Fundamental Theorem of Calculus

Let $f$ be a real valued continuous function on $[a,b]$ . Define the function $F(x) = \int_0^x f(u) du$ . Then $F$ is uniformly continuous on $[a,b]$ and differentiable on $(a,b)$ such that $F'(x) = f(x)$ for all $x \in (a,b)$ .

proof:

Let $x,u\in (a,b)$ . $F(u)-F(x) = \int_x^u f(t) dt = f(c) (u-x)$ for some $c\in (x,u)$ . Hence in the limit, $F'(x) = \lim_{u \rightarrow x} \frac{F(u)-F(x)}{u-x} = \lim_{u\rightarrow x} f(c) = f(x)$

converse:

Let $f$ be a real valued function on a closed interval $[a,b]$ and $F$ a continuous real valued function on $[a,b]$ such that $F'(x) = f(x)$ for all $x\in (a,b)$ . If $f$ is Riemann integrable on $[a,b]$ then $\int_a^b f(x) dx = F(b) - F(a)$ .

proof:

Since $f$ is Riemann integrable, $\int_a^b f(x) dx$ is the limit of Riemann sums. For any subdivision of $[a,b]$ ,

$F(b)-F(a)=\sum_{i=1}^n \left( F(x_{i+1})-F(x_i) \right)= \sum_{i=1}^n f(c_i) (x_{i+1} - x_i)$

Taking the limit, one obtains $F(b)-F(a) = \int_a^b f(t) dt$

Intermediate Value Theorem

Let $f$ be a continuous function on $[a,b]$ such that $a < b$ and suppose that $f(a) < f(b)$ . For any $c\in (f(a) ,f(b))$ there is an $x \in (a,b)$ such that $f(x) = c$ .

proof: Let $G = \left\{x \ge a : f(x) \le c \right\}$ . Note that $G$ is a nonempty subset of the real numbers that has an upper bound. Hence it has a least upper bound $\alpha$ . Either $f(\alpha) < c, f(\alpha) > c$ or $f(\alpha) = c$ .

If $f(\alpha) < c$ then by continuity there is an open interval $I = (\alpha - \epsilon, \alpha + \epsilon)$ such that $f < c$ on $I$ . But this implies that $\alpha$ is not an upper bound of $G$ . This is false.

If $f(\alpha) > c$ , then by continuity there is an open interval $I = (\alpha - \epsilon, alpha + \epsilon)$ such that $f > c$ on $I$ . But this implies that $\alpha$ is not the smallest upper bound of $G$. This is false.

Hence conclude that $f(\alpha) = c$.

Differentiable functions are continuous. Let $\epsilon > 0$ be given. There is a $\delta > 0$ such that $|x - y| < \delta$ implies $| \frac{f(y) - f(x)}{y - x} - f'(x) | < \epsilon/2$ . But this means that $|f(y) - f(x)| < \epsilon/2 + |f'(x)| \delta$ when $|x- y| < \delta$. Choose $\delta_2 < \epsilon/2 / (|f'(x)| + 1) , \delta$ to conclude that |f(y) – f(x)| < \epsilon$ when $|y – x| < \delta_2$. Hence $f$ is continuous at $x$.

Alternatively, since the limit of a product is the product of the limits when the limits exist and since

$\lim_{y\rightarrow x} (fy - fx) = \lim_{y\rightarrow x} \left( \frac{fy - fx}{y - x} (y-x) \right) = \lim_{y\rightarrow x} \frac{fy - fx}{y-x} \lim_{y\rightarrow x} (y - x) = f'(x) \lim_{y\rightarrow x} (y - x) = 0$

the conclusion is reached that $\lim (fy - fx) = 0$ . Since the limit of sum is the sum of limits when the limits exists, $\lim fy = \lim ((fy - fx) + fx) = \lim (fy - fx) + \lim fx = \lim (fy - fx) + fx = fx$ , the conclusion is reached that $\lim_{y\rightarrow x} f(y) = f(x)$ , that is, $f$ is continuous at $x$ .

Mean Value Theorem

Let $f$ be a function differentiable on $(a,b)$ . Then $f(b)-f(a) = f'(c)(b-a)$ for some $c \in (a,b)$ .

case 1: $f(a) = f(b) = 0$ .

Either $f'(x) = 0$ for all $x\in (a,b)$ in which case any $x\in (a,b)$ will suffice, or there is an $x \in (a,b)$ such that $f'(x) \neq 0$ . If $f'(x) < 0$ for some $x \in (a,b)$ then $f'(y) > 0$ for some $y \in (a,b)$ since otherwise $f(b) < f(a)$. Assume that $x < y$. By the IVT there is an $u \in (x,y)$ such that $f'(u) = 0$. Hence

proof:

$latex \phi(t) = f'((1-t)a+tb)(b-a) f(b) – f(a) – f'((1-t)a + tb) (b-a)t

Fundamental Theorem of Calculus

Mean Value Theorem

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