Metode Trapezoida

Dengan \(n=4\) maka :

\(E(X) = \frac{h}{2}(f(x_{0})+2f(x_{1})+2f(x_{2})+2f(x_{3})+f(x_{4})) \text{ , dengan } X=0, 0.25, 0.5, 0.75, 1\)

\[\begin{align} h &= \frac{x_{n}-x_{0}}{n}=\frac{1-0}{4}=0.25 \\ f(x_{0}) &= 15(0)^{5}(1-(0)^{5})^{2}=0 \\ f(x_{1}) &= 15(0.25)^{5}(1-(0.25)^{5})^{2}=0.01461984 \\ f(x_{2}) &= 15(0.5)^{5}(1-(0.5)^{5})^{2}=0.4399109 \\ f(x_{3}) &= 15(0.75)^{5}(1-(0.75)^{5})^{2}=2.070617 \\ f(x_{4}) &= 15(1)^{5}(1-(1)^{5})^{2}=0 \\ \end{align}\]

\[\begin{equation} \begin{split} E(X) &= \int_{0}^{1}15x^{5}(1-x^{5})^{2} dx \\ &= \frac{h}{2}(f(x_{0})+2f(x_{1})+2f(x_{2})+2f(x_{3})+f(x_{4})) \\ &= \frac{0.25}{2}(0+2(0.01461984)+2(0.4399109)+2(2.070617)+0) \\ & = \frac{0.25}{2}(5.050295) \\ & = 0.6312869 \end{split} \end{equation}\]

Metode Simpson

Dengan \(n=4\) maka :

\(E(X) = \frac{h}{3}(f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+f(x_{4})) \text{ , dengan } X=0, 0.25, 0.5, 0.75, 1\)

\[\begin{align} h &= \frac{x_{n}-x_{0}}{n}=\frac{1-0}{4}=0.25 \\ f(x_{0}) &= 15(0)^{5}(1-(0)^{5})^{2}=0 \\ f(x_{1}) &= 15(0.25)^{5}(1-(0.25)^{5})^{2}=0.01461984 \\ f(x_{2}) &= 15(0.5)^{5}(1-(0.5)^{5})^{2}=0.4399109 \\ f(x_{3}) &= 15(0.75)^{5}(1-(0.75)^{5})^{2}=2.070617 \\ f(x_{4}) &= 15(1)^{5}(1-(1)^{5})^{2}=0 \\ \end{align}\]

\[\begin{equation} \begin{split} E(X) &= \int_{0}^{1}15x^{5}(1-x^{5})^{2} dx \\ &= \frac{h}{3}(f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+f(x_{4})) \\ &= \frac{0.25}{3}(0+4(0.01461984)+2(0.4399109)+4(2.070617)+0) \\ & = \frac{0.25}{3}(9.220768) \\ & = 0.7683974 \end{split} \end{equation}\]

Metode Gauss Quadrature

\(\int_{a}^{b}f(x)dx = \int_{-1}^{1} f(\frac{b-a}{2}\xi+\frac{a+b}{2})\frac{dx}{d\xi} d\xi\)

\(\text{dengan }\frac{dx}{d\xi} = \frac{b-a}{2}\)

Pendekatan Gauss Quadrature dengan \(n\) point :

\(\int_{a}^{b}f(x)dx = \frac{b-a}{2} \sum_{i=1}^{n} w_{i} f(\frac{b-a}{2}\xi_{i}+\frac{a+b}{2})\)

Pada metode ini akan digunakan four-point Gauss Quadrature, berikut nilai \(\xi_{i}\) dan \(w_{i}\) untuk \(n=4\) :

\[\begin{align} \xi_{1}&=-\sqrt{\frac{3}{7}-\frac{2}{7}\sqrt{\frac{6}{7}}} = -0.339981 & w_{1} &=\frac{18+\sqrt{30}}{36} = 0.6521452 \\ \xi_{2}&=\sqrt{\frac{3}{7}-\frac{2}{7}\sqrt{\frac{6}{7}}} = 0.339981 & w_{2} &=\frac{18+\sqrt{30}}{36} = 0.6521452 \\ \xi_{3}&=-\sqrt{\frac{3}{7}+\frac{2}{7}\sqrt{\frac{6}{7}}} = -0.8611363 & w_{3} &=\frac{18-\sqrt{30}}{36} = 0.3478548 \\ \xi_{4}&=\sqrt{\frac{3}{7}+\frac{2}{7}\sqrt{\frac{6}{7}}} = 0.8611363 & w_{4} &=\frac{18-\sqrt{30}}{36} = 0.3478548 \end{align}\]

