Mean Value Theorem for Integrals/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 $f$

By the Mean Value Theorem, there therefore exists $k \in \openint a b$ such that:


 * $\map {F'} k = \dfrac {\map F b - \map F a} {b - a}$

As $F$ is differentiable on $\openint a b$ with derivative $f$:


 * $\map {F'} k = \map f k$

We therefore have:

giving:


 * $\displaystyle \int_a^b \map f x \rd x = \paren {b - a} \map f k$

as required.