Graphics Reference
In-Depth Information
which gives
A
(
c
a
)
B
(
b
c
)
h
(
B
A
)
f
A
(
c
a
)
B
(
b
c
)
h
(2
C
B
A
).
If this must hold for all
h
, however small,
f
A
(
c
a
)
B
(
b
c
).
15 The argument of qn 14 shows that a distinct value at an isolated point
does not affect the value of the integral. Questions 12 and 13 establish
the rest of the result.
16 Since
m
f
(
x
),
s
(
x
)
m
defines a lower step function, so
m
(
b
a
)isa
lower sum.
Now
s
(
x
)
f
(
x
)
M
for all
x
,soif
s
(
x
)
A
when
x
x
x
,
A
M
.
So
A
(
x
x
)
M
(
x
x
), and
s
M
(
b
a
).
17 Every lower step function
0, so every lower sum
0. But 0 is a
lower sum, so the lower integral is 0.
18 Since
f
(
x
)
M
,
S
(
x
)
M
defines an upper step function, so
M
(
b
a
)is
an upper sum.
m
f
(
x
)
S
(
x
) for all
x
so, using theargument in qn 16.
S
.
19 Every upper step function
m
(
b
a
)
1, so every upper sum
1. But 1 is an
upper sum, so the upper integral is 1.
20 Since
z
x
for some
i
or
y
for some
j
, the open interval (
z
,
z
)is
contained both within an interval of the type (
x
,
x
), so that
S
is
constant on (
z
,
z
), and within an interval of the type (
y
,
y
), so
that
s
is constant on (
z
,
z
).
If
z
x
z
, then
s
(
x
)
f
(
x
)
S
(
x
), so
s
(
x
)
S
(
x
) on each of the
open intervals (
z
,
z
). So, calculated on the
z
-intervals,
s
S
.
If (
x
,
x
)
(
z
,
z
), then (
x
x
)
A
(
z
z
)
A
(
z
z
)
A
(z
z
)
A
...
(
z
z
)
A
, and so
S
has the same value whichever intervals it is defined on. Likewise for
s
.
So
s
S
for any upper and lower step functions
S
and
s
for the
function
f
.
21
Suppose
f
0. Then there is an upper
step function
S
and a lower step function
s
such that
f
f
and that
f
f
S
f
f
s
f
.
