# Thread: Fundamental Theorem of Calculus

1. ## Fundamental Theorem of Calculus

Theorem. Let $a \in \mathbb{R}$, and let $I$ be an interval containing $a$. Let $I \subseteq K \subseteq \mathbb{R}$. Suppose that the function $f: K \to \mathbb{R}$ is integrable on $[a,x]$ for all $x \in I$ with $x > a$ and on $[x,a]$ for all $x \in I$ with $x < a$. Let $F_a$ be the area accumulation function for $f$ on $I$ based at $a$. Then $F_a$ is uniformly continuous on $I$. If $f$ happens to be continuous on $I$, then $F_a$ is differentiable on $I$ and $F_{a}'(x) = f(x)$ for all $x \in I$.

Proof. Suppose $a \in \mathbb{R}$, $I$ an interval containing $a$, and let $f: I \to \mathbb{R}$ be a function. The area accumulation function $F_a: I \to \mathbb{R}$ is defined by $F_{a}(x) = \int_{a}^{x} f(t) \ dt$. (1) We want to show that $F_a$ is uniformly continuous on $I$. To do this, would we use a sequence approach? Because $F_a$ is not Lipschitz. (2) Suppose $f$ is continuous on $I$. We want to show that $F_a$ is differentiable at some point $c$. So we look at the following:

$\frac{F_{a}(x)-F_{a}(c)}{x-c} = \frac{\int_{a}^{x} f(t) \ dt- \int_{a}^{c} f(t) \ dt}{x-c} = \frac{1}{x-c} \int_{x}^{c} f(t) \ dt$

We want to show that $\lim\limits_{x \to c} \frac{1}{x-c} \int_{c}^{x} f(t) \ dt = f(c)$. And so $f(c) = \frac{1}{x-c} \int_{c}^{x} f(c) \ dt$. The result follows. $\diamond$

Is the second part correct? In the first part, would the easiest way be to use a sequence approach? Or just the regular $\epsilon-\delta$ definition?

2. Originally Posted by manjohn12
Theorem. Let $a \in \mathbb{R}$, and let $I$ be an interval containing $a$. Let $I \subseteq K \subseteq \mathbb{R}$. Suppose that the function $f: K \to \mathbb{R}$ is integrable on $[a,x]$ for all $x \in I$ with $x > a$ and on $[x,a]$ for all $x \in I$ with $x < a$. Let $F_a$ be the area accumulation function for $f$ on $I$ based at $a$. Then $F_a$ is uniformly continuous on $I$. If $f$ happens to be continuous on $I$, then $F_a$ is differentiable on $I$ and $F_{a}'(x) = f(x)$ for all $x \in I$.

Proof. Suppose $a \in \mathbb{R}$, $I$ an interval containing $a$, and let $f: I \to \mathbb{R}$ be a function. The area accumulation function $F_a: I \to \mathbb{R}$ is defined by $F_{a}(x) = \int_{a}^{x} f(t) \ dt$. (1) We want to show that $F_a$ is uniformly continuous on $I$. To do this, would we use a sequence approach? Because $F_a$ is not Lipschitz. (2) Suppose $f$ is continuous on $I$. We want to show that $F_a$ is differentiable at some point $c$. So we look at the following:

$\frac{F_{a}(x)-F_{a}(c)}{x-c} = \frac{\int_{a}^{x} f(t) \ dt- \int_{a}^{c} f(t) \ dt}{x-c} = \frac{1}{x-c} \int_{x}^{c} f(t) \ dt$

We want to show that $\lim\limits_{x \to c} \frac{1}{x-c} \int_{c}^{x} f(t) \ dt = f(c)$. And so $f(c) = \frac{1}{x-c} \int_{c}^{x} f(c) \ dt$. The result follows. $\diamond$

Is the second part correct? In the first part, would the easiest way be to use a sequence approach? Or just the regular $\epsilon-\delta$ definition?

So for (1), would a sequence approach be the best?