Fundamental Theorem of Calculus/First Part/Proof 2

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


Proof

\(\displaystyle D_x F \left({x}\right)\) \(=\) \(\displaystyle \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \left({\int_a^{x + \Delta x} f \left({t}\right) \rd t - \int_a^x f \left({t}\right) \rd t}\right)\) 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.

$\blacksquare$


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