Graphics Reference
In-Depth Information
as close as we like to those used in qn 4. Questions 1 and 2 use step
functions.
from qn 29.
(ii)
a
a
a
(i)
a
a
from qns 28 and 29.
a
c
a
c
a
c
a
/(
n
(iii)
c
/2
/3
...
1).
31
s
(
x
)
0
0
s
(
x
)
s
(
x
)
f
(
x
)
f
(
x
).
s
(
x
)
0
s
(
x
)
0
f
(
x
).
Likewise
f
(
x
)
0
0
f
(
x
)
f
(
x
)
S
(
x
)
S
(
x
).
f
(
x
)
0
f
(
x
)
0
S
(
x
).
So
s
(
x
).
Evidently since
s
and
S
arestep functions, so are
s
(
x
)
f
(
x
)
S
and
S
.Sowe
have established that
s
is a lower step function and
S
is an upper
step function for
f
.
Now clearly, if
S
(
x
)
0, then
s
(
x
)
0, so
S
(
x
)
s
(
x
)
S
(
x
)
s
(
x
)
0. Thus the difference between the upper
sums and lower sums from
S
. So the difference
between upper and lower sums may be made arbitrarily small and so
f
and
s
is integrable.
32
f
(
x
)
0
f
(
x
)
0
f
(
x
)
0
(
f
)
(
x
)
0.
f
(
x
)
0
f
(
x
)
0
f
(
x
)
0
(
f
)
(
x
)
f
(
x
)
(
f
)
(
x
)
f
(
x
).
f
integrable
f
integrable by qn 27.
f
integrable
(
f
)
integrable by qn 31
f
integrable by qn 27.
33 Examine
f
(
x
)
0,
0,
0, to show
f
f
f
.
Integrability of
f
follows from qns 31, 32, 27 and 29.
34 On the interval [0, 1] consider
f
(
x
)
1 when
x
is irrational and
f
(
x
)
1 when
x
is rational.
35 From thedefinitions,
f
f
f
, so thelft-hand sideof the
inequality
f
.
f
f
f
f
f
,so
f
f
f
f
f
.
36
(i)
s
(
x
)
0.
(ii)
f
(
x
)
1/
m
x
p
/
q
, where
q
m
.So
m
1 possiblevalus of
q
.
Now 1
p
q
, so at most (
m
1)
(
m
2)
...
2
1
m
(
m
1) possiblevalus of
x
.
(iii) Take
/
N
, then upper sum
2
N
(1/
m
)(1
2
N
)
.
To avoid overlap, take
min(
(
c
c
),
/
N
), then the
formula for the upper sum is exact.
(iv) Lower sum
0, upper sum
. Function integrable, integral
0.
