Paul's Online Math Notes
Calculus I (Notes) / Applications of Integrals / Average Function Value   [Notes] [Practice Problems] [Assignment Problems]

Notice

On August 21 I am planning to perform a major update to the site. I can't give a specific time in which the update will happen other than probably sometime between 6:30 a.m. and 8:00 a.m. (Central Time, USA). There is a very small chance that a prior commitment will interfere with this and if so the update will be rescheduled for a later date.

I have spent the better part of the last year or so rebuilding the site from the ground up and the result should (hopefully) lead to quicker load times for the pages and for a better experience on mobile platforms. For the most part the update should be seamless for you with a couple of potential exceptions. I have tried to set things up so that there should be next to no down time on the site. However, if you are the site right as the update happens there is a small possibility that you will get a "server not found" type of error for a few seconds before the new site starts being served. In addition, the first couple of pages will take some time to load as the site comes online. Page load time should decrease significantly once things get up and running however.


Paul
August 7, 2018


Calculus I - Notes
Integrals Previous Chapter   Next Chapter Extras
Applications of Integrals (Introduction) Previous Section   Next Section Area Between Curves

 Average Function Value

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.

 

Average Function Value

The average value of a function  over the interval [a,b] is given by,

                                                       

 

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]

 

Solution

(a)  on  

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 right?

 

So, the average value of this function of the given interval is -1.620993.

[Return to Problems]

 

(b)  on  

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 answer.

AvgFcn_Ex1b_G1

[Return to Problems]

 

There is also a theorem that is related to the average function value.

 

The Mean Value Theorem for Integrals

If  is a continuous function on [a,b] then there is a number c in [a,b] such that,

                                                     

 

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]

 

Solution

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.

Applications of Integrals (Introduction) Previous Section   Next Section Area Between Curves
Integrals Previous Chapter   Next Chapter Extras

Calculus I (Notes) / Applications of Integrals / Average Function Value    [Notes] [Practice Problems] [Assignment Problems]

© 2003 - 2018 Paul Dawkins