The first application of integrals that we’ll take a look at
is the average value of a function. The
following fact tells us how to compute this.
To see a justification of this formula see the Proof of Various Integral Properties
section of the Extras chapter.
Let’s work a couple of quick examples.
Example 1 Determine
the average value of each of the following functions on the given interval.
(a) on [Solution]
(b) on [Solution]
There’s really not a whole lot to do in this problem other
than just use the formula.
You caught the substitution needed for the third term
So, the average value of this function of the given
interval is -1.620993.
[Return to Problems]
Again, not much to do here other than use the
formula. Note that the integral will
need the following substitution.
Here is the average value of this function,
So, in this case the average function value is zero. Do not get excited about getting zero
here. It will happen on occasion. In fact, if you look at the graph of the
function on this interval it’s not too hard to see that this is the correct
[Return to Problems]
There is also a theorem that is related to the average
The Mean Value
Theorem for Integrals
Note that this is very similar to the Mean Value Theorem that we saw in the
Derivatives Applications chapter. See
the Proof of Various Integral
Properties section of the Extras chapter for the proof.
Note that one way to think of this theorem is the
following. First rewrite the result as,
and from this we can see that this theorem is telling us
that there is a number such that . Or, in other words, if is a continuous function then somewhere in [a,b]
the function will take on its average value.
Let’s take a quick look at an example using this theorem.
Example 2 Determine
the number c that satisfies the
Mean Value Theorem for Integrals for the function on the interval [1,4]
First let’s notice that the function is a polynomial and
so is continuous on the given interval.
This means that we can use the Mean Value Theorem. So, let’s do that.
This is a quadratic equation that we can solve. Using the quadratic formula we get the
following two solutions,
Clearly the second number is not in the interval and so
that isn’t the one that we’re after.
The first however is in the interval and so that’s the number we want.
Note that it is possible for both numbers to be in the
interval so don’t expect only one to be in the interval.