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  "body": "\n## Proof\n\nIntegrals and derivatives of maps into $\\R^n$ are taken componentwise. We assume that $J$ has more than one point, since otherwise no derivative on $J$ is defined.\n\n1. Let $g(s) = f(s, \\gamma(s))$ for $s \\in J$. The map $s \\mapsto (s, \\gamma(s))$ is continuous from $J$ to $J \\times U$, and $f$ is continuous. So $g$ is [[continuous|0NRQPQX3VA]] on $J$.\n2. First direction. Assume $\\gamma$ is differentiable on $J$ with $\\gamma' = g$ and $\\gamma(t_0) = x_0$. Let $t \\in J$ with $t \\ne t_0$, and let $[c, d]$ be the closed interval with endpoints $t_0$ and $t$. It lies in $J$.\n3. Each component $\\gamma_i$ is continuous on $[c, d]$ and differentiable on $(c, d)$ with derivative $g_i$, which is continuous. By the [[fundamental theorem of calculus|0NHGY0N0JB]], $\\int_c^d g_i(s)\\, ds = \\gamma_i(d) - \\gamma_i(c)$.\n4. In both orders of $t_0$ and $t$ this means $\\gamma(t) - \\gamma(t_0) = \\int_{t_0}^{t} g(s)\\, ds$. With $\\gamma(t_0) = x_0$, this is the integral equation at $t$. At $t = t_0$ it reads $\\gamma(t_0) = x_0$, which holds.\n5. Second direction. Assume the integral equation holds on $J$. At $t = t_0$ the integral is $0$, so $\\gamma(t_0) = x_0$.\n6. Let $\\hat J$ be the open interval obtained from $J$ by adding $(e - 1, e]$ at a left endpoint $e \\in J$, and $[e, e + 1)$ at a right endpoint $e \\in J$.\n7. Let $\\hat g = g$ on $J$, and let $\\hat g$ be constant, equal to $g(e)$, on each added piece beyond an endpoint $e$. The pieces agree at $e$, so $\\hat g$ is continuous on $\\hat J$.\n8. By [[the antiderivative theorem|0NC2A3HWGT]], applied to each component, $G(t) = \\int_{t_0}^{t} \\hat g(s)\\, ds$ is differentiable on $\\hat J$ with $G' = \\hat g$.\n9. On $J$ we have $\\gamma = x_0 + G$, since $\\hat g = g$ on $J$. At an interior point $t$ of $J$, this gives $\\gamma'(t) = \\hat g(t) = g(t)$.\n10. At an endpoint $e \\in J$, the one-sided derivative of $\\gamma$ is the one-sided derivative of $G$, which equals $G'(e) = \\hat g(e) = g(e)$.\n11. So $\\gamma$ is differentiable on $J$, with $\\gamma'(t) = f(t, \\gamma(t))$ for each $t \\in J$, and $\\gamma(t_0) = x_0$.\n\n## Idea\n\nBoth directions are the fundamental theorem of calculus. Integrating the equation gives the integral form, and differentiating the integral form gives back the equation. Continuity of $f$ and $\\gamma$ makes the integrand continuous, which is what both parts of the theorem need.\n",
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