Section 7.5 Average value of a function
Let us try to compute the average value of a function y = f (x), a ≤ x ≤ b. We start by
dividing the interval [a, b] into n equal subintervals, each with length ∆x = (b−a)/n and choose
points x∗i in successive subintervals. Then the average of the numbers f (x∗1 ), f (x∗2 ),...,f (x∗n ), is
f (x∗1 ) + f (x∗2 ) + ... + f (x∗n )
n
Since n = (b − a)∆x,
f (x∗1 ) + f (x∗2 ) + ... + f (x∗n )
b−a
∆x
=
1
(f (x∗1 )∆x + f (x∗2 )∆x + ... + f (x∗n )∆x)
b−a
The limiting value as n → ∞ is
1 ∑
1
f (x∗i )∆x =
lim
n→∞ b − a
b−a
i=1
n
∫b
f (x)dx
a
We define the average value of f on the interval [a, b] as
fave
1
=
b−a
∫b
f (x)dx
a
Example 1. Find the average value of f (x) = sin2 x cos x on [π/4, π/2].
u = sin x
du = cos x
4 ∫1
∫ π/2 2
√
1
√
SOLUTION. fave =
sin x cos x dx = u2 du =
2 =
π/4
2/2
π/2 − π/4
π
π/4 → sin(π/4) =
2 π/2 → sin(π/2) = 1 (√
)
1
4 u3 4
2
=
−1
√
π 3 2/2 3π
4
Example 2. Find the numbers b such that the average value of f (x) = 2 + 6x − 3x2 on
the interval [0, b] is equal to 3.
b
1 ∫b
1
1
2
2
3 SOLUTION. 3 = fave =
(2
+
6x
−
3x
)dx
=
(2x
+
3x
−
x
)
= (2b + 3b2 − b3 ) =
0
b
b
b
0
2
2 + 3b − b
Let us solve the equation 3 = 2 + 3b − b2 for b
b2 − 3b + 1√= 0
√
3± 5
3± 9−4
=
b1,2 =
2
2
Mean value theorem for integrals If f continuous on [a, b], then there exist a number
c in [a, b] such that
∫b
f (x)dx = f (c)(b − a)
a
1
The geometric interpretation of this theorem for positive functions f (x), there is a number c
such that the rectangle with base [a, b] and height f (c) has the same area as a region under the
graph of f from a to b.
Example 3. Find the average value of the function f (x) = 4 − x2 on the interval [0, 2].
Find c (0 ≤ c ≤ 2) such that fave = f (c).
(
)2
1 ∫2
1
x3 8
2
(4 − x )dx =
=
SOLUTION. fave =
4x −
0
2
2
3 0 3
8
f (c) = 4 − c2 =
3
8
2
c =4−
3
4
2
c =
3
2
c= √
3
2