• Pete Clark 's notes on honors calculus provides a handy axiomatization of what properties the definite integral ought to satisfy.
• 1. If $f = C$ is a constant function, then $\int_a^b C = C (b - a)$.
• 2. If $f_1(x) \leq f_2(x)$ for all $x \in [a, b]$, then $\int_a^b f_1(x) \leq \int_a^b f_2(x)$.
• 3. If $a \leq c \leq b$, then $\int_a^b f = \int_a^c f + \int_c^b f$.

§ Proof of fundamental theorem of calculus from the above axiomatization

• Let $f$ be any integrable function over $[a, b]$. For $x \in [a, b]$, we define $F(x) \equiv \int_a^x f$. Then:
• (a) The function $F : [a, b] \to \mathbb R$ is continuous at every $c \in [a, b]$.
• (b) if $f$ is continuous at $c$, then $F$ is differentiable at $c$ and $F'(c) = f(c)$.
• (c) if $f$ is continuous and $F$ is any antiderivative of $f$, that is, $F'(x) = f(x)$, then $\int_a^b f = F(b) - F(a)$.
• Proof:
• First, by coninuity of $f$ and compactness of $[a, b]$, there exists a $M \in \mathbb R$ such that $|f(x)| \leq M$for all $x \in [a, b]$. If $M = 0$, then $f(x) = 0$ and thus from axiom 2 $F = 0$ and everything holds.
• Thus we assume that $M > 0$. For all $\epsilon > 0$, we take $\delta = \epsilon / M$.
• By the third axiom, we see that $F(x) - F(c) = \int_a^x f - \int_a^c F = \int_c^x f$.
• TODO.