# Wednesday, February 27, 2013 Fundamental Theorem of Calculus

Here's an easy result:

Theorem: Let $I = [a,b]$ and $f, g : I \to \mathbb{R}$ integrable then $fg : I \to \mathbb{R}$ is integrable.

Proof: $fg = \frac{1}{2}((f+g)^2 - f^2 - g^2)$ which is a linear combination of integrable functions.

## 1Fundamental Theorem of Calculus¶

### 1.1Second version¶

FTC: Let $I = [a,b]$, $f : I \to \mathbb{R}$ be integrable and let $\displaystyle F(x) = \int_a^bf$ for $x \in [a, b]$. Then $F(x)$ is continuous on $[a,b]$. Moreover, if $f$ is continuous at $c \in [a,b]$ then $F$ is differentiable at $c$ and $F'(c) = f(c)$

Proof: $f$ is integrable, thus $f$ is bounded by some $k \in \mathbb{R}$. Let $x, y \in I$ be such that $x < y$. Then

$$|F(y) - F(x)| = \left|\int_a^y f - \int_a^x f \;\right| = \left| \int_x^y f \;\right| \leq \int_x^y |f| \leq k(y- x) = k|y - x|$$

i.e., $F$ is Lipschitz on $I$ and thus continuous on $I$. If $f$ is continuous at $c \in [a,b]$ then for a given $\epsilon > 0$, there exists some $\delta > 0$ such that if $|x - c| < \delta$ then $|f(x) - f(c)| < \epsilon / 2$. Thus for $0 < |x-c| < \delta$:

\begin{align} \left| \frac{F(x)-F(c)}{x-c} - f(c) \right| &= \left|\frac{1}{x-c} \left[ \int_a^x f - \int_a^c f \right] - f(c) \right| \quad \text{suppose } x > c\\ &= \left| \frac{1}{x-c} \int_c^x f - f(c) \right| \\ &= \left| \frac{1}{x-c} \int_c^x f(t) - f(c) dt \right| \\ &= \left|\frac{1}{x-c}\right| \left| \int_c^x f(t) - f(c) dt \right| \\ &\leq \frac{1}{x-c} \int_c^x |f(t) - f(c)| dt \leq \frac{1}{x-c}\frac{\epsilon}{2}(x-c) < \epsilon \end{align}

The case for $x < c$ follows similarly. Thus $F$ is differentiable at $c$ and $F'(c) = f(c)$.

Corollary: If $I = [a,b]$ and $f : I \to \mathbb{R}$ is continuous on $I$ then $F(c) = \int_a^c f$ is continuous on $[a,b]$, differentiable on $[a,b]$ and $F'(x) = f(x) \; \forall x \in [a,b]$.

### 1.2First version¶

FTC: Let $I = [a,b]$, $f : I \to \mathbb{R}$ integrable on $I$. If $F : I \to \mathbb{R}$ satisfies
(i) $F$ is continuous on $I$
(ii) $F$ is differentiable on $I$ and $F'(x) = f(x) \; \forall x \in (a,b)$
then

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

Proof: Let $\epsilon > 0$ be given. Since $f$ is integrable, there exists a partition $P = [x_0, x_1, \ldots, x_n \}$ such that $U_{F'}(P) - L_{F'}(P) < \epsilon$. Apply Mean Value Theorem to $F$ on $[x_{k-1}, x_k]$. Then there exists some $t_k \in (x_{k-1}, x_k)$ such that

$$\frac{F(x_k) - F(x_{k-1})}{x_k - x_{k-1}} = F'(t_k)$$

or equivalently, $F(x_k) - F(x_{k-1}) = F'(t_k)(x_k - x_{k-1})$. Let $m_k' = \inf \{ F'(x) : x \in [x_{k-1}, x_k]\}$ and $M_k' = \sup \{ F'(x) : x \in [x_{k-1}, x_k]\}$ (these exist as $F'(x) = f(x)$ is bounded on $I$). Thus

$$m_k'(x_k - x_{k-1}) \leq F'(t_k)(x_k - x_{k-1}) \leq M_k' (x_k - x_{k-1})$$

i.e. $m_k'(x_k - x_{k-1}) \leq F(x_k - x_{k-1}) \leq M_k'(x_k - x_{k-1})$. Summing these over $k$ gives us a telescoping sum, and we end up with $L_{F'}(P) \leq F(b) - F(a) \leq U_{F'}(P)$. Conclusion: $L(F') \leq F(b) - F(a) \leq U(F')$, but $L(F') = U(F') = F(b) - F(a)$ and thus

$$\int_a^b F' = \int_a^b f$$