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  "body": "\n## Statement\n\nLet $a \u003c b$ be real numbers, let $f \\colon [a, b] \\to \\R$ be [[continuous|0X3WGX95FE]], and let $n \\ge 1$ be an integer.\nDefine $F_0 = f$ and $F_k(x) = \\int_a^x F_{k-1}(t) \\, dt$ for $k \\ge 1$ and $x \\in [a, b]$.\nThen for all $x \\in [a, b]$,\n$$F_n(x) = \\frac{1}{(n - 1)!} \\int_a^x (x - t)^{n - 1} f(t) \\, dt.$$\n\n## Remarks\n\nThe proof is by induction on $n$.\nThe case $n = 1$ is the definition.\nFor the step, note that both sides vanish at $x = a$.\nDifferentiating the right side under the integral sign gives the same formula with $n - 1$ in place of $n$.\nThe [[fundamental theorem of calculus|0NHGY0N0JB]] then gives equality.\nEquivalently, one writes $F_n$ as an [[iterated integral|0WZCHE1ZF9]] over a simplex and integrates out all variables but the last one.\nExample: for $f = 1$, $a = 0$ and $n = 2$, the right side is $\\int_0^x (x - t) \\, dt = x^2 / 2$, which is the twice repeated integral of $1$.\n\n## Explanations\n\n### L1 human\n\nIntegrating a function $n$ times in a row seems to need $n$ nested integrals.\nCauchy's formula collapses them into one integral with the weight $(x - t)^{n-1} / (n - 1)!$.\n\n### L2 human\n\nThe formula is used in the proof of [[Taylor's theorem|03QHBVW4TT]] with the integral remainder.\nReplacing $(n - 1)!$ by $\\Gamma(n)$ and letting $n$ be any positive real number gives the [[Riemann–Liouville integral|JTKCZ44NN3]], the starting point of [[fractional calculus|T73XK15SA2]].\n",
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