Question

Exercise 2 Use the functions you coded in Exercise 1 to compute the numerical approximation of the integral .1 cos e 30 To this end, write a Matlab/Octave function function [en,et , es] test-integration() =

Use matlab please.

Answer 1

`Hey,

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function [em,et,es]=test_integration()
f=@(x) ((1./(1+x.^2)).*cos((3/2)*exp(-x.^2))-x.^3/30);
Iref=integral(f,-3,1);
em=[];
et=[];
es=[];
v=2:100;
a=-3;
b=1;
for n=2:100
em(n-1)=abs(int_midpoint_rule(f,a,b,n)-Iref);
et(n-1)=abs(int_trapezoidal_rule(f,a,b,n)-Iref);
es(n-1)=abs(int_Simpson_rule(f,a,b,n)-Iref);
  
end
loglog(v,em,v,es,v,et);
legend('Mid point','Trapezoidal','Simpsons');
function I = int_trapezoidal_rule(f,a,b,n)

hval = (b-a)/n;

%Calculate

x = [a+hval:hval:b-hval];

%feval point

xin=sum(feval(f,x));

I = hval/2*(feval(f,a)+2*xin+feval(f,b));

end
%int_Simpson_rule.m

function I = int_Simpson_rule(f,a,b,n)

hval = (b-a)/n;

xval1 = feval(f,a)+feval(f,b);

xval2 = 0;

xval3 = 0;

for i = 1:n-1

x = a+i*hval;

if mod(i,2) == 0

xval3 = xval3+feval(f,x);

else

xval2 = xval2+feval(f,x);

end

end

xival = hval*(xval1+2*xval3+4*xval2)/3;

I=xival;
end
function I = int_midpoint_rule(f,a,b,n)
I=0;

hval = (b-a)/(n+2);

%mid point function

xval = [a+hval:2*hval:b-hval];

I = 2*hval*sum(feval(f,xval));
end
end

Figure 1 Eile Edit Yew Insert 1ools Desktop 岦indow Help 10 -Mid point Trapezoidal Simpsons 101 10 10 10 10 10 10 10 10% 10 10 O Type here to search ENG 4:51 AM IN 03/04/19

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