Metode Trapezoida

• Terlebih dahulu melakukan substitusi nilai \(a\) dan \(b\). \[ f(x|a,b) = abx^{a-1}(1-x^{(a)})^{b-1} \] \[ f(x|5,3) = (5) \cdot (3) \cdot (x)^{5-1}(1-x^{(5)})^{3-1} \] \[ f(x|a,b) = 15x^{4}(1-x^{(5)})^{2} \]

• Menghitung panjang setiap sub interval Diketahui bahwa \(x \in (0,1)\) dan \(n=4\), maka nilai \(h\) adalah: \[h = \frac{b-a}{n}\] \[h = \frac{1-0}{4} = \frac{1}{4}\]

• Selanjutnya, menghitung nilai \(E(X)\) pada masing-masing nilai \(X_i\) \[ E(X) = x \cdot f(x|a,b) \] \[ E(X) = x \cdot 15x^{4}(1-x^{(5)})^{2} \] \[ E(X) = 15x^{5}(1-x^{(5)})^{2} \] Untuk \(X_0\): \[ E(X_0) = E(a+ih) = E(0+(0)\left(\frac{1}{4}\right) = E(0) \] \[ E(0) = 15 (0)^{5}(1-(0)^{(5)})^{2} = 15(0)(1) = 0 \] Untuk \(X_1\): \[ E(X_1) = E(a+ih) = E(0+(1)\left(\frac{1}{4}\right) = E\left(\frac{1}{4}\right) \] \[ E\left(\frac{1}{4}\right) = 15\left(\frac{1}{4}\right) ^{5}\left(1-\left(\frac{1}{4}\right)^{(5)}\right)^{2} = 15(0.000977)(0.998048) = 0.01462 \] Untuk \(X_2\): \[ E(X_2) = E(a+ih) = E(0+(2)\left(\frac{1}{4}\right) = E\left(\frac{1}{2}\right) \] \[ E\left(\frac{1}{2}\right) = 15\left(\frac{1}{2}\right) ^{5}\left(1-\left(\frac{1}{2}\right)^{(5)}\right)^{2} = 15(0.03125)(0.938477) = 0.439911 \] Untuk \(X_3\): \[ E(X_3) = E(a+ih) = E(0+(3)\left(\frac{1}{4}\right) = E\left(\frac{3}{4}\right) \] \[ E\left(\frac{3}{4}\right) = 15\left(\frac{3}{4}\right) ^{5}\left(1-\left(\frac{3}{4}\right)^{(5)}\right)^{2} = 15(0.237305)(0.581704) = 2.070617 \] Untuk \(X_4\): \[ E(X_4) = E(a+ih) = E(0+(4)\left(\frac{1}{4}\right) = E(1) \] \[ E(1) = 15(1) ^{5}\left(1-(1)^{(5)}\right)^{2} = 15(1)(0) = 0 \]

• Bentuk Trapezoida untuk \(n=4\) \[T_4 = \left( \frac{h}{2} \right) \left[ E(X_0) + 2 E(X_1) + 2 E(X_2) + 2 E(X_3) + E(X_4) \right]\] \[T_4 = \left( \frac{\frac{1}{4}}{2} \right) \left[ 0 + 2 (0.01462) + 2 (0.439911) + 2 (2.070617) + 0 \right]\] \[T_4 = \left( \frac{1}{8} \right) (5.050295)\] \[T_4 = 0.631287\] Jadi, \[ \int_{0}^{1} 15x^{5}(1-x^{(5)})^{2} = 0.631287 \] ### Metode Simpson

• Terlebih dahulu melakukan substitusi nilai \(a\) dan \(b\). \[ f(x|a,b) = abx^{a-1}(1-x^{(a)})^{b-1} \] \[ f(x|5,3) = (5) \cdot (3) \cdot (x)^{5-1}(1-x^{(5)})^{3-1} \] \[ f(x|a,b) = 15x^{4}(1-x^{(5)})^{2} \]

