Fundamental Theorem of Calculus/First Part/Proof 2

Proof
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:
 * $\displaystyle \int_x^{x + \Delta x} \map f x \rd x = \map f k \paren {x + \Delta x - x} = \map f k \Delta x$

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

Hence the result, by the definition of primitive.