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- clear all, close all, clc
- N=9; Li=1.0; Ls=4.0; h=(Ls-Li)/N;
- x=Li:h:Ls; y=log(x.^3+sqrt(exp(x)+1));I=0;z=0;
- disp('DIVISAO DE SUBINTERVALOS')
- for i=1:N/3
- A=[i+z,i+1+z,i+2+z,i+3+z]; disp(num2str(A));
- I=I+(3*h/8)*(y(i+z)+3*y(i+1+z)+3*y(i+2+z)+y(i+3+z));
- z=z+2;
- end
- disp(' '); disp(['A integral vale:' num2str(I)]);
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