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  "body": "\n## Statement\n\nLet $f \\colon [a, b] \\times [t_0, t_1] \\to \\mathbb{R}$ be continuous, and suppose that its [[partial derivative|05AXG3QRKR]] $\\partial f / \\partial t$ exists and is [[continuous|0S9EK5MQPE]] on $[a, b] \\times [t_0, t_1]$.\nThen $F(t) = \\int_a^b f(x, t) \\, dx$ is continuously differentiable on $[t_0, t_1]$, and\n$$F'(t) = \\int_a^b \\frac{\\partial f}{\\partial t}(x, t) \\, dx.$$\nIf, in addition, $\\alpha, \\beta \\colon [t_0, t_1] \\to [a, b]$ are continuously differentiable, then\n$$\\frac{d}{dt} \\int_{\\alpha(t)}^{\\beta(t)} f(x, t) \\, dx = \\int_{\\alpha(t)}^{\\beta(t)} \\frac{\\partial f}{\\partial t}(x, t) \\, dx + f(\\beta(t), t) \\, \\beta'(t) - f(\\alpha(t), t) \\, \\alpha'(t).$$\n\n## Remarks\n\nThe second formula follows from the first, the fundamental theorem of calculus and the [[chain rule|0EAAS2EHDR]], applied to $G(u, v, t) = \\int_u^v f(x, t) \\, dx$.\nA measure theory version: let $f(\\cdot, t)$ be integrable for each $t$ in an open set, let $\\partial f / \\partial t$ exist for almost all $x$, and let $|\\partial f / \\partial t (x, t)| \\le g(x)$ for an integrable $g$. Then the same rule holds. Its proof uses the dominated convergence theorem.\nExample: let $F(t) = \\int_0^1 \\frac{x^t - 1}{\\ln x} \\, dx$ for $t \\ge 0$. For $t \u003e 0$ the measure version, with $g = 1$, gives $F'(t) = \\int_0^1 x^t \\, dx = \\frac{1}{t + 1}$. Since $F$ is continuous with $F(0) = 0$, $F(t) = \\ln(1 + t)$.\n\n## Explanations\n\n### L1 human\n\nIf the integrand depends smoothly on a parameter, you can differentiate the integral by differentiating inside it.\nWhen the limits of integration also move, the values of the integrand at the moving limits add extra terms.\n",
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