HP 15c User Manual
Page 50
50
Section 2: Working with
f
50
t
u
t
du
u
e
dx
x
f
1
tan
0
2
tan
0
)
tan
1
(
)
(
2
,
which is calculated readily even with t as large as 10
10
. Using the same substitution with g(x),
values near a = 0 and b = 10
−5
provide good results.
This example involves subdividing the interval of integration. Although a function may have
features that look extreme over the entire interval of integration, over portions of that interval
the function may look more well-behaved. Subdividing the interval of integration works best
when combined with appropriate substitutions. Consider the integral
.
)
1
/(
)
1
(
1
)
1
/(
)
(
1
)
1
/(
)
1
(
)
1
/(
)
1
/(
)
1
/(
)
1
/(
)
1
/(
1
0
8
8
/
55
4
/
1
8
1
1
0
64
64
62
1
0
64
62
1
0
64
62
1
0
64
1
64
1
0
64
0
64
v
dv
v
v
x
dx
x
x
x
dx
x
u
du
u
x
dx
x
dx
x
dx
x
dx
These steps use the substitutions x = 1/ u and x = v
1/8
and some algebraic manipulation.
Although the original integral is improper, the last integral is easily handled by f. In fact,
by separating the constant term from the integral, you obtain (using i8) an answer with
13 significant digits:
1.000401708155 ± 1.2 × 10
−12
.
A final example drawn from real life involves the electrostatic field about an ellipsoidal
probe with principal semiaxes a, b, and c:
0
2
2
2
2
)
)(
)(
(
)
(
x
c
x
b
x
a
x
a
dx
V
for a=100, b =2, and c= 1.
§
Transform this improper integral to a proper one by substituting x = (a
2
− c
2
)/(1 − u
2
) − a
2
:
1
2
2
)
/(
)
1
(
r
du
q
u
u
p
V
where
6
2
2
2
2
10
00060018
.
2
)
)
((
/
2
b
a
c
a
p
3
2
2
2
2
10
001200480
.
3
)
(
/
)
(
b
a
c
b
q
§
From Stratton, J.A., Electromagnetic Theory, McGraw-Hill, New York, 1941, pp.201-217.