Evaluating a double integral using the trapezoidal rule

I am trying to take the double integral of the function using the Trapezoidal rule for G=integral (3*x.^2*y+cos(2*x)*sin(y)+2+4*y.^-2*x+5*y)dxdy with x interval 0 to 2pi and y interval 1 to 10. I found a formula for it but don't know the proper syntax to enter it in. We must use the trapezoidal rule because we are comparing different techniques for evaluating integrals. Here is what I have so far:

x1=0;
x2=2*pi;
y1=1;
y2=10;
N=101
dx=(x2-x1)/(N-1);
dy=(y2-y1)/(N-1);
x=x1:dx:x2;
y=y1:dy:y2;
g(x,y)=3*x.^2*y+cos(2*x)*sin(y)+2+4*y.^-2*x+5*y;
N=101;
for i=1:N+1
for j=1:N+1
    out(i,j)=((dx*dy)/4)*(g(i,j(1,1))+g(i,j(1,N))+g(i,j(N,1))+g(i,j(N,N))+2*(g(i,j(1,2:1:N-1))+g(i,j(N,2:1:N-1))+g(i,j(2:1:N-1,1))+g(i,j(2:1:N-1,N)))+4*(sum(g(i,j(2:1:N-1,2:1:N-1)))));
end
end

The really long formula is from this source: http://www.math.ohiou.edu/courses/math344/lecture24.pdf on page 2. I feel that I could streamline the code by using more summation but I do not know how to do that.

Answer 1

[Andrew Newell]

How about this:

N = 101;
x = linspace(0,2,N)*pi;
y = linspace(1,10,N);
dx = diff(x(1:2));
dy = diff(y(1:2));
[x,y] = meshgrid(x,y);
mat = 3*x.^2.*y+cos(2*x).*sin(y)+2+4*y.^(-2).*x+5.*y;
mat(2:end-1,:) = mat(2:end-1,:)*2;
mat(:,2:end-1) = mat(:,2:end-1)*2;
out = sum(mat(:))*dx*dy/4;

[Robert]

Thank you!

Answer 2

[Mohammed]

∫ 𝑠𝑖𝑛𝑥𝑑𝑥 𝜋 0 = 𝑐𝑜𝑠

Answer 3

[Mohammed]

∫5 1 1/x 𝑑𝑥 = 𝑙𝑛x