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I am trying to calculate the integral of the data. Following this post, I know what to do with a one-dimensional list. But, let's consider a simple function, just to show you, what I want to do.

f[x_, y_]:= x y;

Now, I create a two-dimensional table

L = Table[f[x,y], {x, 0, 5}, {y,0, 5}];

Now on this data, I would like to do the following operation

Integrate[f[x,y,], {x, 0, 5}, {y, 0, 5}]

For one dimension, I would use the Simpson rule from the mentioned post, but is it possible to extend it to calculate the following integral?

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Integrate[f[x, y], {x, 0, 5}, {y, 0, 5}](*625/4*)

integration weights

g = Table[1, {x, 0, 5}, {y, 0, 5}];
g[[All, 1]] = g[[1, All ]] = g[[All, 6]] = g[[6, All ]] = 1/2;
g[[1, 1]] = g[[1, 6]] = g[[6, 1]] = g[[6, 6]] = 1/4;

enter image description here

numerical integration

Flatten[g L] // Total (* 625/4 *)
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  • $\begingroup$ Thank you very much, can you explain, why you take such weights? $\endgroup$
    – blahblah
    Commented Apr 26 at 8:23
  • $\begingroup$ @blahblah Have a look at the grid: Neighborhood of inner points is one, boundary points (except corners) is 1/2 and corners is 1/4. Sum of all weights should be 25 $\endgroup$
    – Ulrich Neumann
    Commented Apr 26 at 9:07
  • $\begingroup$ Ok, I understand, but why the sum should be 25 if the array is 6x6? $\endgroup$
    – blahblah
    Commented Apr 26 at 10:57
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    $\begingroup$ Check the neighbourhood of your new gridpoints: innerpoints .1^2,boundary points .1^2/2, corner points .1^2/4 $\endgroup$
    – Ulrich Neumann
    Commented May 28 at 11:19
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    $\begingroup$ Yes , seems to be correct! $\endgroup$
    – Ulrich Neumann
    Commented May 28 at 11:56

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