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8. INTEGRASI NUMERIK ( Lanjutan )

8. INTEGRASI NUMERIK ( Lanjutan ). 8.8 Metode Newton-Cotes Bentuk umum dari metode Newton-Cotes ditunjukkan pada persamaan berikut. (8.9). n = jumlah pias ( strip ) h = lebar pias = ( b – a )/ n f i = f ( x i ) x i = a + ih α = koefisien

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8. INTEGRASI NUMERIK ( Lanjutan )

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  1. 8. INTEGRASI NUMERIK (Lanjutan)

  2. 8.8 Metode Newton-Cotes Bentukumumdarimetode Newton-Cotes ditunjukkanpadapersamaanberikut. (8.9) n = jumlahpias (strip) h = lebarpias = (b – a)/n fi = f (xi) xi= a + ih α = koefisien β = koefisien E = Galat

  3. Tabel 8.1 Rumus Newton-Cotes

  4. Tabel 8.1 Rumus Newton-Cotes (lanjutan) Tabel 8.1 adalahtujuhdari 10 rumus Newton-Cotes. Rumus 1 sampai 4 masing-masingdidapatdenganaturantrapesium, Simpson 1/3, Simpson 3/8, dan Boole. Rumus 5 danseterusnyadidapatdenganmenggunakanpolinominterpolasiselisihmajuderajat 4, 5, danseterusnya.

  5. Contoh 8.6 Turunkanrumus Newton-Cotes derajat 4 Penyelesaian

  6. = 4f0 + 8 f0 + 20/3 2f0 + 8/3 3f0 +42/135 4f0

  7. f0= f1 – f0 2f0 = f1 – f0=(f2 – f1) – (f1 – f0) = f2 – 2f1+ f0 3f0 = 2f1– 2f0= (f3 – 2f2 + f1) – (f2 – 2f1 + f0) = f3 – 3f2 + 3f1 – f0 4f0 = 3f1– 3f0= (f4 – 3f3 + 3f2 – f1) – (f3 – 3f2 + 3f1 – f0) = f4 – 4f3 + 6f2 – 4f1 + f0

  8. I = 4f0 + 8 f0 + 20/3 2f0 + 8/3 3f0 + 42/135 4f0 = 4 f0 + 8( f1 – f0) + 20/3(f2 – 2f1+ f0)+8/3(f3 –3f2 + 3f1 – f0) +42/135 (f4 – 4f3 + 6f2 – 4f1 + f0) = 4 f0 + 8 f1 – 8 f0 + 20/3 f2 – 40/3f1+ 20/3f0 + 8/3f3 – 8f2 + 8f1 – 8/3f0 + 42/135 f4 –168/135f3 + 252/135f2 –168/135f1 + 42/135f0 = 42/135 f0 +192/135 f1 + 72/135 f2 + 192/135 f3 + 42/135 f4 = 14/45 f0 +64/45 f1 + 24/45 f2 + 64/45 f3 + 14/45 f4 = 28/90 f0 + 128/90 f1 + 48/90 f2 + 128/90 f3 + 28/90 f4 = 4(1/90)(7f0 + 32 f1 + 12/90 f2 + 32/90 f3 + 7/90 f4) n = 4,  = 1/90, α0 = 7, α1 = 32, α2 = 12, α3 = 32, α4 = 7

  9. 8.9 KuadraturGauss Hasilintegrasisejatif (x) darititikasampaititikb adalah (8.10) HasilintegrasisejatiditunjukkanpadaGambar 8.7a. Jikaintegrasidiselesaikandenganmenggunakanmetodetrapesium, maka (8.11) seperti yang ditunjukkanpadaGambar 8.7b. GalatpadametodetrapesiumdapatdiperkecildenganmenggunakanmetodeKuadraturGauss.

  10. f (x) f (x) x x a b a b (b) (a) Gambar 8.7

  11. Jikakaidahtrapesiumditerapkanpadafungsikonstanataufungsi linier, danditulisdalambentukkoefisientaktentu, makaakanmenghasilkannilaisejatidalambentuk Persamaan (8.12) adalahrumusGauss-Legendre 2 titik. Persamaan (8.12) dapatdigeneralisirmenjadirumus Gauss-Legendre untukn titik. (8.12) (8.13)

  12. Gabungkan pers. (810) dengan (8.13) didapat Selanjutnyalakukantransformasibidangxkebidangt. Dari Gambar 8.8 x = a t = –1 x = b  t = 1 Didapat (8.14) (8.15) (8.16)

  13. f (x) F (t) t x –1 t1t21 a x1x2 b (a) (b) Gambar 8.8 Transformasibidangxkebidangt

  14. Dari persamaan (8.18), didapatf (x) = f (mt + c) DefinisikanF(t) = f (mt + c) = f (x) (8.20) (8.17) Substitusi pers. (8.15) – (8.20) ke pers. (8.14), didapat Jika makapersamaan (8.16) menjadi (8.18) (8.21) dandx= m dt (8.19)

  15. Dari persamaan (8.21) didapat Berikutakanditentukannilai (8.22) untuk n = 2 dannilaiF(t) = 1, t, t2 , dant3 (8.23a)

  16. (8.23b) (8.23c) (8.23d)

  17. Persamaan (8.23 a) sampai (8.23d) DidapatC1 = C2 = 1

  18. Contoh 8.7 Diketahuf(x) = 1/x, batasbawah = 3,1, batasatas 3,9. Tentukanhampiranintegrasif(x) denganmenggunakankuadratur Gauss duatitikdengansatu interval. Penyelesaian f(x) = 1/x ; a = 3,1; b = 3,9 ; n = 2

  19. x = mt + c = 0,4 t + 3,5  dx = m dt= 0,4 dt Dari persamaan (8.21)

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