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  "body": "\n## Proof\n\nWe use the sums in the definitions, restricted to partitions that contain the point $b$.\n\n1. The curve $\\gamma$ is continuous, since $\\gamma_1(b) = \\gamma_2(b)$. Join the break points of $\\gamma_1$, the point $b$ and the break points of $\\gamma_2$. This gives a [[partition|069P20KYQT]] $a = p_0 \u003c \\dots \u003c p_m = c$ such that $\\gamma$ is a [[$C^1$ curve|0JF1XXGWB4]] on each $[p_{l-1}, p_l]$.\n2. By the [[extreme value theorem|0WS1A8E2P9]] there is $M \\ge 0$ with $\\norm{\\gamma'} \\le M$ on each piece, with one-sided derivatives at the $p_l$.\n3. Let $s \u003c t$ lie in one piece, and let $w = \\gamma(t) - \\gamma(s)$. The [[fundamental theorem of calculus|0NHGY0N0JB]] gives $w = \\int_s^t \\gamma'(u) \\, du$ componentwise. By the [[Cauchy–Schwarz inequality|0RFWW56Y22]], $\\norm{w}^2 = \\int_s^t w \\cdot \\gamma'(u) \\, du \\le \\norm{w} M (t - s)$, so $\\norm{w} \\le M (t - s)$.\n4. For any $s \u003c t$ in $[a, c]$, insert the points $p_l$ between them. The [[triangle inequality|03HF5PMY5T]] and step 3 give $\\norm{\\gamma(t) - \\gamma(s)} \\le M (t - s)$. So every inscribed polygon of $\\gamma$ has length at most $M (c - a)$.\n5. Hence $\\gamma$ is [[rectifiable|0C28RT5BG7]], and so are its restrictions $\\gamma_1$ and $\\gamma_2$. So the six line integrals in the statement exist.\n6. Let $I$, $I_1$ and $I_2$ be the vector line integrals of $F$ over $\\gamma$, $\\gamma_1$ and $\\gamma_2$. Fix $\\varepsilon \u003e 0$.\n7. By the definition of the [[vector line integral|0YS8RX87F4]] there is $\\delta \u003e 0$ with the following property. Every sum for $\\gamma$, $\\gamma_1$ or $\\gamma_2$ over a partition of mesh less than $\\delta$ lies within $\\varepsilon$ of $I$, $I_1$ or $I_2$ respectively.\n8. Choose a partition $P_1$ of $[a, b]$ and a partition $P_2$ of $[b, c]$, both of mesh less than $\\delta$, with points chosen in their subintervals. Their union $P$ is a partition of $[a, c]$ of mesh less than $\\delta$.\n9. Each subinterval of $P$ is a subinterval of $P_1$ or of $P_2$. On $[a, b]$ we have $\\gamma = \\gamma_1$, and on $[b, c]$ we have $\\gamma = \\gamma_2$. Let $S_1$ be the sum for $\\gamma_1$ and $P_1$, and $S_2$ the sum for $\\gamma_2$ and $P_2$. Then the sum $S$ for $\\gamma$ and $P$ is $S_1 + S_2$.\n10. By step 7, $\\abs{I - I_1 - I_2} \\le \\abs{I - S} + \\abs{S_1 - I_1} + \\abs{S_2 - I_2} \u003c 3 \\varepsilon$.\n11. Since $\\varepsilon$ is arbitrary, $\\int_\\gamma F \\cdot d\\mathbf{r} = \\int_{\\gamma_1} F \\cdot d\\mathbf{r} + \\int_{\\gamma_2} F \\cdot d\\mathbf{r}$.\n12. For the [[scalar line integral|0M50E4WYWJ]], take a subinterval $[t_{i-1}, t_i]$ of $P_1$. The two curves $\\gamma$ and $\\gamma_1$ agree there, so they give the same length $\\Delta s_i$. The same holds for $P_2$ and $\\gamma_2$.\n13. So steps 7 to 11, with these sums, give $\\int_\\gamma f \\, ds = \\int_{\\gamma_1} f \\, ds + \\int_{\\gamma_2} f \\, ds$.\n\n## Idea\n\nA line integral is a limit over all fine partitions, so it can be computed with partitions that contain the joining point $b$. Each such sum splits into a sum for the first curve and a sum for the second curve, and the limits split in the same way.\n",
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