Question

2. We wish to integrate f(u) = 1/[k (1 - "N)], (with constants k = 0.25 and N = 4000) from u= uj = 0.6931 to u= u2 = 7.6009 using Monte Carlo method with the following simple variable transform: u = (u2 – ui) y + ui and du = (u2 – U1)dy. Calculate this integral with 10000 random numbers. (Exact answer: 30.4018)

Include a Matlab code as well

Answer 1

We have solved the given integral using Monte-Carlo method. Calculations are done in MATLAB. (code attached)

het, we are interested in evaluating I-&'+(u) dx . det, us consider n rondom variables xi such that, Xi ~ 160,1) (unitor distribution 7 method, I can be approximated as Ursing Monte-Carlo IⓇ + +(x3) W5 \ § - for the given Limation, we have H6w) = n(ne) M . I = f(u) du. 14 =0. ų = 7.6009 k = 0.25. N = 4000 tet, y u= (14-14) J + 4 i du = (-4,) du ulųų yloli *1=, ali i na wanang du

Value of the integral for different run of the code.

Command Window >> Monte carlo Value of the integral using 10000 random numbers is: 30.3815 >> Monte_carlo Value of the integral using 10000 random numbers is: 30.5073 >> Monte_carlo Value of the integral using 10000 random numbers is: 30.4635 >> Montecarlo Value of the integral using 10000 random numbers is: 30.4008 >> Monte Carlo Value of the integral using 10000 random numbers is: 30.3976

MATLAB code:

function Monte_carlo

u1=0.6931;
u2=7.6009;
k=0.25;
N=4000;
n=10000; % number random numbers;
  
h = @(y) (u2-u1)./(k*(1-(1/N)*exp((u2-u1)*y+u1)));
y=rand(1,n);
Integral=sum(h(y))/n;
fprintf('\n\nValue of the integral using %g random numbers is: %g\n\n',n,Integral);
end