• Menghitung panjang setiap sub interval Diketahui bahwa \(x \in (0,1)\) dan \(n=4\), maka nilai \(h\) adalah: \[h = \frac{b-a}{n}\] \[h = \frac{1-0}{4} = \frac{1}{4}\]

• Selanjutnya, menghitung nilai \(E(X)\) pada masing-masing nilai \(X_i\) \[ E(X) = x \cdot f(x|a,b) \] \[ E(X) = x \cdot 15x^{4}(1-x^{(5)})^{2} \] \[ E(X) = 15x^{5}(1-x^{(5)})^{2} \] Untuk \(X_0\): \[ E(X_0) = E(a+ih) = E(0+(0)\left(\frac{1}{4}\right) = E(0) \] \[ E(0) = 15 (0)^{5}(1-(0)^{(5)})^{2} = 15(0)(1) = 0 \] Untuk \(X_1\): \[ E(X_1) = E(a+ih) = E(0+(1)\left(\frac{1}{4}\right) = E\left(\frac{1}{4}\right) \] \[ E\left(\frac{1}{4}\right) = 15\left(\frac{1}{4}\right) ^{5}\left(1-\left(\frac{1}{4}\right)^{(5)}\right)^{2} = 15(0.000977)(0.998048) = 0.01462 \] Untuk \(X_2\): \[ E(X_2) = E(a+ih) = E(0+(2)\left(\frac{1}{4}\right) = E\left(\frac{1}{2}\right) \] \[ E\left(\frac{1}{2}\right) = 15\left(\frac{1}{2}\right) ^{5}\left(1-\left(\frac{1}{2}\right)^{(5)}\right)^{2} = 15(0.03125)(0.938477) = 0.439911 \] Untuk \(X_3\): \[ E(X_3) = E(a+ih) = E(0+(3)\left(\frac{1}{4}\right) = E\left(\frac{3}{4}\right) \] \[ E\left(\frac{3}{4}\right) = 15\left(\frac{3}{4}\right) ^{5}\left(1-\left(\frac{3}{4}\right)^{(5)}\right)^{2} = 15(0.237305)(0.581704) = 2.070617 \] Untuk \(X_4\): \[ E(X_4) = E(a+ih) = E(0+(4)\left(\frac{1}{4}\right) = E(1) \] \[ E(1) = 15(1) ^{5}\left(1-(1)^{(5)}\right)^{2} = 15(1)(0) = 0 \]

• Bentuk Simpson untuk \(n=4\) \[S_4 = \left( \frac{h}{3} \right) \left[ E(X_0) + 4 E(X_1) + 2 E(X_2) + 4 E(X_3) + E(X_4) \right]\] \[S_4 = \left( \frac{\frac{1}{4}}{3} \right) \left[ 0 + 4 (0.01462) + 2 (0.439911) + 4 (2.070617) + 0 \right]\] \[S_4 = \left( \frac{1}{12} \right) (9.220768)\] \[S_4 = 0.768397\] Jadi, \[ \int_{0}^{1} 15x^{5}(1-x^{(5)})^{2} = 0.768397 \]

four-point Gauss Quadrature

• Dari kedua jawaban sebelumnya, telah diketahui fungsi \(f(x)\) yang dalam hal ini adalah nilai \(E(X)\). \[ E(X) = 15x^{5}(1-x^{(5)})^{2} \]

• Melakukan transormasi sehingga domain integral menjadi \((-1,1)\). \[ x = \frac{1}{2} [t(b-a)+a+b]\] \[ x = \frac{1}{2} [t(1-0)+0+1] \] \[ x = \frac{1}{2} [t(1)+1] \] \[ x = \frac{t+1}{2} \] dan \[ dx = \frac{1}{2} (b-a) dt \] \[ dx = \frac{1}{2} (1-0) dt \] \[ dx = \frac{1}{2} dt \] Sehingga, integralnya menjadi: \[ \int_{0}^{1} 15x^{5}(1-x^{(5)})^{2} = \int_{-1}^{1} \left(\frac{1}{2}\right) 15\left(\frac{t+1}{2}\right)^{5}\left(1-\left(\frac{t+1}{2}\right)^{(5)}\right)^{2}\]

