Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 2

Proof
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.

Let $F : \closedint a b \to \R$ be a real function defined by:


 * $\displaystyle \map F x = \int_a^x \map f x \rd x$

We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$.

From Fundamental Theorem of Calculus: First Part, we have:


 * $F$ is continuous on $\closedint a b$
 * $F$ is differentiable on $\openint a b$ with derivative $f$

Note that:

for all $x \in \openint a b$.

We therefore have, by Real Function with Positive Derivative is Increasing:


 * $F$ is increasing on $\closedint a b$.

However, by hypothesis:

So, it must be the case that:


 * $\map F x = 0$ for all $x \in \closedint a b$.

We therefore have, from Derivative of Constant:


 * $\map {F'} x = \map f x = 0$ for all $x \in \closedint a b$

as required.