Integración númerica con regla de Simpson.
γ
=
∫
a
b
f
(
x
)
d
x
≈
∫
a
b
f
2
(
x
)
d
x
{\displaystyle \gamma =\int _{a}^{b}f(x)\,dx\approx \int _{a}^{b}f_{2}(x)\,dx}
donde la función
f
(
x
)
{\displaystyle f(x)}
f
2
(
x
)
=
(
x
−
x
1
)
(
x
−
x
2
)
(
x
0
−
x
1
)
(
x
0
−
x
2
)
f
(
x
0
)
+
(
x
−
x
0
)
(
x
−
x
2
)
(
x
1
−
x
0
)
(
x
1
−
x
2
)
f
(
x
1
)
+
(
x
−
x
0
)
(
x
−
x
1
)
(
x
2
−
x
0
)
(
x
2
−
x
1
)
f
(
x
2
)
{\displaystyle f_{2}(x)={\frac {(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}}f(x_{0})+{\frac {(x-x_{0})(x-x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2})}}f(x_{1})+{\frac {(x-x_{0})(x-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}}f(x_{2})}
con
x
0
=
a
,
x
1
=
a
+
b
2
,
x
2
=
b
{\displaystyle x_{0}=a,x_{1}={\frac {a+b}{2}},x_{2}=b}
Ejercicio [ ] Mostrar que:
f
2
(
x
0
)
=
f
(
x
0
)
{\displaystyle f_{2}(x_{0})=f(x_{0})}
f
2
(
x
1
)
=
f
(
x
1
)
{\displaystyle f_{2}(x_{1})=f(x_{1})}
f
2
(
x
2
)
=
f
(
x
2
)
{\displaystyle f_{2}(x_{2})=f(x_{2})}
Que
f
2
{\displaystyle f_2}
f
2
(
x
)
=
A
x
2
+
B
x
+
C
{\displaystyle f_{2}(x)=Ax^{2}+Bx+C}
f
2
m
(
x
)
=
0
{\displaystyle f_{2}^{m}(x)=0}
Calcular
∫
a
b
f
2
(
x
)
d
x
{\displaystyle \int _{a}^{b}f_{2}(x)\,dx}
x
0
=
0
,
x
1
=
1
/
2
,
x
2
=
1
{\displaystyle x_0=0, x_1=1/2, x_2=1}
Con
x
=
x
0
{\displaystyle x = x_0}
f
2
(
x
0
)
=
(
x
0
−
x
1
)
(
x
0
−
x
2
)
(
x
0
−
x
1
)
(
x
0
−
x
2
)
1
f
(
x
0
)
+
(
x
0
−
x
0
)
(
x
0
−
x
2
)
(
x
1
−
x
0
)
(
x
1
−
x
2
)
f
(
x
1
)
+
(
x
0
−
x
0
)
(
x
0
−
x
1
)
(
x
2
−
x
0
)
(
x
2
−
x
1
)
f
(
x
2
)
{\displaystyle f_{2}(x_{0})={{\cancelto {1}{\frac {(x_{0}-x_{1})(x_{0}-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}}}f(x_{0})}+{{\cancel {\frac {(x_{0}-x_{0})(x_{0}-x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2})}}}f(x_{1})}+{{\cancel {\frac {(x_{0}-x_{0})(x_{0}-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}}}f(x_{2})}}
∴
f
2
(
x
0
)
=
f
(
x
0
)
{\displaystyle \therefore f_{2}(x_{0})=f(x_{0})}
Con
x
=
x
1
{\displaystyle x=x_{1}}
f
2
(
x
1
)
=
(
x
1
−
x
1
)
(
x
1
−
x
2
)
(
x
1
−
x
1
)
(
x
1
−
x
2
)
f
(
x
0
)
+
(
x
1
−
x
0
)
(
x
1
−
x
2
)
(
x
1
−
x
0
)
(
x
1
−
x
2
)
1
f
(
x
1
)
+
(
x
1
−
x
0
)
(
x
1
−
x
1
)
(
x
1
−
x
0
)
(
x
1
−
x
1
)
f
(
x
2
)
{\displaystyle f_{2}(x_{1})={{\cancel {\frac {(x_{1}-x_{1})(x_{1}-x_{2})}{(x_{1}-x_{1})(x_{1}-x_{2})}}}f(x_{0})}+{{\cancelto {1}{\frac {(x_{1}-x_{0})(x_{1}-x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2})}}}f(x_{1})}+{{\cancel {\frac {(x_{1}-x_{0})(x_{1}-x_{1})}{(x_{1}-x_{0})(x_{1}-x_{1})}}}f(x_{2})}}
∴
f
2
(
x
1
)
=
f
(
x
1
)
{\displaystyle \therefore f_{2}(x_{1})=f(x_{1})}
Con
x
=
x
2
{\displaystyle x = x_2}
f
2
(
x
2
)
=
(
x
2
−
x
1
)
(
x
2
−
x
2
)
(
x
2
−
x
1
)
(
x
2
−
x
2
)
f
(
x
0
)
+
(
x
2
−
x
0
)
(
x
2
−
x
2
)
(
x
2
−
x
0
)
(
x
2
−
x
2
)
f
(
x
1
)
+
(
x
2
−
x
0
)
(
x
2
−
x
1
)
(
x
2
−
x
0
)
(
x
2
−
x
1
)
1
f
(
x
2
)
{\displaystyle f_{2}(x_{2})={{\cancel {\frac {(x_{2}-x_{1})(x_{2}-x_{2})}{(x_{2}-x_{1})(x_{2}-x_{2})}}}f(x_{0})}+{{\cancel {\frac {(x_{2}-x_{0})(x_{2}-x_{2})}{(x_{2}-x_{0})(x_{2}-x_{2})}}}f(x_{1})}+{{\cancelto {1}{\frac {(x_{2}-x_{0})(x_{2}-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}}}f(x_{2})}}
∴
f
2
(
x
1
)
=
f
(
x
1
)
{\displaystyle \therefore f_{2}(x_{1})=f(x_{1})}
