The Fundamental Theorem of Calculus

Average value of a Function

Definition 1. If $f$ is integrable on $[a, b]$, then its average value on $[a, b]$ which is also called its mean, is $$\text{av}(f) = \frac{1}{b-a} \int_a^b f(x) dx.$$

Mean Value Theorem for Definite Integrals

Theorem 1. If $f$ is continuous on $[a, b]$, then at some point $c$ in $[a, b]$, $$f(c) = \frac{1}{b-a} \int_{a}^{b} f(x) dx.$$

Proof. By the property (6) of definite integrals, we have $min f \leq \text{av}(f) \leq max f$. Then by the Intermediate Value Theorem, there is a point $c \in (a, b)$ such that $$f(c) = \text{av}(f) = \frac{1}{b-a} \int_{a}^{b} f(x) dx. \blacksquare$$  

The Fundamental Theorem of Calculus, Part 1

Theorem 2. If $f$ is continuous on $[a, b]$, then $F(x) = \int_a^x f(t) dt$ is continuous on $[a, b]$ and differentiable on $(a, b)$ and its derivative is $f(x)$: $$F'(x) = \frac{d}{dx} \int_a^x f(t) dt = f(x).$$

Proof. By the definition of the derivative, when $x$ and $x+h$ are in $(a, b)$, we can write $$F'(x) = \lim_{h \rightarrow 0} \frac{\int_a^{x+h} f(t) dt - \int_a^x f(t) dt}{h} = \lim_{h \rightarrow 0} \frac{1}{h} \int_x^{x+h} f(t) dt$$ and by Theorem 1, there is a point $c \in [x, x+h]$ such that $$F'(x) = \lim_{h \rightarrow 0} f(c).$$ Since $c$ goes to $x$ as $h$ approaches $0$, we obtain $$F'(x) = f(x), \forall x \in (a, b)$$ and therefore $F$ is differentiable at such $x$, which we desired to prove. 
To complete the proof, we just have to show that $F$ is also continuous at $x = a, b$. To do this, except that at $x = a$ we need only consider $h \rightarrow 0^+$, and similarly at $x = b$ we need only consider $h \rightarrow 0^-$. This shows that $F$ has a one-sided derivative at $x=a$ and at $x=b$, which means that $F$ is continuous at those two points. $\blacksquare$

Definite integral의 정의에는 미분이나 도함수에 관한 얘기가 전혀 없다. 그러나 실제로 적분은 미분과 역연산의 관계에 있다, 따라서 모든 연속함수는 항상 antiderivative를 갖는다는 놀라운 사실을 Fundamental Theorem of Calculus, 줄여서 F.T.C, 는 말해주고 있다. 크게 두 개의 파트로 나뉘는 이 정리는 part 1에서는 위의 사실을, part 2에서는 definite integral을 미분과 적분의 관계를 이용하여 손쉽게 계산하는 방법을 말해준다. 

The Fundamental Theorem of Calculus, Part 2

Theorem 3. If $f$ is continuous over $[a, b]$ and $F$ is any antiderivative of $f$ on $[a, b]$, then $$\int_a^b f(x) dx = F(b) - F(a).$$

Proof. From F.T.C part 1 we know that there is an antiderivative of $f$ exists, namely $$G(x) = \int_a^x f(t) dt.$$ Thus, by Collorary (2), if $F$ is any antiderivative $f$, then $F(x) = G(x) + C, \forall x \in (a, b)$ where $C$ is a constant. Since both $F$ and $G$ are continuous on $[a, b]$, we see that this equality also holds when $x=a$ and $x=b$ by taking one-side limits ($x \rightarrow a^+$ and $x \rightarrow b^-$).
Hence, we have $$F(b) - F(a) = [G(b) + C] - [G(a) + C] = G(b) - G(a) \\ = \int_a^b f(x) dx - \int_a^a f(x) dx = \int_a^b f(x) dx. \blacksquare$$