• Nilai Bobot \(C_i\) dan titik gauss \(x_i\) untuk \(n=4\) adalah:  
  \(C_1 = 0.3478548\); \(x_1 = -0.86113631\)  
  \(C_2 = 0.6521452\); \(x_2 = -0.33998104\)  
  \(C_3 = 0.6521452\); \(x_3 = 0.33998104\)  
  \(C_4 = 0.3478548\); \(x_4 = 0.86113631\)  
  

• Selanjutnya, melakukan integrasi dengan menggunakan persamaan berikut: \[\int_{0}^{1} f(x)dt \approx C_1 \cdot f(x_1) + C_2 \cdot f(x_2) + C_3 \cdot f(x_3) + C_4 \cdot f(x_4)\] Karena dalam hal ini fungsi \(f(x)\) adalah nilai \(E(X)\), maka terlebih dahulu dihitung nilai Ekspektasi untuk setiap \(X_i\):  
  Untuk \(E(X_1)\) \[ E(-0.86113631) = \frac{15}{2} \left( \frac{-0.86113631 + 1} {2}\right)^5 \left(1 - \left(\frac{-0.86113631+1}{2} \right)^{(5)}\right)^2\] \[ = \left( \frac{15}{2} \right) (1.61359 \cdot 10^{-6}) (0.999997) = 1.21019 \cdot 10^{-5}\] Untuk \(E(X_2)\) \[ E(-0.33998104) = \frac{15}{2} \left( \frac{-0.33998104 + 1} {2}\right)^5 \left(1 - \left(\frac{-0.33998104+1}{2} \right)^{(5)}\right)^2\] \[ = \left( \frac{15}{2} \right) (0.003914101) (0.992187) = 0.029126408\] Untuk \(E(X_3)\) \[ E(0.33998104) = \frac{15}{2} \left( \frac{0.33998104 + 1} {2}\right)^5 \left(1 - \left(\frac{0.33998104+1}{2} \right)^{(5)}\right)^2\] \[ = \left( \frac{15}{2} \right) (0.135002959) (0.74822) = 0.757589236\] Untuk \(E(X_4)\) \[ E(0.86113631) = \frac{15}{2} \left( \frac{0.86113631 + 1} {2}\right)^5 \left(1 - \left(\frac{0.86113631+1}{2} \right)^{(5)}\right)^2\] \[ = \left( \frac{15}{2} \right) (0.697816015) (0.091315) = 0.477908863\] Setelah didapatkan nilai ekspektasi untuk setiap \(X_i\), maka dapat dilakukan integrasi pada persamaan sebelumnya. \[\int_{0}^{1} f(x)dt \approx C_1 \cdot f(x_1) + C_2 \cdot f(x_2) + C_3 \cdot f(x_3) + C_4 \cdot f(x_4)\] \[\approx (0.3478548 \cdot 1.21019 \cdot 10^{-5}) + (0.6521452 \cdot 0.029126408 + \] \[ (0.6521452 \cdot 0.757589236) + (0.3478548 \cdot 0.477908863)) \] \[ \approx 4.2097 \cdot 10^{-6} + 0.01899465 + 0.49405818 + 0.16624289\] \[ \approx 0.67929993 \] Jadi, \[ \int_{0}^{1} 15x^{5}(1-x^{(5)})^{2} = 0.67929993 \]

