Graphics Reference
In-Depth Information
a lower sum evaluated on the given subdivision. Since the
subdivision is arbitrary, theonly numbr satisfying this
condition for all possible upper step functions and all possible
lower step functions is the integral itself.
52
F
(
x
)
2
x
sin(1/
x
)
cos(1/
x
) when
x
0, and
F
(0)
0 from qn 8.22.
Notethat
F
is not continuous at 0. Nonethelss,
F
is integrable
by qn 29, because
x
sin(1/
x
) gives a continuous function, which is
necessarily integrable, and cos(1/
x
) gives an integrable function like
g
in qn 44. Thus
F
x
sin(1/
x
) provided
x
0.
53 When
x
0,
F
(
x
)
2
x
sin(1/
x
)
(2/
x
)cos(1/
x
), which is
unbounded near
x
0.
F
(
x
)
F
(0)
x
0
x
sin
1
F
(
x
)
F
(0)
x
0
x
,
,so
x
x
and
F
(0)
0. See qn 8.25(i).
54
(i) By the Mean Value theorem for integrals, qn 45.
(iii) Since
F
is continuous at
x
,
f
(
x
h
)
f
(
x
)as
h
0. So
F
(
x
h
)
F
(
x
)
h
f
(
x
)as
h
0,
and this implies that
F
(
x
)
f
(
x
).
55
F
(
x
)
G
(
x
)
f
(
x
)
(
F
G
)
(
x
)
0
(
F
G
)(
x
)
constant by
qn 9.17.
(
F
G
)(
b
)
(
F
G
)(
a
)
F
(
b
)
F
(
a
)
G
(
b
)
G
(
a
).
Since(
F
G
)(
x
)
constant,
G
(
x
)
F
(
x
)
c
.
56 Given
0, there exists
such that
0
(
x
a
)
h
f
(
x
)
L
L
f
(
x
)
L
1
h
F
(
a
h
)
F
(
a
)
h
L
f
L
L
L
,
F
(
a
h
)
F
(
a
)
h
so lim
L
using the Mean Value Theorem for integrals, qn 45.
57 (
f
·
g
)
f
·
g
f
·
g
, so, since
f
·
g
f
·
g
is continuous, theFundamental
Theorem of Calculus gives [
f
·
g
]
(
f
·
g
f
·
g
)
f
·
g
f
·
g
.
58
f
(
x
)(
b
x
)
dx
[
f
(
x
)(
b
x
)]
f
(
x
)(
1)
dx
f
(
a
)(
b
a
)
f
(
x
)
dx
f
(
a
)(
b
a
)
f
(
b
)
f
(
a
).
