use the MATLAB function trapez to evaluate the intergral that arisesin electrica

Question

use the MATLAB function trapez to evaluate the intergral that arisesin electrical field theory

for the following values x and r

function trapez(f,a,b,n)

% Compute the integral of a f from a to b using the trapezoid rule

h=(b-a)/n;

disp('_________________________________________________________')

disp([' i xi f(xi) h=',num2str(h) ])

disp('_________________________________________________________')

S=feval(f,a);

fprintf(' %2.0f %12.4f %14.6f ',0,a,S);

for i=1:n-1

   x=a+h*i;

   g=feval(f,x);

   S=S+2*g;

   fprintf(' %2.0f %12.4f %14.6f ',i,x,g);

end

S=S+feval(f,b);

fprintf(' %2.0f %12.4f %14.6f ',n,b,feval(f,b));

INT=h*S/2;

fprintf(' The integral of f(x) is =%16.8f ',INT);

Explanation / Answer

%%Without funcyion

x=input('Enter value x=');
r=input('Enter value r=');
f=@(phi) sqrt(1-((x/r)^2*(sin(phi))^2));

a=0;
b=2*pi;
n=100;
h=(b-a)/n;
phi=a:h:b;
for i=1:n+1
    y(i)=f(phi(i));
end
disp('_________________________________________________________')

disp([' i         xi          f(xi)         h=',num2str(h) ])

disp('_________________________________________________________')
for i=1:n+1

or

     fprintf(' %2.0f %12.4f %14.6f ',i,phi(i),f(phi(i)));
end

s=0;
for i=2:n
    s=s+y(i);
end
Int=(h/2)*(y(1)+s+y(n+1))

fprintf('            The integral of f(phi) is =%16.8f ',Int);

%or

% with function

clc;
clear all;
x=input('Enter value x=');
r=input(' Enter valuer=');
a=input('Enter value a=');
b=input('Enter value b=');
n=input('Enter value n=');
f=@(phi) sqrt(1-((x/r)^2*(sin(phi))^2));
trapiz(f,a,b,n)

% note 1 : this programm and below program you should save in one folder

%note 2 : save below program with name ' trapiz.m' otherwise it will not work above program. only you should run %above program

function Int=trapiz(f,a,b,n)
h=(b-a)/n;
phi=a:h:b;
for i=1:n+1
    y(i)=f(phi(i));
end
disp('_________________________________________________________')

disp([' i         xi          f(xi)         h=',num2str(h) ])

disp('_________________________________________________________')
for i=1:n+1
     fprintf(' %2.0f %12.4f %14.6f ',i,phi(i),f(phi(i)));
end

s=0;
for i=2:n
    s=s+y(i);
end

% result for n=10

The integral of f(phi) is =      3.05041254

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