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  "body": "\n## Proof\n\nWe use the mean value theorem and the uniform continuity of $\\partial f / \\partial t$.\n\n1. Write $f_t = \\partial f / \\partial t$ for the [[partial derivative|05AXG3QRKR]] in the second variable. Let $K = [a, b] \\times [t_0, t_1]$.\n2. The set $K$ is closed and bounded, so it is compact by the [[Heine–Borel theorem|07PFYV65MB]]. By the [[Heine–Cantor theorem|06B8BNJCD6]], $f_t$ is uniformly continuous on $K$.\n3. Fix $\\varepsilon \u003e 0$. By step 2 there is $\\delta \u003e 0$ such that $\\abs{f_t(x, s) - f_t(x, s')} \u003c \\varepsilon$ for all $x \\in [a, b]$ and all $s, s' \\in [t_0, t_1]$ with $\\abs{s - s'} \u003c \\delta$.\n4. For each fixed $t$, the functions $x \\mapsto f(x, t)$ and $x \\mapsto f_t(x, t)$ are [[continuous|0S9EK5MQPE]] on $[a, b]$. So they are [[Riemann integrable|06SESV7GQY]], and $F(t) = \\int_a^b f(x, t) \\, dx$ is defined.\n5. Fix $t \\in [t_0, t_1]$ and $h \\ne 0$ with $\\abs{h} \u003c \\delta$ and $t + h \\in [t_0, t_1]$. For each $x$, the function $s \\mapsto f(x, s)$ is differentiable on $[t_0, t_1]$ with derivative $f_t(x, s)$.\n6. By the [[mean value theorem|0BARQY52N5]], for each $x$ there is $\\theta$ strictly between $t$ and $t + h$ with $f(x, t + h) - f(x, t) = h \\, f_t(x, \\theta)$.\n7. Since $\\abs{\\theta - t} \u003c \\delta$, step 3 gives $\\abs{\\frac{f(x, t + h) - f(x, t)}{h} - f_t(x, t)} \u003c \\varepsilon$ for every $x \\in [a, b]$.\n8. Integrate this bound over $[a, b]$: $\\abs{\\frac{F(t + h) - F(t)}{h} - \\int_a^b f_t(x, t) \\, dx} \\le \\varepsilon (b - a)$.\n9. Since $\\varepsilon$ is arbitrary, $F'(t)$ exists and equals $\\int_a^b f_t(x, t) \\, dx$. At $t = t_0$ and $t = t_1$ the derivative is one-sided.\n10. For $s, t \\in [t_0, t_1]$ with $\\abs{s - t} \u003c \\delta$, step 3 gives $\\abs{F'(s) - F'(t)} \\le \\int_a^b \\abs{f_t(x, s) - f_t(x, t)} \\, dx \\le \\varepsilon (b - a)$. So $F'$ is continuous, and $F$ is continuously differentiable.\n11. Now let $\\alpha, \\beta \\colon [t_0, t_1] \\to [a, b]$ be continuously differentiable, and let $H(t) = \\int_{\\alpha(t)}^{\\beta(t)} f(x, t) \\, dx$. Here $\\int_u^v = -\\int_v^u$ when $v \u003c u$.\n12. Fix $t$ and $h$ as in step 5. Then $H(t + h) - H(t) = A_h + B_h$, with $A_h = \\int_{\\alpha(t+h)}^{\\beta(t+h)} \\big( f(x, t + h) - f(x, t) \\big) \\, dx$ and $B_h = \\int_{\\alpha(t+h)}^{\\beta(t+h)} f(x, t) \\, dx - \\int_{\\alpha(t)}^{\\beta(t)} f(x, t) \\, dx$.\n13. The interval between $\\alpha(t + h)$ and $\\beta(t + h)$ lies in $[a, b]$. So step 7 gives $\\abs{\\frac{A_h}{h} - \\int_{\\alpha(t+h)}^{\\beta(t+h)} f_t(x, t) \\, dx} \\le \\varepsilon (b - a)$.\n14. Let $G(u) = \\int_a^u f_t(x, t) \\, dx$ for $u \\in [a, b]$. By the [[first fundamental theorem of calculus|063H5KJC69]], $G$ is continuous on $[a, b]$.\n15. The integral in step 13 is $G(\\beta(t + h)) - G(\\alpha(t + h))$. As $h \\to 0$ it tends to $G(\\beta(t)) - G(\\alpha(t)) = \\int_{\\alpha(t)}^{\\beta(t)} f_t(x, t) \\, dx$, since $\\alpha$, $\\beta$ and $G$ are continuous.\n16. With step 13 and $\\varepsilon$ arbitrary, $A_h / h \\to \\int_{\\alpha(t)}^{\\beta(t)} f_t(x, t) \\, dx$ as $h \\to 0$.\n17. Next, $B_h = \\int_{\\beta(t)}^{\\beta(t+h)} f(x, t) \\, dx - \\int_{\\alpha(t)}^{\\alpha(t+h)} f(x, t) \\, dx$, by additivity of the integral over intervals.\n18. By the [[mean value theorem for integrals|0N91N98EG5]], there is $\\xi_h$ between $\\beta(t)$ and $\\beta(t + h)$ with $\\int_{\\beta(t)}^{\\beta(t+h)} f(x, t) \\, dx = \\big( \\beta(t + h) - \\beta(t) \\big) f(\\xi_h, t)$. If $\\beta(t + h) = \\beta(t)$, take $\\xi_h = \\beta(t)$.\n19. As $h \\to 0$, $\\abs{\\xi_h - \\beta(t)} \\le \\abs{\\beta(t + h) - \\beta(t)} \\to 0$. Since $f$ is continuous, $f(\\xi_h, t) \\to f(\\beta(t), t)$.\n20. So the first term of $B_h$, divided by $h$, tends to $\\beta'(t) \\, f(\\beta(t), t)$. In the same way, the second term divided by $h$ tends to $\\alpha'(t) \\, f(\\alpha(t), t)$.\n21. Adding steps 16 and 20, $H'(t) = \\int_{\\alpha(t)}^{\\beta(t)} f_t(x, t) \\, dx + f(\\beta(t), t) \\, \\beta'(t) - f(\\alpha(t), t) \\, \\alpha'(t)$. This is the second formula.\n\n## Idea\n\nThe difference quotient of $f$ in $t$ is a value of $\\partial f / \\partial t$ at a nearby point, by the mean value theorem. Uniform continuity on the compact rectangle makes this approximation good for all $x$ at once, so the limit passes under the integral. When the limits move, the thin strips near the ends add the boundary terms.\n",
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