4-4 Average Value of a Function
A Mathematics
Academy
Production
Let’s get started:
• You already know about functions and how
to take the average of some finite set.
• Today we’re going to take the average over
infinitely many values (those that the
function takes on over some
interval)…which means CALCULUS!
• Before I show how to do this…let’s talk
about WHY we might want to do this?
Consider the following picture:
• How high would the water level be if the waves all
settled?
Okay! So, now that you have
seen that this an interesting
question...
Let’s forget about real life, and...
Do Some Math
Suppose we have a “nice”function and we need to find
its average value over the interval [a,b].
Let’s apply our knowledge of how to find the
average over a finite set of values to this problem:
First, we partition the interval [a,b] into n subintervals of
equal length to get back to the finite situation:
In the above graph, we have n=8
Let us set up our notation:
x  (b  a) / n
xi comes from the i-th interval
Now we can get an estimate for the
average value:
f ( x1)  f ( x 2)  ...  f ( xn)
faverage 
n
Let’s try to clean this up a little:
f ( x1)  f ( x 2)  ...  f ( xn)
faverage 
n
x

[ f ( x1)  f ( x 2)  ... f ( xn)]
ba
Since x  (b  a) / n
In a
more condensed form, we now get:
n
1
faverage 
f ( xi )x

b  a i 1
But we want to get out of the finite,
and into the infinite!
How do we do this?
Take Limits!!!
In this way, we get the average value
of f(x) over the interval [a,b]:
b
1
faveragelim n 1  f ( xi ) 
f ( x ) dx

baa
b  a i 1
n
So, if f is a “nice” function (i.e. we
can compute its integral) then we have
a precise solution to our problem.
Let’s look back at our graph:
Remember the original MVT?
If f is continuous on  a, b then at some point c in (a, b),
f (b)  f (a )
f (c) 
ba
When looking at anti-derivatives and definite
integrals, we write it another way:
Average Value Theorem (for definite integrals)
1 b
f c 
f  x  dx

ba a
So we just say that:
Average Value of f (x)
F (b)  F (a )
f (c ) 
ba


b
a
f ( x) dx
ba
Average value of a function
The average value of function f on the interval [a, b] is
defined as
b
f ave
1

f ( x)dx

ba a
Note: For a positive function, we can think of this
definition as saying area/width = average height
Example: Find the average value of f(x)=x3 on [0,2].
f ave
2
1 2 3
1 x 
1 24

x dx       2

20 0
2  4 0 2 4
4
So we’ve solved our problem!
If I give you the equation f ( x)  3x 2  1
and ask you to find it’s average value
over the interval [0,2], you’ll all say
1
faverage 
ba
b

f ( x)dx
a
2
1
2

3
x

1
dx

20 0
2
1 3
 [x  x ]
2
0
1
 (10  0)  5
2
Now we can answer our fish tank question!
(That is, if the waves were described by an integrable function)
The Mean Value Theorem
for Integrals
Theorem: If f is continuous on [a, b], then there exists
a number c in [a, b] such that
f (c)  f ave
that is,

b
a
1 b

f ( x)dx

ba a
f ( x)dx  f (c)(b  a )