Use python.

The value for where R is the square [0,1] x [0,1] is approximately 0.946083.

Calculate the integral using Riemann sum so that the deviation is less than 0.003.

This is what I have done but there is something missing because it doesn't work:

import math

def f(x,y): math.cos(x*y) n = 100

sum = 0

for i in range(n): for j in range(n): sum += 1 * 1 * f((i*1), (j*1))

print(sum) print(float(0,946083 - sum))

R cos(xy)dA R cos(xy)dA