The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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February 12, 2008 - Posted by | Analysis, Calculus

7 Comments »

  1. This theorem is just a reformulation of the differential MVT if you take into account the fundamental theorem of calculus (exercise).

    Comment by Michael Livshits | February 13, 2008 | Reply

  2. Which theorem I have not yet covered. Patience, Michael. I know you know far better than I how I should be covering this material, but in the absence of your definitive weblog on the subject, I’m going through it in my own way.

    Comment by John Armstrong | February 13, 2008 | Reply

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  5. Great post! Very powerful theorems.

    Comment by Daily Calculus | April 9, 2008 | Reply

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