Its volume is
R
r

r
π
(
√
r
2

x
2
)
2
dx
=
π
R
r

r
(
r
2

x
2
)
dx
=
π
(
r
2
x

1
3
x
3
)

r

r
=
π
(
2
3
r
3


2
3
r
3
) =
4
3
π
r
3
.
47
Integration techniques
8.2: Volumes of revolution and average value
Average value
Definition
If
y
1
, . . . ,
y
n
is a finite sequence of real numbers, then the average or
mean
value of the sequence is
μ
=
y
1
+
· · ·
+
y
n
n
What would be meant by the average value of a function over some
interval? For example, what is the average value of
f
(
x
) =
x
2
over the
interval from 1 to 3?
48
Integration techniques
8.2: Volumes of revolution and average value
Average value, imprecisely
If
f
(
x
) is a continuous function defined on some interval [
a
,
b
], we can
define the
average value
of
f
(
x
) from the notion of the average of a finite
sequence.
Fix a number
n
and sample consider the sequence
f
(
x
1
)
, . . . ,
f
(
x
n
) where
a
≤
x
1
≤
x
2
≤ · · · ≤
x
n
≤
b
is a sequence of points between [
a
,
b
] chosen
with a “uniform” distribution.
49
Integration techniques
8.2: Volumes of revolution and average value
Average value, precisely
To make the notion of “uniform” precise, we could break [
a
,
b
] into
n
intervals each of length Δ
x
=
b

a
n
and choose
x
i
in the
i
th
interval.
The average of the sequence is then
μ
=
f
(
x
1
) +
· · ·
+
f
(
x
n
)
n
=
(
f
(
x
1
) +
· · ·
+
f
(
x
n
))
1
n
=
1
b

a
(
f
(
x
1
) +
· · ·
+
f
(
x
n
))
b

a
n
=
1
b

a
(
f
(
x
1
) +
· · ·
+
f
(
x
n
))Δ
x
1
b

a
[Riemann sum of
f
(
x
) with respect to
n
and
x
1
, . . . ,
x
n
]
50
Integration techniques
8.2: Volumes of revolution and average value
Definition of average value
As the Riemann sums approach the integral
R
b
a
f
(
x
)
dx
, we may
define
the
average value of a continuous function over an interval in terms of its
integral.
Definition
Let
f
(
x
) be a continuous function on the interval [
a
,
b
] we define the
average value
or
mean
of
f
(
x
) over [
a
,
b
] to be
μ
=
1
b

a
Z
b
a
f
(
x
)
dx
51
Integration techniques
8.2: Volumes of revolution and average value
The example of
f
(
x
) =
x
2
We compute that the average value of
f
(
x
) =
x
2
over the interval from
x
= 1 to
x
= 3 is
μ
=
1
3

1
Z
3
1
x
2
dx
=
1
2
Z
3
1
x
2
dx
=
1
2
(
1
3
x
3
)

3
1
=
1
6
(27

1)
=
13
3
=
4
1
3
52
Integration techniques
8.2: Volumes of revolution and average value
An average with a more difficult integral
What is the average value of
f
(
x
) = 3
x
+ 1 +
xe

x
2
over the interval

1
to 1?
53
Integration techniques
8.2: Volumes of revolution and average value
Solution
The average is
μ
=
1
1

(

1)
Z
1

1
(3
x
+ 1 +
xe

x
2
)
dx
=
1
2
(3
Z
1

1
xdx
+
Z
1

1
dx
+
Z
1

1
xe

x
2
dx
)
=
1
2
(
3
2
x
2

1

1
+
x

1

1
+
Z
1

1
xe

x
2
dx
)
=
1
2
(0 + 2 +
Z
1

1
xe

x
2
dx
)
=
1 +
1
2
Z
1

1
xe

x
2
dx
54
Integration techniques
8.2: Volumes of revolution and average value
Computation of
1
2
R
1

1
xe

x
2
dx
If
g
(
x
) =
xe

x
2
, then
g
(

x
) = (

x
)
e

(

x
)
2
=

xe

x
2
=

g
(
x
). Thus,
R
0

1
xe

x
2
dx
=

R
1
0
xe

x
2
dx
and
1
2
R
1

1
xe

x
2
dx
= 0 yielding
μ
= 1.
55
Integration techniques
8.2: Volumes of revolution and average value
A symbolic solution
Take
u
=

x
2
so that
du
=

2
xdx
to compute
Z
xe

x
2
dx
=
Z

1
2

2
xe

x
2
dx
=

1
2
Z
e

x
2

2
xdx
=

1
2
Z
e
u
du
=

1
2
e
u
+
C
=

1
2
e

x
2
+
C
Hence,
Z
1

1
xe

x
2
dx
=

1
2
e

1


1
2
e

1
= 0 .