# Mean Value Theorem for Integrals

## 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:

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

### Generalization

Let $f$ and $g$ be continuous real functions on the closed interval $\closedint a b$ such that:

$\forall x \in \closedint a b: \map g x \ge 0$

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

$\displaystyle \int_a^b \map f x \map g x \rd x = \map f k \int_a^b \map g x \rd x$

## Proof 1

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

By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that:

$\displaystyle \map f m = \min_{x \mathop \in \closedint a b} \map f x$
$\displaystyle \map f M = \max_{x \mathop \in \closedint a b} \map f x$

Then, from Upper and Lower Bounds of Integral:

$\displaystyle \map f m \paren {b - a} \le \int_a^b \map f x \rd x \le \map f M \paren {b - a}$

Dividing all terms by $\paren {b - a}$ gives:

$\displaystyle \map f m \le \frac 1 {b - a}\int_a^b \map f x \rd x \le \map f M$

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

 $\displaystyle \frac 1 {b - a} \int_a^b \map f x \rd x$ $=$ $\displaystyle \map f k$ $\displaystyle \leadsto \ \$ $\displaystyle \int_a^b \map f x \rd x$ $=$ $\displaystyle \map f k \paren {b - a}$

$\blacksquare$

## Proof 2

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:

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

giving:

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

as required.

$\blacksquare$