`Hey,
Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries
clc
clear all
f=@(x) exp(-5*x)*sin(1/x)*sin(1/sin(1/x));
a=0.19;
b=0.29;
If=0.02115910278025609;
n=5;
disp('Mid point inegral is');
I=0;
h=(b-a)/n;
for i=1:n
I=I+h*f(a+(2*i-1)*h/2);
end
disp(I)
disp('Error is');
abs(I-If)
disp('Trapezoidal inegral is');
I=trapezoid(f,a,b,n)
disp('Error is');
abs(I-If)
disp('Simpson inegral is');
I=SimpsonComp(f,a,b,n)
disp('Error is');
abs(I-If)
disp('Boole inegral is');
I=(b-a)/90*(7*f(a)+32*f((3*a+b)/4)+12*f((a+b)/2)+32*f((a+3*b)/4)+7*f(b))
disp('Error is');
abs(I-If)
function I = trapezoid(func,a,b,n)
% I = trap(func,a,b,n):
% multiple-application trapezoidal rule.
% input:
% func = name of function to be integrated
% a, b = integration limits
% n = number of segments
% output:
% I = integral estimate
x = a;
h = (b - a)/n;
s = feval(func,a);
for i = 1 : n-1
x = x + h;
s = s + 2*feval(func,x);
end
s = s + feval(func,b);
I = (b - a) * s/(2*n);
end
function int=SimpsonComp(f,a,b,n)
%takes in parameter as function handel f, starting interval a and
ending
%interval b for integration and number of intervals as n
h=(b-a)/n;%step of integration
fa=f(a);%value of function at a
fb=f(b);%value of function at b
ff=0;%initialize ff to 0
for i=2:2:n%traversing on even numbers i
x=(a+(i-1)*h);%accessing even points in the intervals divided
fx=feval(f,x);%value of function at x
ff=ff+4*fx;%add 4*fx to ff
end
for i=3:2:n%traversing on odd numbers i
x=(a+(i-1)*h);%accessing odd points in the intervals divided
fx=feval(f,x);%value of function at x
ff=ff+2*fx;%add 4*fx to ff
end
int=(h/3)*(fa+fb+ff);%finally add first and last points tto all the
sum and divide by 3
end
Kindly revert for any queries
Thanks.
Throughout this homework, rz)-e-5s sin 릎 sin siIT, unless stated otherwise. Q1 The Boole's quadr...