The mean value theorem for integrals

We have seen that the average value μ is a height lying between the minimum and the maximum of f. The mean value theorem (for integrals) states something stronger: for a continuous function, this average height is actually reached by the function at at least one point.

   If f is continuous on [a, b], then there exists c in [a, b] such that:
        f(c) = μ = 1/(b − a) · ∫ from a to b f(x) dx

Geometrically: there exists (at least) one abscissa c where the curve passes exactly at the height of the average rectangle.

   f(x)
    |      __--‾‾--__
    |    _/     •c    \_       <- at c, the curve touches height μ
    |μ__/______________\__
    |
    +--+------+--------+--> x
       a      c        b

The proof relies on the intermediate value theorem: since μ is squeezed between min(f) and max(f), and a continuous f takes every intermediate value, it must pass through μ somewhere.

Why is continuity essential? Without it, a function could "jump" over the value μ without ever reaching it. The theorem then fails.

   f continuous    ->  the average is reached (∃ c, f(c) = μ)
   f discontinuous ->  the average may never be reached

This theorem is the integral counterpart of the mean value theorem for derivatives. It guarantees that the average is not just a computational artifact: it is a value actually taken by the function. Remember the statement and its key ingredient — continuity.