# Mean Value Theorem for Integrals/Proof 2

## Theorem

Let $f$ be a continuous real function on the closed interval $\closedint a b$.

Then there exists a real number $k \in \closedint a b$ such that:

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

## 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:

$\ds \map F x = \int_a^x \map f x \rd x$

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:

 $\ds \map f k$ $=$ $\ds \frac {\map F b - \map F a} {b - a}$ $\ds$ $=$ $\ds \frac 1 {b - a} \paren {\int_a^b \map f x \rd x - \int_a^a \map f x \rd x}$ $\ds$ $=$ $\ds \frac 1 {b - a} \int_a^b \map f x \rd x$ Definite Integral on Zero Interval

giving:

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

as required.

$\blacksquare$