Metode Trapezoida

trapezoid <- function(ftn, a, b, x) {
  t <- data.frame(n=NULL, hasil=NULL, error=NULL) #hasil trapezoid
  e = 1 #nilai error
  n = 4
  exact <- integrate(ftn, a, b) #nilai exact
  while (e>x){
  h <- (b-a)/n
  x.vec <- seq(a, b, by = h)
  f.vec <- sapply(x.vec, ftn) # ftn(x.vec)
  trpzd <- h*(f.vec[1]/2 + sum(f.vec[2:n]) + f.vec[n+1]/2)
  e <- abs(trpzd - exact$value)
  t <- rbind(t, data.frame(n=n, hasil=trpzd, error=e))
  n = n + 1
  }
  t
}

Metode Simpson

simpson <- function(ftn, a, b, x) {
  s <- data.frame(n=NULL, hasil=NULL, error=NULL) #hasil simpson
  e = 1 #nilai error
  n = 4
  exact <- integrate(ftn, a, b) #nilai exact
  while (e>x){
  nb <- max(c(2*(n %/% 2), 4))
  h <- (b-a)/nb
  x.vec1 <- seq(a+h, b-h, by = 2*h) #ganjil
  x.vec2 <- seq(a+2*h, b-2*h, by = 2*h) #genap
  f.vec1 <- sapply(x.vec1, ftn) #ganjil
  f.vec2 <- sapply(x.vec2, ftn) #genap
  smpsn <- h/3*(ftn(a) + ftn(b) + 4*sum(f.vec1) + 2*sum(f.vec2))
  e <- abs(smpsn - exact$value)
  s <- rbind(s, data.frame(n=n, hasil=smpsn, error=e))
  n = n+1
  }
  s
}

Metode four-point Gauss Quadrature

library(pracma)
fpgauss <- gaussLegendre(n=4,a=-1,1)
data.frame(i = c(1:4), C_i = fpgauss[[2]], x_i = fpgauss[[1]])

##   i       C_i        x_i
## 1 1 0.3478548 -0.8611363
## 2 2 0.6521452 -0.3399810
## 3 3 0.6521452  0.3399810
## 4 4 0.3478548  0.8611363

fpgaussianq <- function(ftn, a, b, x) {
  g <- data.frame(n=NULL, hasil=NULL, error=NULL) #hasil gaussian
  e = 1 #nilai error
  n = 2
  exact <- integrate(ftn, a, b) #nilai exact
  while (e>x){
    Lq <- gaussLegendre(n, a, b)
    xi = Lq$x
    wi = Lq$w 
    gaussian <- sum(wi * ftn(xi))
    e <- abs(gaussian - exact$value)
    g <- rbind(g, data.frame(n=n, hasil=gaussian, error=e))
    n = n + 1
  }
  g
}

Perbandingan

EX <- function(x){
  15*x^5*(1-x^5)^2
} #fungsi f(x) yang dalam hal ini adalah nilai E(X)
a <- 0
b <- 1
x <- 1*10^{-4} #nilai toleransi

MT <- trapezoid(EX, a, b, x)
MS <- simpson(EX, a, b, x)
MG <- fpgaussianq(EX, a, b, x)

comparison <- cbind(
  Metode = c('Metode Trapezoid', 'Metode Simpson', 'Metode Gauss Quadrature'), 
  rbind(tail(MT, n=1), tail(MS, n=1), tail(MG, n=1), make.row.names = F)
)
comparison

##                    Metode  n     hasil        error
## 1        Metode Trapezoid 23 0.7101278 9.950361e-05
## 2          Metode Simpson 34 0.7103096 8.227772e-05
## 3 Metode Gauss Quadrature  6 0.7101422 8.504999e-05

Terlihat bahwa metode Simpson memiliki nilai error yang paling kecil dibandingkan kedua metode lainnya, sehingga dapat kita simpulkan bahwa metode terbaik untuk menghitung nilai ekspektasi dari distribusi Kumaraswamy adalah dengan menggunakan metode Simpson.