\[\begin{equation} \begin{split} E(X) &= \int_{0}^{1}15x^{5}(1-x^{5})^{2} dx \\ &= \frac{b-a}{2}(w_{1}f(\frac{b-a}{2}\xi_{1}+\frac{a+b}{2})+w_{2}f(\frac{b-a}{2}\xi_{2}+\frac{a+b}{2})+w_{3}f(\frac{b-a}{2}\xi_{3}+\frac{a+b}{2})+w_{4}f(\frac{b-a}{2}\xi_{4}+\frac{a+b}{2})) \\ &= \frac{1-0}{2}(w_{1}f(\frac{1-0}{2}\xi_{1}+\frac{0+1}{2})+w_{2}f(\frac{1-0}{2}\xi_{2}+\frac{0+1}{2})+w_{3}f(\frac{1-0}{2}\xi_{3}+\frac{0+1}{2})+w_{4}f(\frac{1-0}{2}\xi_{4}+\frac{0+1}{2})) \\ &= \frac{1}{2}(w_{1}f(\frac{1}{2}\xi_{1}+\frac{1}{2})+w_{2}f(\frac{1}{2}\xi_{2}+\frac{1}{2})+w_{3}f(\frac{1}{2}\xi_{3}+\frac{1}{2})+w_{4}f(\frac{1}{2}\xi_{4}+\frac{1}{2})) \\ &= \frac{1}{2}(0.6521452f(0.3300095)+0.6521452f(0.6699905)+0.3478548f(0.06943185)+0.3478548f(0.9305682)) \\ &= \frac{1}{2}(0.6521452(0.05825283)+0.6521452(1.515178)+0.3478548(0.000024)+0.3478548(0.9558169)) \\ &= \frac{1}{2}(1.358599) \\ & = 0.6792997 \end{split} \end{equation}\]

Metode Trapezoida

trapezoid <- function(ftn, a, b, n = 100) {
     h <- (b-a)/n
     x.vec <- seq(a, b, by = h)
     f.vec <- sapply(x.vec, ftn)     # ftn(x.vec)
     Trap <- h*(f.vec[1]/2 + sum(f.vec[2:n]) + f.vec[n+1]/2)
     return(Trap)
}


tol <-  0.0001
err <- 1
n = 2


while(err>tol){
  res_trap <- trapezoid(f,0,1,n = n)
  
  err <- abs(res_trap-exact_value)
  
  #cat("n=",n,", result=",res_trap,", error=",err,"\n",sep = "")
  
  n=n+1
  if(n==1000){
    break
  }
  
}

n_trap <- n
err_trap <- err
res_trap

## [1] 0.7101278

Metode Simpson

simpson_n <- function(ftn, a, b, n = 100) {
    n <- max(c(2*(n %/% 2), 4))
    h <- (b-a)/n
    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
    S <- h/3*(ftn(a) + ftn(b) + 4*sum(f.vec1) + 2*sum(f.vec2))
    return(S)
}

tol <-  0.0001
err <- 1
n <- 2
f <- function(x){15*(x^5)*((1-(x^5))^2)}

while(err>tol){
  res_simp <- simpson_n(f,0,1,n = n)
  
  err <- abs(res_simp-exact_value)
  
  #cat("n=",n,", result=",res_simp,", error=",err,"\n",sep = "")
  
  n=n+1
  if(n==1000){
    break
  }
  
}

n_simp <- n
err_simp <- err
res_simp

## [1] 0.7103096

Metode Gauss Quadrature

library(polynom) 
library(orthopolynom)
library(gaussquad)
library(pracma)

## 
## Attaching package: 'pracma'

## The following object is masked from 'package:polynom':
## 
##     integral

f <- function(x){15*(x^5)*((1-(x^5))^2)}

tol <-  0.0001
err <- 1
n <- 2

while(err>tol){
  
  gL <- gaussLegendre(n = n,a=0,b=1)

 Ci <- gL$w # koefisien
 xi <- gL$x # gauss point
  
  res_gl <- sum(Ci * f(xi))

  err <- abs(res_gl-exact_value)
  
  #cat("n=",n,", result=",res_gl,", error=",err,"\n",sep = "")
  
  n=n+1
  if(n==1000){
    break
  }
  
}

n_gl <- n
err_gl <- err
res_gl

## [1] 0.7101422

Rangkuman Hasil Penghitungan Nilai Harapan

Metode <- c('Trapezoida','Simpson','Gauss Quadrature')
Hasil <- c(res_trap,res_simp,res_gl)
Error <- c(err_trap,err_simp,err_gl)
n <- c(n_trap,n_simp,n_gl)
res_fin <- data.frame(Metode,Hasil,Error,n)
res_fin

Dari perbandingan 3 metode di atas dapat disimpulkan metode Simpson menghasilkan error yang paling kecil.