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Aproximael valorde laintegral de lasiguiente funciónparan=4 utilizandoel métodode Romberg
∫ (𝑥)4
1
−1 dx
h1 = b- a = 1+1 = 2
h2 = b-a / 2 = 1+1 /2 = 1
h3= b-a/4=1+1/4 = 0.5
h4= b-a/8= 1+1/8 =0.25
I (h1) = 1+1 /2 [(−1)4 +(1)4] = 2
I (h2) = 1+1 /4 [(−1)4 + 2(0)4 + (1)4] = 1
I(h3) = 1+1 / 8 [(−1)4+ 2(−0.5)4 + 2(0)4 + 2(0.5)4 + (1)4]= 0.5625
I(h4) = 1+1 / 16 [(−1)4 + 2(−0.75)4 + 2(−0.5)4 +
2(−0.25)4+2(0)4+2(0.25)4+2(0.5)4+2(0.75)4 + (1)4]= 0.44140625
Primernivel
Trapecio
Segundonivel Tercer nivel Cuarto nivel
2
(4/3 (1) )– (1/3(2))=
0.6666666666666666
1 (16/15 (0.4166666666666667) )–
(1/15(0.6666666666666666))=
0.4
(4/3 (0.5625) )–(1/3(1))=
0.4166666666666667
(64/63 (0.39999999999999997) )–(1/63(0.4))=
0.39999999999999997
0.5625 (16/15 (0.40104166666666663) )–
(1/15(0.4166666666666667))=
0.39999999999999997
(4/3 (0.44140625))–(1/3(0.5625))=
0.40104166666666663
0.44140625
Valoraproximadode laintegral =0.39999999999999997
Valorreal de la integral = 2/5 = 0.4

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133467 p3a4

  • 1. Aproximael valorde laintegral de lasiguiente funciónparan=4 utilizandoel métodode Romberg ∫ (𝑥)4 1 −1 dx h1 = b- a = 1+1 = 2 h2 = b-a / 2 = 1+1 /2 = 1 h3= b-a/4=1+1/4 = 0.5 h4= b-a/8= 1+1/8 =0.25 I (h1) = 1+1 /2 [(−1)4 +(1)4] = 2 I (h2) = 1+1 /4 [(−1)4 + 2(0)4 + (1)4] = 1 I(h3) = 1+1 / 8 [(−1)4+ 2(−0.5)4 + 2(0)4 + 2(0.5)4 + (1)4]= 0.5625 I(h4) = 1+1 / 16 [(−1)4 + 2(−0.75)4 + 2(−0.5)4 + 2(−0.25)4+2(0)4+2(0.25)4+2(0.5)4+2(0.75)4 + (1)4]= 0.44140625 Primernivel Trapecio Segundonivel Tercer nivel Cuarto nivel 2 (4/3 (1) )– (1/3(2))= 0.6666666666666666 1 (16/15 (0.4166666666666667) )– (1/15(0.6666666666666666))= 0.4 (4/3 (0.5625) )–(1/3(1))= 0.4166666666666667 (64/63 (0.39999999999999997) )–(1/63(0.4))= 0.39999999999999997 0.5625 (16/15 (0.40104166666666663) )– (1/15(0.4166666666666667))= 0.39999999999999997 (4/3 (0.44140625))–(1/3(0.5625))= 0.40104166666666663 0.44140625 Valoraproximadode laintegral =0.39999999999999997 Valorreal de la integral = 2/5 = 0.4