Lecture 4 - Computer Science Department - University of Kentucky
Lecture 4 - Computer Science Department - University of Kentucky
Lecture 4 - Computer Science Department - University of Kentucky
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Similarly, we have<br />
A<br />
An Example<br />
2 2<br />
2<br />
2<br />
1<br />
3<br />
( )<br />
(1 )<br />
⎤<br />
1 2 1<br />
2 ⎢<br />
⎡ = ∫ l x dx = − = − =<br />
3 ⎥<br />
⎣ ⎦<br />
−<br />
− ∫ x dx x x<br />
−<br />
−2<br />
2<br />
=<br />
2<br />
2 1 2<br />
1 ⎡1<br />
3 1 2 ⎤<br />
A = ∫ l<br />
( x<br />
) dx =<br />
+ 1) =<br />
x(<br />
x<br />
dx<br />
x + x<br />
2 2 2<br />
2<br />
2<br />
2<br />
⎢<br />
⎣<br />
3<br />
2<br />
⎥ − ∫ −<br />
⎦<br />
So the quadrature formula defined on the interval [‐2,2] and using the<br />
node ‐1, 0, 1, is<br />
2<br />
∫ − 2<br />
8<br />
4<br />
8<br />
f ( x)<br />
dx≈<br />
f ( −1)<br />
− f (0) +<br />
3 3 3<br />
f (1)<br />
It can be verified that this formula gives exact values for the three<br />
functions<br />
2<br />
f ( x)<br />
= 1, x,<br />
x<br />
−2<br />
8<br />
3<br />
4<br />
3<br />
50