Mean Value Theorem for Integrals/Proof 2
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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.
$\blacksquare$