Fundamental Theorem of Calculus/First Part

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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:

$\displaystyle F \left({x}\right) = \int_a^x f \left({t}\right) \rd t$


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


Corollary

$\displaystyle \frac \d {\d x} \int_a^x f \left({t}\right) \rd t = f \left({x}\right)$


Proof 1

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:

$\displaystyle \int_a^x f \left({t}\right) \rd t + \int_x^\xi f \left({t}\right) \rd t = \int_a^\xi f \left({t}\right) \rd t$

That is:

$\displaystyle F \left({x}\right) + \int_x^\xi f \left({t}\right) \rd t = F \left({\xi}\right)$

So:

$\displaystyle F \left({x}\right) - F \left({\xi}\right) = - \int_x^\xi f \left({t}\right) \rd t = \int_\xi^x f \left({t}\right) \ \rd t$


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

$\Box$


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:

\(\displaystyle \frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} - f \left({\xi}\right)\) \(=\) \(\displaystyle \frac 1 {x - \xi} \left({F \left({x}\right) - F \left({\xi}\right) - \left({x - \xi}\right) f \left({\xi}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {x - \xi} \left({\int_\xi^x f \left({t}\right) \rd t - \left({x - \xi}\right) f \left({\xi}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {x - \xi} \int_\xi^x \left({f \left({t}\right) - f \left({\xi}\right)}\right) \rd t\) Integral of Function plus Constant, putting $c = f \left({\xi}\right)$


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:

\(\displaystyle \left\vert{\frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} - f \left({\xi}\right)}\right\vert\) \(=\) \(\displaystyle \frac 1 {\left\vert{x - \xi}\right\vert} \left\vert{\int_\xi^x \left({f \left({t}\right) - f \left({\xi}\right)}\right) \rd t}\right\vert\)
\(\displaystyle \) \(<\) \(\displaystyle \frac 1 {\left\vert{x - \xi}\right\vert} \epsilon \left\vert{x - \xi}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

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:

$\forall x \in \left[{a \,.\,.\, b}\right]: F' \left({x}\right) = f \left({x}\right)$

$\blacksquare$


Proof 2

\(\displaystyle D_x F \left({x}\right)\) \(=\) \(\displaystyle \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \left({\int_a^{x + \Delta x} f \left({t}\right) \rd t - \int_a^x f \left({t}\right) \rd t}\right)\) Definition of Derivative of Real Function at Point
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \left({\int_x^a f \left({t}\right) \rd t + \int_a^{x + \Delta x} f \left({t}\right) \rd t}\right)\) because $\displaystyle \int_a^b f \left({x}\right) \rd x = - \int_b^a f \left({x}\right) \rd x$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \left({\int_x^{x + \Delta x} f \left({t}\right) \rd t}\right)\) 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 $\left[{a \,.\,.\, b}\right]$.

By hypothesis, $f$ is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$

Thus it is also continuous on the closed interval $\left[{x \,.\,.\, x + \Delta x}\right]$.

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

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

$\displaystyle \int_x^{x + \Delta x} f \left({x}\right) \rd x = f \left({k}\right) \left({x + \Delta x - x}\right) = f \left({k}\right) \Delta x$

Then:

\(\displaystyle \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \left({\int_x^{x + \Delta x} f \left({t}\right) \rd t}\right)\) \(=\) \(\displaystyle \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} f \left({k}\right) \Delta x\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\Delta x \mathop \to 0} f \left({k}\right)\)

By the definition of $k$:

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

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

So:

\(\displaystyle \lim_{\Delta x \mathop \to 0} f \left({k}\right)\) \(=\) \(\displaystyle \lim_{k \mathop \to x} \ f \left({k}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f \left({x}\right)\) because $f$ is continuous

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

Hence the result, by the definition of primitive.

$\blacksquare$