Average Value of a Function
We all have good grasp of what is meant by an average. You
should have no problems calculating the average of 5 numbers.
But, suppose you are required to determine the average depth
across a pond. In this case you have an infinite number of points
to try to calculate the average of. This problem is solved by
breaking the pond into many narrow rectangles and adding up
their heights and dividing by their number.
This should look familiar, the adding up of infinitely many
rectangles.
Average (Mean) Value of a Function
If f is integrable on [a, b], then its average (mean) value on [a,
b] is
1 b
av f 
f 
x dx
a
b a 
Example
Find the average of f 
x 3 x 2 on [0, 4] {note: you will have to
use nInt on your calculator to evaluate the integral}. Does f
actually take on this value at some point in the given interval?
Solution:
1 b
av f 
f 
x dx
a
b a 
1 4

3 x 2 
dx


0
4 0
1  28 
2
 
 ; via nINT(3 x , x ,0,4)
4 
3
 
7

3
7
The function assumes this value when 3 x 2  .
3
That is :
7
3 x 2 
3
2
9 3 x 7
3x 2 16
x 
4
3
4
is in the interval [0,4], the function does assume its
3
average value in the given interval.
Since x 
The average value appearing in the interval in the above
example is not a coincidence, but rather an important theorem of
integral calculus.
Average (Mean) Value of a Function
If f is continuous of [a,b], there is a number c 
a , b for which
1 b
f 
c 
f 
x dx .
a
b a 
Example:
Find the average value of the function y x 1 on the interval
2
[0, 3]. At what point(s) in the interval does the function assume
its average value?
Solution:
1 b
av f 
f 
x dx
a
b a 
1 3
2

x 1dx


3 0 0
3
1 1 3 2 2

 
x

x

x

c

3
3
2


0
1
 
3
;using NINT
3
1
Now,
x 11
2
x 2 2x 1 1
x 2 2x 0
x
x 20
x 0, x 2
The points are (0, 1) and (2, 1).
Example: (AB- 1998)
The temperature outside a house during a 24 hour period is
given by
t 
F 
t 80 10 cos  
, 0 t 24
12 
where F(t) is measured in degrees Fahrenheit and t is measured
in hours.
a) Sketch the graph.
b) Find the average temperature, to the nearest degree,
between t=6 and t=14
c) The air-conditioned cools the house whenever the outside
temperature was at or above 78 degrees. For what values of
t was the air-conditioned cooling the house.
d) The cost of cooling the house accumulates at the rate of
$0.05 per hour for each degree the outside temperature
exceeds 78 degrees. What was the total cost, to the nearest
cent, to cool the house for this 14 hour period?
Solution:
t 
F 
t 80 10 cos  
12 
 
10 cos  t 80
12 
Inverted cosine curve, shifted up 10 with a period of
2
24

12
b) Ave=
14 
1
t  1
80

10cos
dt  
697.2957795



14 6 6 
12 


 8
87.162
87
c)
t 
80 10 cos  78
12 
t 
2 10 cos  
12 
1
t 
cos  
5
12 

t
1.369 or t=5.231
12
t
4.914 ot t=18.770
12
Therefore 5.231 t 18.770
d)
18.770 
t  
C 0.05  
80 10cos  78 dt
5.231
12  

5.096
Therefore the total cost is $5.10
The Integral Mean Value Theorem is sometimes applied this way
on AP exams. Look for it.
Suppose f 
x is continuous on [a, b]. Then there exists some
c 
a, bso that
b
1
f 
c 
f 
x dx

b a a
b
f 
x dx  f 
c 
b a 

a
Here the definite Integral equals the average x interval.
g 
b g 
a
since we are given the function
b a
to average. To use the difference quotient we would have
to find g, the anti-derivatives first. Which is the average
value definition.
Note: We don’t use
Suppose that s 
t and v 
t are the position and velocity
functions of a particle. Using the Average Value Formula to
calculate the average velocity.
1
vave 
t1 t0
1
t0
1 
t dt 
v 
t
t1
t0
s 
t1 s 
t0 
t1 t0
t dt 
s
t1
t0
This states that the average rate of change of s on [t0, t1 ]
is equal to the average value of its derivative on [t0, t 1].
Max-Min Inequality
Recall:
If f is integrable on [a, b], its average (mean) value on [a, b] is
1 b
av f 
f 
x dx
a
b a 
If M is the maximum value of f, and if m is the minimum value
of f, both on [a, b], then
m
f 
x dx M 
b a 
b a 
a
b
Example
Show that the value of
x dx
1 cos 
1
0
is less than
3
.
2
Solution:
The max/min Inequality says that m 
b a is a lower bound for
x dx
f 
b
a
and that M 
b a is the upper bound.
The maximum value of
1 cos 
x on [0, 1] is
2
x dx  2 
1 0

1 cos 
3
x dx 
1 cos 
2
1
0
1
0
Example
Use the max/min inequality to find the upper and lower bounds
for the value of
1
1
dx

0 1 x 4
Solution:
When x=0 then M=1
1
When x=1 then m=
2
m
x dx M 
b a f 
b a 
b
a
1
1
1

dx 1
1 0 
1 0 
0 1 x 4
2
1
1
1
 4 dx 1
0
2
1x
Example (1988 BC6)
Let f be a differentiable function defined for all x 0 such that
f 
0 5 and f 
3 1 . Suppose that for any number b>0 the
f 
0 f 
b
average value of f(x) on the interval 0 x b is
.
2
x dx .
f 
3
a) Find
0
b) Prove that f 

x 
f 
x 5
x 0
x
c) Using part b), find f 
x
Solution:
a) The hint to this question is that you are given an expression
for the average value.
f 
0 f 
b
1 3

f 
x dx
0
2
3 0 
5 f 
3
1 3
 
f 
x dx
2
3 0
3
5 1
3 f 
x dx
0
2
3
f 
x dx 6

0
f av 
b)
Since we are given an f 
x and a –5, let’s try with 0 t x
t dt f 
0 f 
x
f 

x
0
x
2
5 x xf 
x
f

t
dt


0
2
1
f 
x  
5 xf 
x f 
x 
,


2
f 
x 5 xf 
x

f 
x 5
f
x 
,
x 0

x
x
do derivative of both sides
c)
From b)
f 
x 5
x
dy y 5

dx
x
1
1
dy  dx
y 5
x
1
1
dx
y 5 dy 
x
ln y 5 ln x C
f
x 

We are given two points. 
0,5 
, 3, 1, but (0,5) is not useable.
ln 1 5 ln 3 C Therefore C=ln(2)
So
ln y 5 ln x ln 
2
ln y 5 ln 2 x
y 5 2 x
3, 1only satisfies

Therefore y 5 2 x
y 2 x 5
y 5 2 x