# Fundamental Theorem of Calculus/First Part

## 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 $\closedint a b$.

### Corollary

- $\ds \frac \d {\d x} \int_a^x \map f t \rd t = \map f x$

## Proof 1

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 Functionsâ€Ž, 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:

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

$\blacksquare$

## Proof 2

\(\ds \dfrac \d {\d x} \map F x\) | \(=\) | \(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_a^{x + \Delta x} \map f t \rd t - \int_a^x \map f t \rd t}\) | Definition of Derivative of Real Function at Point | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^a \map f t \rd t + \int_a^{x + \Delta x} \map f t \rd t}\) | because $\ds \int_a^b \map f x \rd x = -\int_b^a \map f x \rd x$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}\) | Sum of Integrals on Adjacent Intervals for Continuous Functions |

Suppose $\Delta x > 0$ and both $x$ and $x + \Delta x$ are in the closed interval $\closedint a b$.

By hypothesis, $f$ is continuous on the closed interval $\closedint a b$

Thus it is also continuous on the closed interval $\closedint x {x + \Delta x}$.

Thus the conditions of the Mean Value Theorem for Integrals are fulfilled.

So, by the Mean Value Theorem for Integrals, there exists some $k \in \closedint x {x + \Delta x}$ such that:

- $\ds \int_x^{x + \Delta x} \map f x \rd x = \map f k \paren {x + \Delta x - x} = \map f k \Delta x$

Then:

\(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}\) | \(=\) | \(\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \map f k \Delta x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{\Delta x \mathop \to 0} \map f k\) |

By the definition of $k$:

- $x \le k \le x + \Delta x$

which means that $k \to x$ as $\Delta x \to 0$.

So:

\(\ds \lim_{\Delta x \mathop \to 0} \map f k\) | \(=\) | \(\ds \lim_{k \mathop \to x} \map f k\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f x\) | because $f$ is continuous |

For $\Delta x < 0$, consider $k \in \closedint {x + \Delta x} x$, and the argument is similar.

Hence the result, by the definition of primitive.

$\blacksquare$