Graphics Reference
In-Depth Information
65
If
I
(
b
)
f
and lim
I
(
b
)
L
, then we write
f
L
.
dx
(
x
)
[
2
(
x
)]
2
(
a
)
2
2as
a
0
.
66 [
2
(1
x
)]
2
(1
a
)
2
2as
a
1
.
0
x
1
1/
(1
x
)
1
1/
(1
x
)
1/
(1
x
).
dx
(1
x
)
dx
(1
x
)
for 0
a
1.
Thus 0
dx
(1
x
)
increases as
a
1
, and is bounded above by 2.
Now
I
(
a
)
dx
(1
x
)
exists, as in qn 64.
So
67
f
1
...
(
1)
/
N
. This series is convergent by the
alternating series test (qn 5.62) to log 2, by qn 9.41.
f
1
...
1/
N
. This harmonic series is divergent, qn
5.30.
68
1
r
x
dx
1
r
.
Integral
2
(
n
1)
2, using the Fundamental Theorem.
2
(
n
1)
2
1
r
2
(
n
1)
1
1
(
n
1)
.
Now dividethrough by
n
and obtain thelimit 2 on both sids as
n
.
See qn 5.59.
