Fundamental Theorem of Calculus

First Part
Let $f$ be a real function which is continuous on the closed interval $\left[{a. . b}\right]$.

Let $F$ be a real function which is defined on $\left[{a. . b}\right]$ by:
 * $\displaystyle F \left({x}\right) = \int_a^x f \left({t}\right) \mathrm d t$

Then $F$ is a primitive of $f$ on $\left[{a. . b}\right]$.

Second Part
Let $f$ be a real function which is continuous on the closed interval $\left[{a. . b}\right]$.

Then:
 * $f$ has a primitive on $\left[{a . . b}\right]$
 * If $F$ is any primitive of $f$ on $\left[{a . . b}\right]$, then:
 * $\displaystyle \int_a^b f \left({t}\right) \mathrm d t = F \left({b}\right) - F \left({a}\right) = \left[{F \left({t}\right)}\right]_a^b$

Proof of First Part
To show that $F$ is a primitive of $f$ on $\left[{a. . b}\right]$, we need to establish the following:


 * $F$ is continuous on $\left[{a . . b}\right]$
 * $F$ is differentiable on the open interval $\left({a . . b}\right)$
 * $\forall x \in \left[{a . . b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$.

Proof that F is Continuous
Since $f$ is continuous on $\left[{a. . b}\right]$, it follows from the Continuity Property that $f$ is bounded on $\left[{a. . b}\right]$.

Suppose that:
 * $\forall t \in \left[{a . . b}\right]: \left|{f \left({t}\right)}\right| < \kappa$

Let $x, \xi \in \left[{a. . b}\right]$.

From Sum of Integrals on Adjacent Intervals‎, we have that:
 * $\displaystyle \int_a^x f \left({t}\right) \mathrm d t + \int_x^\xi f \left({t}\right) \mathrm d t = \int_a^\xi f \left({t}\right) \mathrm d t$

That is:
 * $\displaystyle F \left({x}\right) + \int_x^\xi f \left({t}\right) \mathrm d t = F \left({\xi}\right)$

So:
 * $\displaystyle F \left({x}\right) - F \left({\xi}\right) = - \int_x^\xi f \left({t}\right) \mathrm d t = \int_\xi^x f \left({t}\right) \mathrm d t$

From the corollary to Upper and Lower Bounds of Integral:
 * $\left|{F \left({x}\right) - F \left({\xi}\right)}\right| < \kappa \left|{x - \xi}\right|$

Thus it follows that $F$ is continuous on $\left[{a. . b}\right]$.

Proof that F is Differentiable and f is its Derivative

 * Second, that $F$ is differentiable on $\left({a . . b}\right)$ and that $\forall x \in \left[{a . . b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$:

Let $x, \xi \in \left[{a. . b}\right]$ such that $x \ne \xi$.

Then:

Now, let $\epsilon > 0$.

If $\xi \in \left({a . . b}\right)$, then $f$ is continuous at $\xi$.

So for some $\delta > 0$, $\left|{f \left({t}\right) - f \left({\xi}\right)}\right| < \epsilon$ provided $\left|{t - \xi}\right| < \delta$.

So provided $\left|{x - \xi}\right| < \delta$ it follows that $\left|{f \left({t}\right) - f \left({\xi}\right)}\right| < \epsilon$ for any $t$ in an interval whose endpoints are $x$ and $\xi$.

So from the corollary to Upper and Lower Bounds of Integral, we have:

provided $0 < \left|{x - \xi}\right| < \delta$.

But that's what this means:


 * $\displaystyle \frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} \to f \left({\xi}\right)$ as $x \to \xi$

So $F$ is differentiable on $\left({a . . b}\right)$, and:
 * $\forall x \in \left[{a . . b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$

Proof of Second Part
Let $G$ be defined on $\left[{a. . b}\right]$ by:
 * $\displaystyle G \left({x}\right) = \int_a^x f \left({t}\right) \mathrm d t$

We have:
 * $\displaystyle G \left({a}\right) = \int_a^a f \left({t}\right) \mathrm d t = 0$ from Integral on Zero Interval
 * $\displaystyle G \left({b}\right) = \int_a^b f \left({t}\right) \mathrm d t$ from the definition of $G$ above.

From Sum of Integrals on Adjacent Intervals, we have:
 * $\displaystyle \int_a^b f \left({t}\right) \mathrm d t = \int_a^a f \left({t}\right) \mathrm d t + \int_a^b f \left({t}\right) \mathrm d t$

Thus:
 * $\displaystyle \int_a^b f \left({t}\right) \mathrm d t = G \left({b}\right) - G \left({a}\right)$

By the first part, $G$ is a primitive of $f$ on $\left[{a. . b}\right]$.

By Primitives which Differ by a Constant‎, we have that any primitive $F$ of $f$ on $\left[{a. . b}\right]$ satisfies $F \left({x}\right) = G \left({x}\right) + c$, where $c$ is a constant.

Thus:
 * $\displaystyle \int_a^b f \left({t}\right) \mathrm d t = \left({G \left({b}\right) + c}\right) - \left({G \left({a}\right) + c}\right)$

and so $\displaystyle \int_a^b f \left({t}\right) \mathrm d t = F \left({b}\right) - F \left({a}\right)$.

Alternate proof of Second Part
Let $f$ have a primitive $F$ continuous on the closed interval $\left[{a..b}\right]$.

$[a..b]$ can be divided into any number of subintervals of the form $[x_{k-1}..x_k]$ where

$ a = x_0 < x_1 ... < x_{k-1} < x_k = b $

By repeatedly adding and subtracting like quantities,

$F(x_k) \underbrace{- F(x_{k-1}) + F(x_{k-1})}_{0}... \underbrace{- F(x_1)+ F(x_1)}_{0} - F(x_0)$

$\implies$

$(A) \quad F(b) - F(a) = \sum_{i=1}^{k} F(x_i) - F(x_{i-1}) $

Because $F = f'$, $F$ is differentiable. Because $F$ is differentiable, $F$ is continuous. By the mean value theorem, in every subinterval $[x_{k-1}..x_k]$ there is some $c_i$ where $F'(c_i) = \frac{F(x_i)-F(x_{i-1})}{\Delta x_i}$

where $\Delta x_i \equiv x_{i} - x_{i-1}$

If I multiply both sides by $\Delta x_i$ I get

$F'(c_i)\Delta x_i = F(x_i) - F(x_{i-1})$

Substituting $F'(c_i)\Delta x_i$ into $(A)$ we get

$F(b) - F(a) = \sum_{i=1}^{k} F'(c_i)\Delta x_i$

Because $F' = f$

$F(b) - F(a) = \sum_{i=1}^{k} f(c_i)\Delta x_i$

Because $f$ and $F$ are both continuous, we can take the limit $||\Delta||→0$ of both sides, where $||\Delta||$ is the magnitude of the largest subinterval $[x_{k-1}..x_k]$

$\lim_{||\Delta|| \to 0} F(b) - F(a) = \lim_{||\Delta|| \to 0} \sum_{i=1}^{k} f(c_i)\Delta x_i$

The $LHS$ is a constant. The $RHS$ is the definition of the integral.

$F(b) - F(a) = \int_{a}^{b}f(x)\mathrm{d}{x}$