Fundamental Theorem of Calculus/First Part

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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 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 the corollary to Upper and Lower Bounds of Integral:

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


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

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

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.

$\blacksquare$

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