The Mean Value Theorem for integrals of continuous functions

To get to the mean value theorem for integrals of continuous functions, we first prove the following preliminary, but basic and intuitively clear result:

• Consider the following hypotheses:
  • $a$ and $b$ are given real numbers, with $a < b$.
  • $f$ is continuous on the closed interval $[a,b]$.
  • $f \ge 0$ on $(a, b)$.
    
  Then either $f = 0$ on $[a,b]$, or $\int_a^b f(t) dt > 0$.

Intuitively this is saying that that area under the graph of a non-negative continuous function can only be zero if the function is everywhere zero.
This is intuitively correct, because if the function is ever non-zero, then it must be strictly positive in some region and then the area under the graph should be also strictly positive.

Proof:

Let $m$ and $M$ be the minimum and maximum values of $f$ on the interval $[a,b]$.
These exist and are values of $f$ somewhere on the interval, by the Extreme Value Theorem.
So we have $0 \le m \le f(t) \le M$ for all $t$ in $[a,b]$.

• If $M = 0$, then $f = 0$ identically and we are done.

So now assume that $M > 0$.

• If $m = M$, then $f(t) = M$ for all $t$ and $\int_a^b f(t) dt = M(b -a) > 0$ and we are done.

So now assume that $m < M$.

• Then the number $A = \frac{1}{2}(M + m)$ is positive and strictly smaller than $M$.
  Let $f(c) = M$, with $c$ in $[a,b]$.
  Since $f$ is continuous on $[a,b]$, we can make $f(t)$ as close as we like to $M$, if we get sufficiently close to $t = c$, so in particular we can make $f$ larger than $A$ if we are close enough to $t = c$, so there is a neighbourhood $[p, q]$ containing $c$, such that $f(t) \ge A$ on that interval, where we have: $a\le p < q\le b$ and $p \le c\le q$.
  Then, since $f(t) \ge 0$ on $[a,b]$, we have $\int_a^p f(t) dt \ge 0$ and $\int_q^b f(t) dt \ge 0$, so we get:
  $$\int_a^b f(t) dt = \int_a^p f(t) dt + \int_p^q f(t) dt + \int_q^b f(t) dt$$
  $$\ge \int_p^q f(t) dt \ge \int_p^q A dt = A(q - p) > 0.$$
  So $\int_a^b f(t) dt > 0$ and we are done.

Our first corollary is the first version of the Mean Value Theorem for integrals:

• Consider the following hypotheses:
  • $a$ and $b$ are given real numbers, with $a \ne b$.
  • $f$ is continuous on the closed interval joining $a$ and $b$.
  • $\int_a^b f(t) dt = 0$.
  Then $f$ must vanish somewhere on the open interval joining $a$ and $b$.

Proof:

• By interchanging $a$ and $b$ if necessary, we may assume that $a < b$.
• Now suppose that $f$ never vanishes on $(a, b)$.
  Then since $f$ is continuous, $f$ cannot change sign on the interval $(a, b)$, since otherwise if $c$ and $d$ exist in $(a, b)$ with $f(c)$ and $f(d)$ of opposite signs, then by the Intermediate Value Theorem, $f$ vanishes somewhere on the interval joining $c$ and $d$.
  By replacing $f$ by $-f$ if necessary, we can assume that $f > 0$ on the interval $(a, b)$.
  But now we can apply the previous theorem and we conclude that the integral $\int_a^b f(t) dt > 0$, contradicting the hypothesis that $\int_a^b f(t) dt = 0$.
  So $f$ must vanish somewhere on $(a, b)$ and we are done.

Our second corollary is the complete Mean Value Theorem for integrals:

• Consider the following hypotheses:
  • $a$ and $b$ are given real numbers.
  • $f$ is continuous on the closed interval joining $a$ and $b$.
  Then we conclude:
  • Some $\xi$ exists in the closed interval joining $a$ and $b$, such that:
    $$f(\xi)(b - a) = \int_a^b f(t) dt.$$

  • Further, if $a \ne b$, then $\xi$ may be taken to lie in the open interval joining $a$ and $b$.

Proof:

• If $a = b$, then both sides of the desired relation are zero, so the result holds with $\xi = a$.

So now we assume that $a \ne b$

• Make the following definitions:
  • Put $A(f) = \frac{1}{b -a} \int_a^b f(t) dt$.
  • Put $g(t) = f(t) - A(f)$.
  Then we have:
  • $g$ is continuous on the interval joining $a$ and $b$.
  • The integral of $g$ on the interval joining $a$ and $b$ vanishes:
    $$\int_a^b g(t) dt = \int_a^b (f(t) - A(f)) dt = \int_a^b f(t) dt - \int_a^b A(f) dt$$
    $$= \int_a^b f(t) dt - A(f)(b - a) = \int_a^b f(t)dt - \int_a^b f(t) dt = 0.$$

  So the previous corollary applies.
  Therefore by the previous corollary, $g$ vanishes at some $\xi$ in the open interval joining $a$ and $b$.
  Then $0 = g(\xi) = f(\xi) - A(f)$, so $f(\xi) = A(f)$ and we are done.

We can see geometrically why this theorem holds in the case that $a < b$ and $f(t) > 0$.
In that case, the integral is the area under the curve from $t = a$ to $t = b$ and the right-hand side is the area of a rectangle of base $b - a$ and height $f(\xi)$.
Denote the this area by $J$.
If we consider a $LRS$ for the area with just one interval, then its value is $m(b - a)$ where $m$ is the minimum of $f$ on $[a,b]$.
If we consider a $URS$ for the area with just one interval, then its value is $M(b - a)$ where $M$ is the maximum of $f$ on $[a,b]$.
Since $LRS \le J\le URS$, we have $m(b -a)\le J\le M(b - a)$.
Now make a rectangle with height $h$ with base the interval $[a,b]$.
As $h$ increases from $m$ to $M$, its area increases continuously from $m(b - a)$ to $M(b - a)$.
So there is a particular height at $h$, between $m$ and $M$, when its area exactly matches the area under the curve $J$ (of course $h = \frac{J}{b - a}$).
But on the interval $[a,b]$, $f$ attains all values between $m$ and $M$, so some $\xi$ in the interval $[a,b]$ exists with $f(\xi) = h$ and then $f(\xi)(b - a) = J$, as required.
Note that one way of visualizing the level $h = f(\xi)$ is that it is the level such that there is an equal amount of area under the curve $f$ and above the level $h$ as there is below, since $\int_a^b (f(t) - h) dt = \int_a^b f(t) dt - h(b - a) = 0$ and geometrically $\int_a^b (f(t) - h) dt$ is the area under the curve $y = f(t)$ above the level $y = h$ minus the area above the curve $y = f(t)$ and below the level $y = h$.