How to evaluate an array filled with symbolic integrals?

I managed to use symbolic math for calculating a bunch of integrals (which have all the same structure but differs only on a few parameters) one by one like this:

syms x y s ;
I = zeros(3,3);

% 1)
t = 2e-3;
s_f = 0.3;
x = 0.42;
y = -0.19 + s;
I(1,1) = int((t*y^2), s, 0, s_f);
I(2,1) = int((t*x^2), s, 0, s_f);
I(3,1) = int((t*x*y), s, 0, s_f);

% 2)
t = 2e-3;
s_f = 0.6;
x = 0.42 - s;
y = 0.11;
I(1,2) = int((t*y^2), s, 0, s_f);
I(2,2) = int((t*x^2), s, 0, s_f);
I(3,2) = int((t*x*y), s, 0, s_f);

% 3)
t = 1e-3;
s_f = 0.3;
x = -0.18;
y = 0.11 - s;
I(1,3) = int((t*y^2), s, 0, s_f);
I(2,3) = int((t*x^2), s, 0, s_f);
I(3,3) = int((t*x*y), s, 0, s_f);

Result being:

I =

     5.4600e-006    14.5200e-006     2.7300e-006
   105.8400e-006    53.2800e-006     9.7200e-006
   -10.0800e-006    15.8400e-006     2.1600e-006

But now I'd like to make it a bit more concise and elegant. I thought I could define t,s_f,x,y as vectors and build the resulting I array someway; I read docs and tutorials ... but I'm lost. I cannot figure out how to deal with it.

I'd like to come to something like this (which is pseudo-code only):

T    = [2e-3       2e-3      1e-3     ];
X    = [(0.42)     (0.42-ss) (-0.18)  ];
Y    = [(-0.19+ss) (0.11)    (0.11-ss)];
S_F  = [0.3        0.6       0.3      ];

I = int((T*Y^2), S, 0, S_F);

in which I define the parameters I need first, and later build the I array in some concise way (without duplicating the same code over and over ...) , resulting in the same I array as above. But I cannot figure out a way.

Tried several constructs/functions/syntax but without success. I surely have to learn more about that symb-math.

Could you please point me in the correct direction? Is this kind of expressions even thinkable with Matlab?

I am new to Matlab, so please excuse me if I result being naive.

Answer 1

[madhan ravi]

Numerical integration is faster compared to symbolic integration. See matlabFunction() to convert symbolic expressions to function handle thereby achieving to adapt to numerical methods.

syms ss
T    = [2e-3       2e-3      1e-3     ];
X    = [0.42    0.42-ss      -0.18  ];
Y    = [-0.19+ss 0.11    0.11-ss];
S_F  = [0.3        0.6       0.3      ];
func=@(T,Y)(T.*Y.^2);
func=func(T,Y);
func=matlabFunction(func);
I=cell(1,numel(func));    %pre-allocation
for i = 1:numel(func) 
I{i} = integral(@(ss)func(i), 0, S_F(i),'ArrayValued',true);
end
final_result = vpa([I{:}])

command window:

>> COMMUNITY
final_result =
[ 0.00039365999999999999516692161805054, 0.0000072599999999999982526911052049812, 0.00023762999999999995931476703958651]
>> 

[Roberto Inzerillo]

Computation efficiency is not a priority now, and I'm trying to rely on the symbolic tools for learning purposes.

I will look into the numerical approach also (as you're suggesting) nonetheless.

[madhan ravi]

so remove the line with matlabFunction integral and replace it with int() , it would be the same approach that i showed you

[Roberto Inzerillo]

Thank you. Will look into it :-)

Answer 2

[YT]

I don't know if this is exactly what you're looking for, but I would do something like this

clear all;
syms ss;

T    = [2e-3       2e-3      1e-3];
X    = [(0.42)     (0.42-ss) (-0.18)];
Y    = [(-0.19+ss) (0.11)    (0.11-ss)];
S_F  = [0.3        0.6       0.3];

I = zeros([3 length(T)]); %3-by-amount-of-data-columns
for i = 1:length(T)
    I_ty = int((T(i)*Y(i)^2), ss, 0, S_F(i));
    I_tx = int((T(i)*X(i)^2), ss, 0, S_F(i));
    I_xy = int((T(i)*X(i)*Y(i)), ss, 0, S_F(i));
    
    I(:,i) = [I_ty; I_tx; I_xy];
end

If you for some reason ever want to add more data to T, X, Y, S_F, you can just simply add it to those columns and it still works fine:

T    = [2e-3       2e-3      1e-3      4e-3];
X    = [(0.42)     (0.42-ss) (-0.18)   (-0.20)];
Y    = [(-0.19+ss) (0.11)    (0.11-ss) (0.33+ss)];
S_F  = [0.3        0.6       0.3      0.4];

[Roberto Inzerillo]

I thought about using "for" loops too. It's just that it doesn't look "elegant" enough. I was fishing for a more Mathlab-ish solution if you know what I mean :-)

But you're on point. Those few lines do the job I was expecting.