[proofplan] We prove two claims: first, that $F$ is continuous on $[a,b]$ (via a Lipschitz bound from the boundedness of $f$), and second, that $F$ is differentiable at every continuity point $c$ of $f$ with $F'(c) = f(c)$ (via an $\varepsilon$-$\delta$ estimate on the difference quotient). [/proofplan] [step:Show that $F$ is Lipschitz continuous on $[a,b]$] Since $f$ is Riemann [integrable](/page/Integral) on $[a,b]$, it is bounded: there exists $M > 0$ such that $|f(t)| \le M$ for all $t \in [a,b]$. For any $x, y \in [a,b]$: \begin{align*} |F(x) - F(y)| = \left|\int_y^x f(t) \, dt\right| \le M|x - y|. \end{align*} This shows $F$ is Lipschitz [continuous](/page/Continuity) with constant $M$, and in particular continuous on $[a,b]$. [guided] Why does the Lipschitz bound follow? Since $f$ is Riemann [integrable](/page/Integral) on $[a,b]$, it is bounded: there exists $M > 0$ such that $|f(t)| \le M$ for all $t \in [a,b]$. By additivity of the integral, \begin{align*} F(x) - F(y) = \int_a^x f(t)\,dt - \int_a^y f(t)\,dt = \int_y^x f(t)\,dt. \end{align*} Taking absolute values and applying the estimate $\left|\int_y^x f(t)\,dt\right| \le \int_{\min(x,y)}^{\max(x,y)} |f(t)|\,dt \le M|x-y|$ (since $|f| \le M$ on $[a,b]$), we obtain the Lipschitz condition \begin{align*} |F(x) - F(y)| \le M|x - y|. \end{align*} Any Lipschitz function is [continuous](/page/Continuity): given $\varepsilon > 0$, choose $\delta = \varepsilon / M$, and $|x - y| < \delta$ implies $|F(x) - F(y)| \le M \cdot \delta = \varepsilon$. [/guided] [/step] [step:Show that $F'(c) = f(c)$ at every continuity point $c$ of $f$] Assume $f$ is continuous at $c \in [a,b]$. Given $\varepsilon > 0$, by continuity of $f$ at $c$ there exists $\delta > 0$ such that $|t - c| < \delta$ implies $|f(t) - f(c)| < \varepsilon$. For $0 < |h| < \delta$ with $c + h \in [a,b]$: \begin{align*} \frac{F(c+h) - F(c)}{h} - f(c) = \frac{1}{h}\int_c^{c+h} f(t)\,dt - f(c) = \frac{1}{h}\int_c^{c+h} \bigl(f(t) - f(c)\bigr)\,dt. \end{align*} Taking absolute values: \begin{align*} \left|\frac{F(c+h) - F(c)}{h} - f(c)\right| \le \frac{1}{|h|}\int_{\min(c,c+h)}^{\max(c,c+h)} |f(t) - f(c)|\,dt < \frac{1}{|h|} \cdot \varepsilon \cdot |h| = \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\lim_{h \to 0} \frac{F(c+h) - F(c)}{h} = f(c)$, so $F$ is [differentiable](/page/Derivative) at $c$ with $F'(c) = f(c)$. [guided] The key identity is $\frac{F(c+h) - F(c)}{h} = \frac{1}{h}\int_c^{c+h} f(t)\,dt$. We can rewrite this as a deviation from $f(c)$ by using the fact that $f(c) = \frac{1}{h}\int_c^{c+h} f(c)\,dt$ (the average of a constant over any interval equals that constant). Subtracting: \begin{align*} \frac{F(c+h) - F(c)}{h} - f(c) = \frac{1}{h}\int_c^{c+h}\bigl(f(t) - f(c)\bigr)\,dt. \end{align*} Since $|h| < \delta$, every $t$ in the interval $[\min(c, c+h), \max(c, c+h)]$ satisfies $|t - c| \le |h| < \delta$, so $|f(t) - f(c)| < \varepsilon$. The triangle inequality for integrals then gives the bound $\varepsilon$, completing the argument. Note that we only used continuity of $f$ at $c$, not on all of $[a,b]$. The differentiability of $F$ is a pointwise statement that depends only on the local behaviour of $f$. [/guided] [/step]