Texas Instruments TI-89 Tip List page 267

substitute these expressions in the function definition
and solve for y
s
then take the derivative
or
Now since
so
Replace [2A] in [1A]
or
or simply
which is the relation we wanted. In this splice application, we scale only x, not y, so r = 1. In terms of
the actual scaling variables x
so by inspection
then
Scaling the splice integral
Given the scaled splice polynomial s
¶
x sb
I s = s s (x s )d xs
x sa
but we really want the integral of the unscaled function:
¶
x b
I =
s ( x ) dx
x a
Using the definition of the definite integral:
x s −q
dy s
= r $ d
f
p
dx s dx s
dy s
= r p $ d
f ( x s − q )
dx s dx s
x s = p $ x + q
x s + dx s = p $ x + p $ dx + q = p $ x + q + p $ dx
x s + dx s = x s + p $ dx
dx s = p $ dx
dy s
d
= r p $
f ( x s − q )
dx s p$dx
x s −q
dy s
= r p $ d
f
p
dx s dx
dy s
= r p $ d
f ( x )
dx s dx
and h, the scaling is
2
xs = 1 $ ( x − x2 ) = 1
x −
h h
p = 1
h
dy s
= h $ d
f ( x )
dx s dx
(x ), we can easily find the definite integral
s s
y s −s x s −q
= f
r p
x s −q
y s = r $ f
p
x 2
h
∞
✟
Is = lim
s s ( x s ) $ ✁x s m
nd∞
m=1
+ s
[1A]
[2A]
[1B]
6 - 109