The code below shall be used to calculate numerically the integral of f(x), using rectangles. The while loop ' error> er_user_max' seems not to evaluate the logical operation properly and changes the row dimension of matrix 'save_area'.

 clear all

 %define function
 f= @(x)x.^2; 
 %length of function domain
 L=10;
 %initial integration step
 dx=1;
 %introducing a parameter for the total sum underneath the 
 %curve.
 sum_area=0;
 %max refinment passes within one dx element
 ref_max=10;

 %define the allowed error, calculated by the equidistantly 
 %split  rectangle
 %in two parts to calculate each area of both split parts  
 %separatly. 
 % Used formular is written below @error calculation
 %For comparision, the two parts are summed up and divided by the 
 %former,
 %unsplit rectangle.
 er_user_min=0.995;
 er_user_max=1.2499999;

 %____________________________________    
 for i=1:1 %for each dx 
    a=(i-1)*dx;
    b=i*dx;
    %introduce a running number for counting the partitions 
    %within a
    %initial bar element with the length of dx.
    pn=1;
    rn=pn+1;
    %max refinment passes within one dx element
    %ref_max=10;
    %initialize storage matrix of created partitions
    save_area= zeros(ref_max,3);
    %introducing a parameter for the partial areas within a dx  
    %element.
    area_part= 0;
    %introducing a parameter for the sum of rectangular pieces 
    %within dx
    area_dx=0;

    %calculate initial area of specific dx element.
        area=((b-a))*f((i-1)*dx+(b-a)/2);

    %intial error calculation
    b_new=(i-1)*dx+(b-a)/2;
    area_l = ((b_new-a))*f((i-1)*dx+(b_new-a)/2);
    area_r = ((b-b_new))*f(b_new+(b-b_new)/2);

    error = (area_l+area_r)/area;

    %save area of dx to add to sum_area, if error< er_user.
    area_comp= area_l+area_r;

    %calculate the total area of all partitions within one 
    %initial bar
    %element with the lengt of dx, with respect to the error 
    %evaluation.
   % while rn>1

    %split dx in partitions to grant a certain grade of accuracy  
    %due to
    %limitations of maximum possible refinment passes.
        while  error> er_user_max

            %calculate new boundaries of the left side rectangle.
            b_new=(i-1)*dx+(b-a)/2;
            area_l = ((b_new-a))*f((i-1)*dx+(b_new-a)/2);
            area_r = ((b-b_new))*f(b_new+(b-b_new)/2);

            %store boundaries and area of right side rectangle.
            a_r= b_new;
            b_r= b;
            %save_area(pn,:)=[a_r,b_r,area_r];

            save_area(pn,1)=a_r;
            save_area(pn,2)=b_r;
            save_area(pn,3)=area_r;

            %update the latest considered area
            area_part= area_l;

            %Without if condition, the row dimension as defined 
            %of ref_max
            %of matrix 'save_area' will be largly overexceeded!

             %if pn < 10
             %add information that one refinment has been done.
             %pn= pn+1;
             %rn=rn+1;
             %adjust new boundaries
             %b= b_new;
             %a=a;
             %else
                % break
             %end

            %calculate new error
            b_new=(i-1)*dx+(b-a)/2;
            area_l = ((b_new-a))*f((i-1)*dx+(b_new-a)/2);
            area_r = ((b-b_new))*f(b_new+(b-b_new)/2);

            area_comp=area_l+area_r;
            error = area_comp/area_part;

        end

        %add error evaluated left-side part to the sum of the 
        %area of dx.
        area_dx=area_dx+ area_comp;

        %...further coding, does not display the mentioned 
        %problem with the coding%
        %...continuing relevant coding

%Sum up the calculated and error evaluated partial rectangles   
%underneath
%the curve
%sum_area=sum_area+area_dx;

 end
 disp(sum_area)
 disp(save_area)

Answer 1

[dpb]

Way too much code instead of concentrating on the concern and put the question in the body of the text and make the title just that...but, that aside--

...
error = (area_l+area_r)/area;
...
%limitations of maximum possible refinment passes.
while  error> er_user_max
  ...

As defined your error term is a ratio, not a relative nor absolute and the test as written requires the estimates always be high in comparison to the reference to fail; if the initial guess is low instead, nothing will happen.

Probably you intend something like

error = 1-(area_l+area_r)/area;

and

while  abs(error)>er_user_max

Also note that error is a Matlab builtin function; suggest using some other name for the variable ( err or similar isn't taken).