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A caveat...
This substitution process has one implication which you need to be wary of
and so it is worth examining the process in more detail...
S
1
x
∫
−
=
2
x
1
dx
0
=
S
=
3
The potential problem lies with the second line, where the substitution of zero
results in the second bracket disappearing. This will not always happen.
For example...
S
1
(
)
∫
−
4
=
x
2
dx
0
=
(
S
=
The final constant of 6.4 comes from substituting zero into the expression
and should not be there if we were doing this with the aim of finding an
indefinite integral. On the other hand, we all know that the answer should
have a constant of integration, so perhaps this extra constant will help you to
remember the '+c'!
In addition to this, there are strict limits to what the hp 39g+ can integrate.
For example, if you try to evaluate
not be able to do it. Essentially, beyond polynomials forget it.
The hp 40g Computer Algebra System
Owners of the hp 40g will be aware that it has a CAS which allows it to
perform the most complex algebra, including complex integrations, with ease.
More information on the CAS can be found in the supplementary appendix on
the hp 40g. See the Table of Contents.
=
x S
1
3
−
x
3
=
x
0
3
3
S
1
0
−
−
−
S
1
0
3
3
3
1
−
S
1
=
x S
1
(
)
−
5
x
2
5
=
x
0
(
)
(
)
−
5
−
5
S
1 2
0 2
−
5
5
)
−
5
1 2
− ⋅
6 4
5
∫
sin
using the calculator, it will
2
x
.cos
x dx
82
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