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In-Depth Information
37
(i) The function is not bounded above, so there is no upper integral.
(ii)
[
x
]
.
38 By qns 7.31, 7.32 and 7.34, the maximum-minimum theorem.
39
(i) Yes, as in qn 38.
(ii)
s
(
x
)
m
when
x
x
x
, gives a lower step function, and
, gives an upper step function.
(iii) Choose
n
so that (
b
a
)/
n
S
(
x
)
M
when
x
x
x
, then
b
a
b
a
n
(
M
m
)
·
a
.
n
n
b
40 No. Step functions and monotonic functions are integrable whether
they are continuous or not.
41 If
f
(
x
)
0,
s
(
x
)
0 is a lower step function and so 0 is a lower sum.
Since
f
0, there must be arbitrarily small upper sums. However, if
f
(
c
)
0 for some
c
, there is a neighbourhood of
c
on which
f
(
x
)
f
(
c
)
f
(
c
). So for some
,
x
c
f
(
x
)
f
(
c
). So any
upper sum
f
(
c
), which is not arbitrarily small. This
contradiction shows that
f
(
x
)
2
·
0 for all
x
. In qns 25 and 36,
discontinuous
f
satisfies the conditions.
42 Theconclusion holds if both
f
and
g
arecontinuous functions, using qn
44. Otherwise we could take
f
as in qn 25 and
g
(
x
)
0.
43
(i)
m
inf
f
(
x
)
x
x
x
,
and
M
sup
f
(
x
)
x
x
x
.So
m
f
(
x
)
M
, and
m
(
x
x
)
f
(
x
)(
x
x
)
M
(
x
x
).
Thus
lower sum
f
(
x
)(
x
x
)
upper sum.
(
b
a
)/
n
.
(ii) We proved in qn 39 that the difference between these upper and
lower sums may be made arbitrarily small for suMciently large
n
,
and it therefore follows that each of these tends to the integral as
n
a
(
i
/
n
)(
b
a
) and
x
x
x
.
Denoting the sum in question by
, lower sum
upper
sum
0
lower sum
upper sum
lower sum, and now,
.
(iii) This summation gives a minimal upper sum for the given
subdivision when the function is monotonic increasing.
by a sandwich theorem,
tends to the integral as
n
