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8. INTEGRASI NUMERIK

8. INTEGRASI NUMERIK. 8.4 Aturan Titik Tengah Gambar berikut adalah sebuah persegi panjang dari x = x 0 sampai x = x 1 dan titik tengah x = x 1/2 = x 0 + h /2. y. f ( x ). h. x. a=x 0 x 1/2 b=x 1. O. Gambar 8.4 Aturan titik tengah. y. …. Gambar 8.3

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8. INTEGRASI NUMERIK

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  1. 8. INTEGRASI NUMERIK

  2. 8.4 AturanTitik Tengah Gambarberikutadalahsebuahpersegipanjangdari x = x0sampaix = x1dantitiktengahx = x1/2 = x0 + h/2 y f (x) h x a=x0x1/2 b=x1 O Gambar 8.4 Aturantitiktengah

  3. y … Gambar 8.3 nbuahpersegipanjangdenganpanjangmasing-masingf (xn+h/2) x O a=x0x1x2 … xn-1 b=xn n = (xn – x0)/h (8.7)

  4. Luasn buahtitiktengahadalah (8.8) Persamaan (8.8) adalahhampiranintegrasif(x) denganmetodetitiktengah.

  5. Contoh 8.3 Dari tabelberikut, gunakanmetodetitiktengahdengann = 8 untukmengevaluasi integral

  6. Penyelesaian n = 8 ; xn = x8 = 2,0 ; x0 = 0,4 Dari persamaan (8.5) didapat h = (xn – x0)/n = (2,0 – 0,4) / 8 = 0,2

  7. Dari tabeldidapat: f (x0+h/2) = f (0,4+0,1) = f (0,5) = 4,4266 f (x1+h/2) = f (0,6+0,1) = f (0,7) = 3,4166 f (x2+h/2) = f (0,8+0,1) = f (0,9) = 3,0700 f (x3+h/2) = f (1,0+0,1) = f (1,1) = 3,0534 f (x4+h/2) = f (1,2+0,1) = f (1,3) = 3,2476 f (x5+h/2) = f (1,4+0,1) = f (1,5) = 3,5993 f (x6+h/2) = f (1,6+0,1) = f (1,7) = 4,0806 f (x7+h/2) = f (1,8+0,1) = f (1,9) = 4,6757

  8. 8.5 Aturan Simpson 1/3 Aturansimpson 1/3 adalahaturan yang mencocokkanpolinomialderajat 2 padatigatitik data diskrit yang mempunyaijarak yang sama. p2(x) y f (x) x x0 = 0 x1 = hx2 = 2h Gambar 8.4 Aturan Simpson 1/3

  9. Dari persamaan (8.5) s = (x – x0)/h ataux = x0 + sh Sehingga Untukx = x0, makas = 0 Untukx = x2, makas = 2

  10. ataudapatditulisdalambentuk (8.9)

  11. Contoh 8.4 Selesaikan denganmenggunakanmetode: Trapesium Titiktengah Simpson 1/3 Bandingkanhasilmasing-masingmetodedengansolusisejatinya. Penyelesaian h = 0,10

  12. a. MetodeTrapesium n = (xn – x0)/h = (1 – 0) / 0,1 = 10

  13. I  0,05(0,5000 + 1,0476 + 1,0909 + 1,1304 + 1,1666 + 1,2000 + 1,2308 + 1,2592 + 1,2856 + 1,3104+0,6666  0,59438

  14. b. Metodetitiktengah: f (x0+h/2) = f (0,0+0,05) = f (0,05) = 0,5122 f (x1+h/2) = f (0,1+0,05) = f (0,15) = 0,5348 f (x2+h/2) = f (0,2+0,05) = f (0,25) = 0,5556 f (x3+h/2) = f (0,3+0,05) = f (0,35) = 0,5744 f (x4+h/2) = f (0,4+0,05) = f (0,45) = 0,5918 f (x5+h/2) = f (0,5+0,05) = f (0,55) = 0,6078 f (x6+h/2) = f (0,6+0,05) = f (0,65) = 0,6226 f (x7+h/2) = f (0,7+0,05) = f (0,75) = 0,6364 f (x8+h/2) = f (0,8+0,05) = f (0,85) = 0,6491 f (x9+h/2) = f (0,9+0,05) = f (0,95) = 0,6610

  15. c. Metode Simpson 1/3 n = (xn – x0)/h = (1 – 0) / 0,1 = 10

  16. Solusisejati

  17. 8.6 Aturan Simpson 3/8 Aturansimpson 3/8 adalahaturan yang mencocokkanpolinomialderajat 3 padaempattitik data diskrit yang mempunyaijarak yang sama. y f (x) pn(x) x1 = h x2 = 2h x3 =3h x0 = 0 x Gambar 8.5 Aturan Simpson 3/8

  18. Dari persamaan (8.5) s = (x – x0)/h ataux = x0 + sh Sehingga Untukx = x0, makas = 0 Untukx = x2, makas = 2

  19. ataudapatditulisdalambentuk Selesaikandenganmetode Simpson 3/8 Contoh 8.5 (8.9) dengann = 9. Bandingkanhasilnyadenganmetodetrapesium, titiktengah, Simpson 1/3 dandengansolusisejati. Penyelesaian

  20. Untukn = 9  h = ( 1 – 0)/9 = 1/9

  21. Itotal 3h/8{ f0 + 3( f1 + f2 + f4 + f5 + f7 + f8 )+ 2(f3 + f6 ) + f6 }  (3/8)(1/9){ 1,16666 + 10,70931+ 2,39286}  0,59453

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