# Fundamental Theorem of Calculus/First Part/Proof 1

## Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Let $F$ be a real function which is defined on $\closedint a b$ by:

- $\ds \map F x = \int_a^x \map f t \rd t$

Then $F$ is a primitive of $f$ on $\closedint a b$.

## Proof

To show that $F$ is a primitive of $f$ on $\closedint a b$, we need to establish the following:

- $F$ is continuous on $\closedint a b$
- $F$ is differentiable on the open interval $\openint a b$
- $\forall x \in \closedint a b: \map {F'} x = \map f x$.

### Proof that $F$ is Continuous

We have that $f$ is continuous on $\closedint a b$.

It follows from Continuous Image of Closed Interval is Closed Interval that $f$ is bounded on $\closedint a b$.

Suppose that:

- $\forall t \in \closedint a b: \size {\map f t} < \kappa$

Let $x, \xi \in \closedint a b$.

From Sum of Integrals on Adjacent Intervals for Continuous Functions, we have that:

- $\ds \int_a^x \map f t \rd t + \int_x^\xi \map f t \rd t = \int_a^\xi \map f t \rd t$

That is:

- $\ds \map F x + \int_x^\xi \map f t \rd t = \map F \xi$

So:

- $\ds \map F x - \map F \xi = -\int_x^\xi \map f t \rd t = \int_\xi^x \map f t \rd t$

From Darboux's Theorem: Corollary:

- $\size {\map F x - \map F \xi} < \kappa \size {x - \xi}$

Thus it follows that $F$ is continuous on $\closedint a b$.

$\Box$

### Proof that $F$ is Differentiable and $f$ is its Derivative

It is now to be shown that that $F$ is differentiable on $\openint a b$ and that:

- $\forall x \in \closedint a b: \map {F'} x = \map f x$

Let $x, \xi \in \closedint a b$ such that $x \ne \xi$.

Then:

\(\ds \frac {\map F x - \map F \xi} {x - \xi} - \map f \xi\) | \(=\) | \(\ds \frac 1 {x - \xi} \paren {\map F x - \map F \xi - \paren {x - \xi} \map f \xi}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {x - \xi} \paren {\int_\xi^x \map f t \rd t - \paren {x - \xi} \map f \xi}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {x - \xi} \int_\xi^x \paren {\map f t - \map f \xi} \rd t\) | Definite Integral of Function plus Constant, putting $c = \map f \xi$ |

Now, let $\epsilon > 0$.

If $\xi \in \openint a b$, then $f$ is continuous at $\xi$.

So for some $\delta > 0$:

- $\size {\map f t - \map f \xi} < \epsilon$

provided $\size {t - \xi} < \delta$.

So provided $\size {x - \xi} < \delta$ it follows that:

- $\size {\map f t - \map f \xi} < \epsilon$

for any $t$ in an interval whose endpoints are $x$ and $\xi$.

So from Darboux's Theorem: Corollary:

\(\ds \size {\frac {\map F x - \map F \xi} {x - \xi} - \map f \xi}\) | \(=\) | \(\ds \frac 1 {\size {x - \xi} } \size {\int_\xi^x \paren {\map f t - \map f \xi} \rd t}\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \frac 1 {\size {x - \xi} } \epsilon \size {x - \xi}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \epsilon\) |

provided $0 < \size {x - \xi} < \delta$.

But that is what this means:

- $\dfrac {\map F x - \map F \xi} {x - \xi} \to \map f \xi$ as $x \to \xi$

So $F$ is differentiable on $\openint a b$, and:

- $\forall x \in \closedint a b: \map {F'} x = \map f x$

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.12$