# The Unapologetic Mathematician

## The Integral Mean Value Theorem

Okay: time to get back on track. Today, we’ll see a theorem about integrals that’s similar to the Differential Mean Value Theorem. Specifically, it states that if we have a continuous function $f:\left[a,b\right]\rightarrow\mathbb{R}$ then there is some $c\in\left[a,b\right]$ so that

$\displaystyle f(c)=\frac{1}{b-a}\int\limits_a^bf(x)dx$

Let’s consider the Darboux sums we use to define the integral. We know that if we choose a partition, then its upper Darboux sum is greater than any Riemann sum of any refinement of that partition. So let’s take the absolute coarsest possible partition: the one where we just have partition points $a$ and $b$. Then the upper Darboux sum is $(b-a)M$, where $M$ is the maximum value of $f$ on the interval $\left[a,b\right]$. Similarly, the lower Darboux sum on this interval is $(b-a)m$ (where $m$ is the minimum value of $f$), and it’s the lowest possible Darboux sum. Then we can divide everything in sight by $b-a$ to get the inequality

$\displaystyle m\leq\frac{1}{b-a}\int\limits_a^bf(x)dx\leq M$

Now the Intermediate Value Theorem tells us that $f$ must take every value between $m$ and $M$ at some point between $a$ and $b$. And thus there must exist a $c\in\left[a,b\right]$ so that

$\displaystyle f(c)=\frac{1}{b-a}\int\limits_a^bf(x)dx$

just as we wanted.