Theorem – Fundamental theorem of Calculus

Contents
1. Theorem 1 – Proof of uniform convergence
2. Theorem 2 – FTC part 1
3. Theorem 3 – FTC part 2

Theorem 1, Proof of uniform convergence

Let $f$ be a function which is (Riemann) integrable on $[a,b]$ . Then, $F(x)$ is uniformly continuous on $[a,b]$ .

Proof

First, if, for any given $x,y\in[a,b]$ , we have, for a given $\epsilon \in \mathbb{R}^+$ ,

$\begin{aligned} |F(x)-F(y)| &< \epsilon \\ \Leftrightarrow |\int\limits_a^x f(t)dt-\int\limits_a^y f(t)dt| &< \epsilon \end{aligned}$

from the definition of the integral of a function. Now, if $x = y$ , then clearly $0<\epsilon$ , so that uniform convergence holds. We consider the case that $x<y$ , as the remaining case follows almost the same argument and can be performed following the same script. We have,

$\begin{aligned} |\int\limits_a^x f(t)dt-\int\limits_a^y f(t)dt| &< \epsilon \\ \Leftrightarrow |\int\limits_a^x f(t)dt-\int\limits_a^x f(t)dt-\int\limits_x^y f(t)dt| &<\epsilon \\ \Leftrightarrow |-\int\limits_x^y f(t)dt| &< \epsilon \\ \Leftrightarrow |y-x|\cdot m < |-\int\limits_x^y f(t)dt| &< \epsilon\end{aligned}$ .

Where $M = min\{f(t)|~t\in[a,b]\}$ . Then, we have that for each $\epsilon>0$ , there exists a $\sigma = \frac{\epsilon}{m}$ such that $|y-x|<\sigma$ . This proves the result.

Theorem 2, Fundamental theorem of calculus I

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . Let $c\in(a,b)$ . Then,

$F'(c) = f(c)$

Proof

Recall that $F'(c)$ is defined to be

$\lim_{x\rightarrow c} \frac{F(x)-F(c)}{x-c}$ .

Thus, let $\epsilon>0$ be given, so that

$\begin{aligned} |\frac{F(x)-F(c)}{x-c} - f(c)| &=|\frac{1}{x-c} (\int\limits_a^x f(t)dt - \int\limits_a^c f(t)dt)|\\ &= |\frac{1}{x-c}\int\limits_c^x f(t)dt - f(c)| \\ &= |\frac{1}{x-c}\int\limits_c^x f(t)dt - \frac{1}{x-c}\int\limits_c^x f(t)dt| \\ &= |\frac{1}{x-c}\int\limits_c^x f(t) -f(c)dt |\end{aligned}$

But, we have that

$|\frac{1}{x-c}\int\limits_c^x f(t) -f(c)dt| < \frac{1}{|x-c|}\int\limits_c^x |f(t) -f(c)|dt < \frac{1}{|x-c|}\int\limits_c^x \epsilon~dt = \frac{\epsilon}{|x-c|} |x-c| = \epsilon$

Where we have introduced the variable $\epsilon$ as (from theorem 1) $F(x)$ is uniformly continuous on $[a,b]$ . Thus, we have shown that if $F$ is uniformly continuous (which it is, from theorem 1), then via the final inequality, it also follows that

$F'(c) = f(c)\quad\square$

Theorem 3, Fundamental theorem of calculus II

Let $f$ be integrable on $[a,b]$ , and differentiable on $(a,b)$ . Then,

$\int_a^b f(t)dt = F(a) - F(b)$

Proof

Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , by the mean value theorem, there exists some $c\in (a,b)$ such that

$f'(c) = \frac{f(b)-f(a)}{b-a}$

Which is equivalent to $f'(c)(b-a) = f(b)-f(a)$ . Then, for any partition of $[a,b]$ , we have that for each subinterval $[x_{i+1},x_i]$ , we have that there is some $x_i$ such that

$f'(c_i)(x_{i+1}-x_i)= f(x_{i+1})-f(x_i) \\\\ \Longrightarrow \sum\limits_{i=1}^n f'(c_i)(x_{i+1}-x_i) = f(b)-f(a)$

Now, since, for each subinterval $[x_{i+1},x_i]$ , $f'(x)$ lies between the minimum and maximum values of $f( [x_{i+1},x_i])$ , we have that

$m_i \leq f'(x) \leq M_i$

Which implies that

$\sum\limits_{i=1}^n m_i(x_{i+1}-x_i) \leq \sum\limits_{i=1}^n f'(c_i)(x_{i+1}-x_i) \leq \sum\limits_{i=1}^n M_i(x_{i+1}-x_i)$

and then, we have,

$L(P,f) \leq \sum\limits_{i=1}^n f'(c_i)(x_{i+1}-x_i) \leq U(P, f)$

where $U$ and $L$ are the upper and lower sums. As this inequality is true for any partition, then even though $U(f)$ and $L(f)$ may be defined for different partitions, no matter which partition we choose, the inequality above will be true, and so,

$L(f) \leq \sum\limits_{i=1}^n f'(c_i)(x_{i+1}-x_i) \leq U(f) \\\\ \Longrightarrow L(f) \leq f(b)-f(a) \leq U(f)$

But, as $f$ is integrable, $L(f) = U(f)$ , and so

$\int\limits_a^b f(x)~dx = f(b)-f(a) \quad \square$