HP 40gs HP 39gs_40gs_Mastering The Graphing Calculator_English_E_F2224-90010.p - Page 73

Integration: The algebraic indefinite integral

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Integration: The algebraic indefinite integral Algebraic integration is also possible (for simple functions), in the following fashions: • If done in the SYMB view of the Function aplet, then the integration must be done using the symbolic variable S1 (or S2, S3, S4 or S5). If done in this manner then the results are satisfactory, except that there is no constant of integration 'c'. The screenshot right shows the results of defining ( ) ∫ F1( X ) = X 2 −1 and then F 2(X ) = 0, S1, X 2 −1, X , together with the results of the same thing after pressing the key (the result is placed in F3(X) only for convenience of viewing). All that is now necessary is to read '-S1+S1^3/3' as −x + x3 , 3 or as it should be read as x3 − x + c . 3 • If done in the HOME view, then S1 must again be used as the variable of integration. ∫ i.e. x2 −1 dx is entered as ∫ ( 1, S1, X2 - 1, X ). This is shown above, together with the results of highlighting the answer and pressing . The result may seem odd but is caused by calculator assuming that X itself may be a function of some other variable and integrating accordingly as a 'partial integration'. While mathematically correct, this is not what most of us want. The way to simplify this answer to a better form is to highlight it, it, and press ENTER again, giving the result shown right. This is the result of the calculator performing the substitutions implied in the previous expression. 73

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Integration: The algebraic indefinite integral
Algebraic integration is also possible (for simple functions), in the following fashions:
If done in the
SYMB
view of the Function aplet, then the integration must be done using the symbolic
variable
S1
(or
S2
,
S3
,
S4
or
S5
). If done in this manner then the results are satisfactory, except that
there is no constant of integration ‘c’.
The screenshot right shows the results of defining
1(
)
=
S
X
2
1,
X
)
, together
F
X
X
2
1
and then
F
2(
X
)
=
(
0, 1,
with the results of the same thing after pressing the
key
(the result is placed in
F3(X)
only for convenience of viewing).
3
All that is now necessary is to read ‘
-S1+S1^3/3
’ as
+
x
,
x
3
3
or as it should be read as
x
+
.
x
c
3
If done in the
HOME
view, then
S1
must again be used
as the variable of integration.
2
i.e.
x
1
dx
is entered as
( 1, S1, X
2
- 1, X )
.
This is shown above, together with the results of highlighting the answer and pressing
. The
result may seem odd but is caused by calculator assuming that
X
itself may be a function of some
other variable and integrating accordingly as a ‘partial integration’.
While mathematically correct, this is not what most of us want.
The way to simplify this answer to a better form is to highlight it,
it, and press
ENTER
again, giving the result shown
right. This is the result of the calculator performing the
substitutions implied in the previous expression.
73