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  "body": "\n## Proof\n\nWe write the difference quotient of $F$ as an average of $f$.\n\n1. Let $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be [[Riemann integrable|0N1JQXZXHC]] with integral $I$. We first show that $f$ is bounded.\n   Take $\\delta \u003e 0$ such that each tagged partition of mesh less than $\\delta$ has [[Riemann sum|0HBY48G8G1]] within $1$ of $I$.\n2. By the [[Archimedean property|0A3GWZ86AY]], fix a partition $(x_0, \\dots, x_n)$ of $[a, b]$ with all subintervals shorter than $\\delta$, and fix tags $t$.\n   For $x \\in [x_{i-1}, x_i]$, replace $t_i$ by $x$. The two Riemann sums differ by $(f(x) - f(t_i))(x_i - x_{i-1})$, which has absolute value less than $2$.\n3. So $|f(x)| \u003c |f(t_i)| + \\frac{2}{x_i - x_{i-1}}$ on the $i$-th subinterval. Let $K \u003e 0$ be larger than all these $n$ bounds. Then $|f(x)| \\le K$ on $[a, b]$.\n4. Now we check that $F$ is well defined. $F(a) = 0$ by the convention for the [[definite integral|021WEM680W]]. For $a \u003c x \u003c b$, $f$ is integrable on $[a, x]$ by the [[additivity of the integral|05RCDKHZM6]].\n5. Let $a \\le x \u003c y \\le b$. We claim that $f$ is integrable on $[x, y]$ and $F(y) - F(x) = \\int_x^y f(t) \\, dt$.\n   For $x = a$ this is the definition of $F$. For $x \u003e a$, apply additivity to $[a, y]$ split at $x$. This gives $F(y) = F(x) + \\int_x^y f(t) \\, dt$.\n6. Each Riemann sum of a constant $k$ over $[x, y]$ equals $k (y - x)$. So the constant $k$ is integrable over $[x, y]$ with integral $k (y - x)$.\n   Since $-K \\le f \\le K$, the [[monotonicity of the integral|0MBWQKM8QR]] gives $|F(y) - F(x)| \\le K (y - x)$.\n7. Let $\\varepsilon \u003e 0$ and $\\delta' = \\varepsilon / K$. If $x, y \\in [a, b]$ and $|x - y| \u003c \\delta'$, then $|F(x) - F(y)| \u003c \\varepsilon$ by step 6. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], $F$ is continuous on $[a, b]$.\n8. Now let $c \\in (a, b)$, and let $f$ be continuous at $c$. Let $\\varepsilon \u003e 0$. By the epsilon-delta form, there is $\\eta \u003e 0$ with $|f(t) - f(c)| \u003c \\varepsilon$ for $t \\in [a, b]$ with $|t - c| \u003c \\eta$.\n   Making $\\eta$ smaller, we may assume $(c - \\eta, c + \\eta) \\subseteq (a, b)$.\n9. Let $0 \u003c h \u003c \\eta$. By step 5, by step 6 for the constant $f(c)$, and by the [[linearity of the integral|080RWVVSB6]],\n   $$F(c + h) - F(c) - f(c) h = \\int_c^{c + h} \\big(f(t) - f(c)\\big) \\, dt.$$\n10. On $[c, c + h]$ the integrand lies in $[-\\varepsilon, \\varepsilon]$. By monotonicity and step 6, the right side of step 9 has absolute value at most $\\varepsilon h$. Dividing by $h$ gives $\\big| \\frac{F(c + h) - F(c)}{h} - f(c) \\big| \\le \\varepsilon$.\n11. Let $-\\eta \u003c h \u003c 0$. In the same way, $F(c) - F(c + h) - f(c)(-h) = \\int_{c + h}^{c} \\big(f(t) - f(c)\\big) \\, dt$, and its absolute value is at most $\\varepsilon |h|$.\n    Dividing by $-h \u003e 0$ gives again $\\big| \\frac{F(c + h) - F(c)}{h} - f(c) \\big| \\le \\varepsilon$.\n12. So for each $\\varepsilon \u003e 0$ there is $\\eta \u003e 0$ with $\\big| \\frac{F(c + h) - F(c)}{h} - f(c) \\big| \\le \\varepsilon$ whenever $0 \u003c |h| \u003c \\eta$.\n    By the [[epsilon-delta form of limits|0GBZDAFPX8]], the difference quotient tends to $f(c)$ as $h \\to 0$.\n13. By the definition of the [[derivative|0F1WR9C83K]], applied to $F$ on the open interval $(a, b)$, $F$ is differentiable at $c$ and $F'(c) = f(c)$.\n\n## Idea\n\nThe difference $F(c + h) - F(c)$ is the integral of $f$ over a short interval, so the difference quotient is the average of $f$ there. A bounded $f$ makes $F$ change at most linearly, so $F$ is continuous. If $f$ is continuous at $c$, its averages over shrinking intervals at $c$ tend to $f(c)$.\n",
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