Applications of the Integral
Average Value of a Function
It is not entirely obvious what is meant by the average (or mean) value of a function on an interval. We know how to find the mean of a finite collection of numbers (their sum divided by their number). Needless to say, we run into problems when we want to talk about the mean of all the values of a function on a particular interval, since they are infinite in number.
To find our way out of this conundrum, we recall the definition of the n -th (upper) Riemann sum for the function f on the interval [a, b] :
U
n(f, a, b) =
M
i
|
Note that U n(f, a, b) is equal to to the product of b - a (the length of the interval) and the mean of the values of f at n more or less evenly-spaced points in the interval. Clearly this is a reasonable approximate mean of the function f on the interval [a, b] .
Naturally, the same is true for the
n
th lower Riemann sum. As
n
gets larger and larger, we might imagine the upper and lower Riemann
sums to approach (one from above, one from below) the product of
b - a
and some "true" mean of the function
f
on
[a, b]
. Indeed, this
indicates precisely how we will define the average value, denoted
. We set
|
= |
U
n(f, a, b) |
|
| = |
L
n(f, a, b) |
||
| = |
f (x)dx
|
There is a way of seeing graphically that this definition makes sense. An easy
computation shows that the integral of the constant
from
a
to
b
is
equal to that of the function
f (x)
:
dx
|
= |
|a
b
|
|
| = |
(b - a) |
||
| = |
f (x)dx
|
Thus,
is the height of a rectangle of length
b - a
that will have the same area as the region below the graph of
f (x)
from
a
to
b
. In physical terms, if
f (t)
represents the velocity
of a moving object, then another object moving with velocity
will travel the same distance between the moments
t = a
and
t = b
.
M
i
U
n(f, a, b)
f (x)dx





