6.5: Average Value of a Function
(Dated: October 10, 2011)
Let f be a function on an interval [a, b]. Dividing [a, b]
b−a
and choosing
into n subintervals of equal width ∆x =
n
a sample point from each interval yields the average of the
numbers f (x 1∗ ), f (x 2∗ ), · · · , f (x n∗ ):
(b) Find c such that f ave = f (c).
Solution. (a) By definition,
Z 5
1
(x − 3)2 d x
5−2 2
Z
1 5 2
=
(x − 6x + 9) d x
3 2
·
¸5
1 x 3 6x 2
=
−
+ 9x
3 3
2
2
f ave =
n
f (x 1∗ ) + f (x 2∗ ) + · · · + f (x n∗ )
1X
f (x i∗ ) =
.
n i =1
n
Here x i∗ is a sample point from the i th subterval [x i −1 , x i ].
b−a
, the average can be expressed
Using the relation ∆x =
n
as
n
n
1 X
∆x X
f (x i∗ ) =
f (x i∗ )∆x.
b − a i =1
b − a i =1
Passing to the limit n → ∞ yields
f ave =
1
b−a
Z
b
f (x) d x,
a
the average value of f on the interval [a, b].
Theorem 1 (Mean Value Theorem for Integrals) If f is continuous function on [a, b], then there exists a number c in
[a, b] such that
f (c) = f ave =
1
b−a
Z
b
f (x) d x.
a
In other words,
Z
b
a
f (x) d x = f (c)(b − a).
Example 1 (Exercise 6.5.9 in the text) Let f (x) = (x − 3)2 on
[2, 5].
(a) Find the average value of f on the given interval.
= 1.
(b) Let c be a number in [2, 5] such that f (c) = f ave = 1. Then
(c −1)2 = 1 or equivalently c −1 = ±1 and c = 1±1 = 2, 0. Since
0 is not in [2, 5], it must be the case that c = 2.
Example
R 3 2 (Exercise 6.5.13 in the text) If f is continuous
and 1 f (x) d x = 8, show that f takes on the value 4 at least
once on teh interval [1, 3].
Solution. By the mean value theorem, there is a number c in
[1, 3] such that
f (c) = f ave =
1
3−1
3
Z
1
f (x) d x =
1
· 8 = 4.
2
Example 3 (Imitation of exercise 6.5.14 in the text)
Find the numbers b such that the average value of
f (x) = 6x 2 + 10x − 8 on the interval [0, b] is equal to 4,
where b > 0. Solution. The average value of f on the interval
[0, b] is given by
f ave =
1
b −0
b
Z
0
(6x 2 + 10x − 8) d x = 2b 2 + 5b − 8.
Thus f ave = 4 if and only if 2b 2 + 5b − 8 = 4 or 2b 2 + 5b − 12 =
3
(2b − 3)(b + 4) = 0. This implies that either b = .
2