Mean Value Theorem for Integrals/Proof 1

From ProofWiki
Jump to navigation Jump to search


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:

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


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:

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

Then, from Upper and Lower Bounds of Integral:

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

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

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

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

\(\ds \frac 1 {b - a} \int_a^b \map f x \rd x\) \(=\) \(\ds \map f k\)
\(\ds \leadsto \ \ \) \(\ds \int_a^b \map f x \rd x\) \(=\) \(\ds \map f k \paren {b - a}\)