Math 104 – Rimmer
6.5 Average Value of a Function
Average of n numbers:
x1 + x2 + x3 + xn
n
Take the n numbers to be sample points for a function:
( )
( )
( )
( )
f x1* + f x2* + f x3* + f xn*
n
Partition the interval [ a, b ] into n subintervals of equal length.
Here n = 8.
a
b
What is the width of each subinterval?
We now have:
b−a
n
∆x =
n
( )
( )
( ) = ∑ f ( x ) ∆x
( )
b−a
b−a
⇒n=
n
∆x
*
f x1* + f x2* + f x3* + f xn*
b−a
∆x
Call this ∆x
i
i =1
b−a
n
Math 104 – Rimmer
6.5 Average Value of a Function
lim ∑ f xi ∆x
( )
*
Taking the limit as n → ∞, we get
n →∞
i =1
b−a
b
1
f ( x ) dx
Average value of f ( x ) on an interval [ a, b ] =
b − a ∫a
Find the average value of the function
Average value
6
1
3
3
dr
= ∫
2
of f ( x ) =
on
1,
6
[
]
5
2
1 (1 + r )
(1 + r )
=
3
(1 + r )
2
on the interval [1, 6] .
u =1+ r r = 1⇒ u = 2
du = dr r = 6 ⇒ u = 7
7
7
7
1 3
3 −2
3  −1  = 3  − 1 − − 1 
du
=
u
du
=

2 
5 ∫2 u 2
5 ∫2
5  u  2 5  7
3  −2 7 
3 5 
3
=  +  =   =
5  14 14 
5 14 
14
1
Math 104 – Rimmer
6.5 Average Value of a Function
Let f be continuous on [ a, b ] ,
then there is a value c in [ a, b ] such that
b
1
f (c) =
f ( x ) dx
b − a ∫a
or
b
f ( c )( b − a ) = ∫ f ( x ) dx
a
area of the
rectangle
area under the
curve
=
Math 104 – Rimmer
6.5 Average Value of a Function
Proof:
Let f be continuous on [ a, b ] ,
f ( x)
By the Extreme Value Theorem,
there is a m and M such that m ≤ f ( x ) ≤ M .
b
Then m ( b − a ) ≤ ∫ f ( x ) dx ≤ M ( b − a )
a
b
⇒ m≤
1
f ( x ) dx ≤ M
b − a ∫a
By the Intermediate Value Theorem,
b
there is a c in [ a, b ] with f ( c ) =
1
f ( x ) dx.
b − a ∫a
2
Math 104 – Rimmer
6.5 Average Value of a Function
(a) Find the average value of f ( x ) = x on the interval [0, 4].
(b) Find c such that f ave = f ( c ) .
(c) Sketch the graph of f and a rectangle whose area
is the same as the area under the graph of f .
Average value
( a ) of f
( x)
on [ a, b ]
b
=
4
Average value
1
1 1/ 2
f
x
dx
⇒
(
)
x dx
=
b − a ∫a
of f ( x ) = x on [ 0, 4] 4 ∫0
4
=
4
4
16
(b ) f ( c ) = ⇒ c = ⇒ c =
3
3
9
=
1
1  2 3/ 2 
x  = 43/ 2

4 3
0 6
(
1 1/ 2
4
6
3
( )
=
8
=
6
)
4
3
(c)
3