N$ for all $t \\in [a, b]$.\n8. Fix $\\varepsilon \u003e 0$. By step 6 there is $\\delta \u003e 0$ such that $\\abs{s - t} \u003c \\delta$ implies $\\norm{\\gamma'(s) - \\gamma'(t)} \u003c \\varepsilon$, $\\abs{f(\\gamma(s)) - f(\\gamma(t))} \u003c \\varepsilon$ and $\\norm{F(\\gamma(s)) - F(\\gamma(t))} \u003c \\varepsilon$.\n9. Take a partition $a = t_0 \u003c \\dots \u003c t_k = b$ with $\\max_i (t_i - t_{i-1}) \u003c \\delta$, and points $\\tau_i \\in [t_{i-1}, t_i]$. Write $I_i = [t_{i-1}, t_i]$ and $\\Delta t_i = t_i - t_{i-1}$.\n10. Let $\\Delta s_i$ be the length of $\\gamma$ on $I_i$. The partition $\\set{t_{i-1}, t_i}$ shows that $\\Delta s_i \\ge \\norm{\\gamma(t_i) - \\gamma(t_{i-1})}$.\n11. By step 2, $\\gamma(t_i) - \\gamma(t_{i-1}) = \\Delta t_i \\, \\gamma'(t_i) + \\int_{I_i} (\\gamma'(t) - \\gamma'(t_i)) \\, dt$. With steps 3 and 8, its norm is at least $\\Delta t_i \\norm{\\gamma'(t_i)} - \\varepsilon \\Delta t_i$.\n12. Also $\\int_{I_i} \\norm{\\gamma'(t)} \\, dt \\le \\Delta t_i \\norm{\\gamma'(t_i)} + \\varepsilon \\Delta t_i$, by step 8. With steps 10 and 11, $\\Delta s_i \\ge \\int_{I_i} \\norm{\\gamma'} \\, dt - 2 \\varepsilon \\Delta t_i$.\n13. Step 5 gives $\\Delta s_i \\le \\int_{I_i} \\norm{\\gamma'} \\, dt$. So $\\abs{\\Delta s_i - \\int_{I_i} \\norm{\\gamma'} \\, dt} \\le 2 \\varepsilon \\Delta t_i$.\n14. Let $S = \\sum_i f(\\gamma(\\tau_i)) \\, \\Delta s_i$ and $I = \\int_a^b f(\\gamma(t)) \\norm{\\gamma'(t)} \\, dt$. Then $S - I = \\sum_i f(\\gamma(\\tau_i)) \\big( \\Delta s_i - \\int_{I_i} \\norm{\\gamma'} \\, dt \\big) + \\sum_i \\int_{I_i} \\big( f(\\gamma(\\tau_i)) - f(\\gamma(t)) \\big) \\norm{\\gamma'(t)} \\, dt$.\n15. By steps 7, 8 and 13, $\\abs{S - I} \\le 2 N \\varepsilon (b - a) + \\varepsilon M (b - a)$.\n16. Since $\\varepsilon$ is arbitrary, the sums of the [[scalar line integral|0M50E4WYWJ]] tend to $I$. So $\\int_\\gamma f \\, ds = \\int_a^b f(\\gamma(t)) \\norm{\\gamma'(t)} \\, dt$.\n17. For the vector field, let $T = \\sum_i F(\\gamma(\\tau_i)) \\cdot (\\gamma(t_i) - \\gamma(t_{i-1}))$. By step 2 and linearity, $T = \\sum_i \\int_{I_i} F(\\gamma(\\tau_i)) \\cdot \\gamma'(t) \\, dt$.\n18. Hence $T - \\int_a^b F(\\gamma(t)) \\cdot \\gamma'(t) \\, dt = \\sum_i \\int_{I_i} \\big( F(\\gamma(\\tau_i)) - F(\\gamma(t)) \\big) \\cdot \\gamma'(t) \\, dt$.\n19. By the Cauchy–Schwarz inequality and steps 7 and 8, each integrand has absolute value at most $\\varepsilon M$. So the difference in step 18 is at most $\\varepsilon M (b - a)$ in absolute value.\n20. Since $\\varepsilon$ is arbitrary, the sums of the [[vector line integral|0YS8RX87F4]] tend to $\\int_a^b F(\\gamma(t)) \\cdot \\gamma'(t) \\, dt$. This is the second formula.\n\n## Idea\n\nOn a short piece of a $C^1$ curve, the chord $\\gamma(t_i) - \\gamma(t_{i-1})$ is the integral of $\\gamma'$, and the arc length is squeezed between the chord length and $\\int \\norm{\\gamma'}$. Uniform continuity makes all the errors small at once, so the defining sums turn into Riemann sums of ordinary integrals in $t$.\n",
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