Evaluating the following integral using comp trap method for 2 trapezoids

% I'm trying to evaulate this integral using the composite trapezoid method
%  The output for the 'I' gives me 8.7797 which matches the solution, yet
%  when I calculate the error percentage, I get  0.45278  when the solution
%  is 0.45484.

% I've used the same code to calculate for n # of trapezoids and dont seem
% to have an issue. 

y = @(x) (10 - 2*cos(x));
b= pi/3;
a = 0;

              
n = 2;                                   % Compute with comp trap method for 2 trapezoids
h = (b-a)/n;
x= a:h:b;
I = h*(y(a)*0.5+y(b)*0.5+y(x(2)));
err = 100*abs((I-integral(y,a,b))/I);

Answer 1

[Alan Stevens]

You need to divide by the true value of the integral in calculating the error:

y = @(x) (10 - 2*cos(x));
b= pi/3;
a = 0;

              
n = 2;                                   % Compute with comp trap method for 2 trapezoids
h = (b-a)/n;
x= a:h:b;
I = h*(y(a)*0.5+y(b)*0.5+y(x(2)))

I = 8.7797

Itrue = 10*(b-a) - 2*(sin(b) -sin(a))

Itrue = 8.7399

err = 100*abs((I-Itrue)/Itrue)

err = 0.4548

[Oscar]

thank you but can you explain this true value formula?

[Alan Stevens]

I just integrated ypour y-function by hand - it's easy to do. However, you can replace it by your integral(y,a,b) if you wish (i.e. set Itrue = integral(y,a,b)).