Is MATLAB giving wrong (symbolic) definite integration??

Hello, I am trying to compute a definite integral of symbolic array. I did this in two ways: (1) using function int() (2)using integral(). However I am facing problem in both cases and I am not understanding why this is happening. Here is my code:

syms xlc
ke=[2000000*(0.16*xlc - 0.6)^2, -2000000*(0.32*xlc - 0.8)*(0.16*xlc - 0.6),  2000000*(0.16*xlc - 0.2)*(0.16*xlc - 0.6); -2000000*(0.32*xlc - 0.8)*(0.16*xlc - 0.6),        2000000*(0.32*xlc - 0.8)^2, -2000000*(0.16*xlc - 0.2)*(0.32*xlc - 0.8);2000000*(0.16*xlc - 0.2)*(0.16*xlc - 0.6), -2000000*(0.16*xlc - 0.2)*(0.32*xlc - 0.8),           2000000*(0.16*xlc - 0.2)^2];

%ke is 3x3 symbolic matrix and it is symmetric
a=0; %lower bound of integration
b=5; %upper bound of integration

%Integration of each element in matrix ke, integration ke(i,j) from a to b
 for i=1:1:3
     for j=i:1:3
         %t1=matlabFunction(ke(i,j)); %see COMMENT 1 below
         ke_num(i,j)=double(int(ke(i,j),xlc,a,b));
     end
 end

This is very straightforward code. As ke is symmetric, I expect integration ke_num matrix to be symmetric. However, I get ke_num as upper triangular matrix. What is wrong? I am not able to find. %COMMENT 1: One thing I know is we should vectorize function. How to achieve this in this program? Just using matlabFunction() does not crate that. Any ideas, suggestions? Thanks.

Answer 1

[Roger Stafford]

Your inner for-loop is written

 for j=i:1:3

with the lower value as 'i' rather that 1. This of course gives you an upper triangular matrix because you must always have j>=i. With a symmetric matrix such as yours this is a more efficient method but you must copy the result into both ke_num(i,j) and ke_num(j,i).

[Andrew]

Thanks for your answer. Indeed for symmetric matrix, computing upper triangular is more efficient.

[Andrew]

Just a question, I can start j from 1. But is there any other method or predefined function in MATLAB which can copy upper triangular results into lower triangular part without loops and all?

[Roger Stafford]

M = triu(M,1).'+tril(M,-1);