Fundamental Theorem of Calculus/First Part/Proof 2

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:
 * $\displaystyle F \left({x}\right) = \int_a^x f \left({t}\right) \ \mathrm d t$

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

Proof
Suppose $\Delta x > 0$.

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:
 * $\displaystyle \int_x^{x + \Delta x} f \left({x}\right) \ \mathrm d x = f \left({k}\right) \left({x + \Delta x - x}\right) = f \left({k}\right) \Delta x$

Then:

By the definition of $k$:


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

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

So:

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.