Mathematics">
Soluciòn Ejercicio 5 - Análisis Numérico
Soluciòn Ejercicio 5 - Análisis Numérico
Soluciòn Ejercicio 5 - Análisis Numérico
Solución
a) De forma analytical
∫ ( 1−e−2 x ) dx
1
4
= ∫¿¿
1
= x
4
4−1=3 - ∫ (−e−2 x ¿ ) dx ¿
1
4
eu
= −∫ ( )du
1 2
e−2 x
-
2
4
∫ ( 1−e−2 x ) dx=¿2.932
1
f ( a )−f (b)
I =( b−a)
2
∫ ( 1−e−2 x ) dx
1
( 0.864 ) +(0.999)
I =( 4−1)
2
1.863
I =( 3)
2
I =2.794
|2.79−2.932|=−0.14
Para n= 2
n−1
∑ f ( x 1 ) +¿ f (x n)
I =( b−a) f ( x0 ) + 2 i=0 ¿
2
3
h¿
2
n−1
∑ f ( x1 ) +¿ f (x n)
I =( 4−1)0.864 +2 i=0 ¿
2n
0.864+0.981+0.999
I =( 3)
4
I =2.8 33
Para n = 4
n−1
∑ f ( x 1 ) +¿ f (x n)
I =( b−a) f ( x0 ) + 2 i=0 ¿
2
3
h¿
4
Imagines de la function
∫ ( 1−e−2 x ) dx
1
f ( 1 ) =0.864 ; f ( 1.75 ) =0.969 ; f ( 2.5 )=¿ 0.993 ; f ( 3.25 )=0.998 f ( x 4 )=¿ 0.999
4−1
∑ f ( x1 ) +¿ 0.999
I =( 3)0.864+ 2 i=1 ¿
2
7.783
I =( 3)
8
I =2.9186
|2.918−2.932|=−0.014
f ( x0)+ 4 f ( xi )+ f ( xn)
I =( b−a)
6
I =2.893
|2.893−2.932|=−0.0385
(b−a)
f ( x0 ) + 4 f ( x1 ) +2 f ( x 2 ) + f (x n)
I=
n
3
∗ [ 1 ]
3
h¿
4
f ( 1 ) =0.864 ; f ( 1.75 ) =0.969 ; f ( 2.5 )=¿ 0.993 ; f ( 3.25 )=0.998 f ( x 4 )=¿ 0.999
|2.932−2.932|=0
3 h f ( x 0 ) +3 f ( x 1 ) +3 f ( x 2 ) + f ( x 3)
I=
8
∗ [ 1 ]
b−a
h=
3
4−1
h=
3
h=1
3 0.864+ 3 ( 0.981 ) +3 ( 0.997 )+ 0.999
I= ∗
8 [ 1 ]
I =2.923
|2.923−2.932|=−0.00825
3 h f ( x 0 ) +3 f ( x 1 ) +3 f ( x 2 ) + f ( x 3)
I=
8
∗ [ 1 ]
b−a
h=
5
4−1
h=
5
3
h=
5
f ( 1 ) =0.864 ; f ( 1.6 )=0.959 ; f ( 2.2 ) =¿ 0.987 ; f ( 2.8 )=0.996 ; f ( 3 . 4 )=¿ 0.998 ; f ( 4 ) =¿0.999
3
3
5 f ( 0.864 )+ 3 ( 0.959 )+ 3 ( 0.987 ) +3 ( 0.9 9 6 ) +3 ( 0.998 ) +0.999
I=
8
∗ [ 1 ]
|2.871−2.932|=−0 .061
2.932−2.794
Incise b ∗100 %=0.047
2.932
2.932−2.833
Incise c ∗100 %=0.033
2.932
2.932−2.898
Incise d ∗100 %=0.011
2.932
2.932−2.932
Incise e ∗100 %=0
2.932
2.932−2.923
Incise f ∗100 %=0.00306
2.932
2.932−2.871
Incise g ∗100 %=0.020
2.932