> [!NOTE] Statement 1 (The Fundamental Theorem of Calculus I) > Let $f$ be a [[Real Function|real function]] that is [[Riemann Integration|Darboux integrable]] on a [[Closed Real Interval|closed real interval]] $[a,b].$ Then the function $F:[a,b]\to \mathbb{R}$ defined by $F:x \mapsto \int_{a}^{x} f(t) \, dt $is [[Continuous Real Function|continuous]]. If $f$ is continuous at some $u\in (a,b)$ then $F$ is [[Fréchet Differentiation|differentiable]] at $u$ and $F'(u)=f(u).$ **Proof that $F$ is continuous**: Let $h>0$ such that $x,x+h\in [a,b].$ Then $F(x+h)-F(x)=\int^{x+h}_{x} f(t) \, dt .$By definition, $f$ is [[Bounded Real Function|bounded]] so there exists $M>0$ such that for all $x\in [a,b],$ $|f(x)|\leq M.$ Then by [[Triangle Inequality for Riemann Integral]], $|F(x+h)-F(x)|= \left\lvert \int_{x}^{x+h} f(t) \, dt \right\rvert \leq \int_{x}^{x+h} |f(t)| \, dt \leq Mh. $Similarly for $|F(x-h)-F(x)|.$ So $F$ is continuous since $Mh$ is arbitrary. **Proof that $F$ is an antiderivative of $f$**: Suppose $f$ is continuous at some $u\in[a,b].$ Let $\varepsilon>0.$ Then there exists $\delta>0$ such that for all $t\in [a,b],$ $|t-u|<\delta \implies |f(t)-f(u)|<\varepsilon.$ Let $0<h<\delta,$ then by [[Darboux Integral of Constant]] and triangle inequality again: $ \begin{aligned} \left|\frac{F(u+h)-F(u)}{h}-f(u)\right| & =\left|\frac{1}{h} \int_u^{u+h} f(t) d t-\frac{1}{h} \int_u^{u+h} f(u) d t\right| \\ & =\left|\frac{1}{h} \int_u^{u+h}(f(t)-f(u)) d t\right| \\ & \leq \frac{1}{h} \int_u^{u+h}|f(t)-f(u)| d t \\ & \leq \frac{1}{h} \int_u^{u+h} \varepsilon d t=\varepsilon . \end{aligned} $Similarly for $h<0.$ Thus the [[Limit of Real Function at a Point|limit]] $\lim_{ h\to 0 } \frac{F(u+h)-F(u)}{h}$ which is the [[Derivative of Real Function|derivative]] $F'(u)$ exists and indeed equals $f(u).$