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  "body": "\n## Proof\n\nWe show that $F(b) - F(a)$ is a Riemann sum of $f$ for every partition.\n\n1. Let $a \u003c b$, let $F$ be continuous on $[a, b]$ and differentiable on $(a, b)$, and let $f \\colon [a, b] \\to \\mathbb{R}$ be [[Riemann integrable|0N1JQXZXHC]] with $f = F'$ on $(a, b)$. Let $I = \\int_a^b f(x) \\, dx$ be its [[integral|0QM4TXT3DR]].\n2. Let $P = (x_0, \\dots, x_n)$ be a [[partition|069P20KYQT]] of $[a, b]$. The sum telescopes: $F(b) - F(a) = \\sum_{i=1}^{n} \\big(F(x_i) - F(x_{i-1})\\big)$.\n3. Fix $i$. The function $F$ is continuous on $[x_{i-1}, x_i]$ and differentiable on $(x_{i-1}, x_i) \\subseteq (a, b)$.\n   By the [[mean value theorem|0BARQY52N5]], there is $c_i \\in (x_{i-1}, x_i)$ with $F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}) = f(c_i)(x_i - x_{i-1})$.\n4. The points $c_i$ are tags for $P$. By steps 2 and 3, $F(b) - F(a) = S(f, P, c)$, the [[Riemann sum|0HBY48G8G1]] of $f$ for $(P, c)$.\n5. Let $\\varepsilon \u003e 0$. By integrability, there is $\\delta \u003e 0$ with $|S(f, P, t) - I| \u003c \\varepsilon$ for every tagged partition of [[mesh|04GM4NYXJB]] less than $\\delta$.\n6. By the [[Archimedean property|0A3GWZ86AY]], take $n$ with $n \\delta \u003e b - a$, and the partition $P$ with points $a + i (b - a)/n$. Its mesh is less than $\\delta$.\n7. With the tags $c$ of step 3 for this $P$, steps 4 and 5 give $|F(b) - F(a) - I| \u003c \\varepsilon$.\n8. This holds for each $\\varepsilon \u003e 0$. So $\\int_a^b f(x) \\, dx = F(b) - F(a)$.\n\n## Idea\n\nOn each piece of a partition, the mean value theorem turns the increase of $F$ into $f$ at some point times the length of the piece. So the total increase $F(b) - F(a)$ is exactly a Riemann sum of $f$, for every partition. Fine partitions make Riemann sums close to the integral, so the two numbers agree.\n",
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