Graphics Reference
In-Depth Information
Answers
1
BEFC
is an inscribed polygon.
ODEGHC
is a circumscribed polygon.
Three inscribed rectangles of width
a
: heights 0,
a
/9, 4
a
/9.
Combined area
/27.
Three circumscribed rectangles of width
a
(
a
/9
4
a
/9)
5
a
a
: heights
a
/9, 4
a
/9,
a
.
Combined area
a
(
a
/9
4
a
/9
a
)
a
· 14
a
/9.
5
a
/8
a
/8
4
a
/8
a
.14
a
/27
5
a
/27
9
a
/27
a
.
1
n
1
2
n
A
1
n
1
2
n
1
1
1
1
.
a
Each of thebounds tends to 1 as
n
, so no number different from
A
a
will satisfy all the inequalities.
4
a
(1
r
)
r
a
(1
r
)
C
.
1
r
1
r
1, then
r
(
r
1)
r
(
r
1)
5
Let
r
1)
C
.
r
(
r
r
1
After division we have
1
r
(
r
r
r
r
r
1)
C
1
.
...
...
Sincethis holds for all
r
1,
C
1/(
k
1), thelimit of each sideof the
inequality as
r
1.
6 Let
h
a
. Inscribed rectangles have area
(
h
1)/
h
(
h
h
)/
h
...
(
h
h
)/
h
n
(
h
1)/
h
.
Circumscribed rectangles have area
(
h
1) · 1
(
h
h
)/
h
...
(
h
h
)/
h
n
(
h
1).
The sequence (
n
(
a
1)) is monotonic decreasing from qn 2.50 so,
from qn 3.78, the closed interval property, its limit
D
its limit.
(i) Since
c
(
a
b
),
b
c
c
a
(
b
a
). Thus thefirst
7
inequality is equivalent to
f
(
a
)
f
(
c
); thethird to
f
(
c
)
f
(
b
); and
the second to these two together.
(ii) Apply the argument of (i) to each of the
n
subintervals.
(iii) From (ii), (
I
) is monotonic increasing. It is bounded above by
.
So it is convergent by qn 4.35.
(iv) From (ii), (
C
C
) is monotonic decreasing. It is bounded below by
I
.
So it is convergent by qn 4.34.
