Fundamental Theorem of Calculus/First Part/Proof 1

Theorem
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 $\left[{a \,.\,.\, b}\right]$.

Proof
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
We have that $f$ is continuous on $\left[{a \,.\,.\, b}\right]$.

It follows from Continuous Image of Closed Interval is Closed Interval 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 for Continuous Functions‎, 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
It is now to be shown that that $F$ is differentiable on $\left({a \,.\,.\, b}\right)$ and that:
 * $\forall x \in \left[{a \,.\,.\, b}\right]: F' \left({x}\right) = f \left({x}\right)$

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:


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

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