Average Values
When we want to find the average or mean of a finite collection of objects we simply sum
the values together, and divide by the number of ob jects in the collection. Thus, the average
of the scores 45, 64, and 90 on a test is
45 + 64 + 90
3=199
3≈66.33
What about a continuous function? Is there an intuitive notion of the average value of a con-
tinuous function? Let’s think about a physical example. Suppose we know the instantaneous
velocity of a person hiking in the mountains over some interval of time. What should the
average velocity of the hiker be? It seems intuitive to define the average velocity as the total
distance traveled divided by the time it took to travel this distance. In this way, another
person who was traveling at the average velocity of the first hiker for the same amount of
time would travel the same distance. We can find the distance traveled by calculating the
integral of the hiker’s velocity over the interval of time the hiker traveled. The time traveled
will simply be the difference of the endpoints of the interval of time.
Mathematically, we say that the average value of a function fover an interval [a, b] is given
by
1
b−aZb
a
f(x)dx
which is in complete analogy with the above thought experiment. The average value of a
function over a given interval also has a useful graphical interpretation. If we integrate the
average value over the interval [a, b], we will find the same area as under the function f.
Thus, the average value gives the height of a rectangular box with width [a, b] that has the
same area as the function f.
Example 1 Find the average value of x3over [0,3].
Solution Using the above formula we know the average value is
1
3−0Z3
0
x3dx =1
3
x3
3
3
0=33
9−0
9= 3
Example 2 Find the average value of sin(x) over [0,2π].
Solution We already know that
Z2π
0
sin(x)dx =−cos(2π) + cos(0) = 0
so the average value is 0.
The above result is a problem in the realm of electrical circuits, where AC currents and
voltages are represented by sinusoidal functions. Over the appropriate intervals of 2πthese
functions have an average value of 0. Even if the average value over an interval is not 0, all
segments of length 2πare cancelled out when we find the average value. If we were to think
