SIMPSON1.M

N=10;Li=2.0;Ls=4.0;h=(Ls-Li)/N;
x=Li:h:Ls;y=(log10(x)+x.^2)./(x+3).^2;I=0;z=0;
disp('DIVISAO DE SUBINTERVALOS')
for i=1:N/2
    A=[i+z,i+1+z,i+2+z];disp(num2str(A));
    I=I+(h/3)*(y(i+z)+4*y(i+1+z)+y(i+2+z));
    z=z+1;
end
disp(' ');disp(['A integral vale:' num2str(I)]);