It is a valid simpsons rule implementation. Let me break the line down.
sr=(h/3)*(f(x(1))+2*sum(f(x(3:2:end-2)))+4*sum(f(x(2:2:end)))+f(x(end)))
This expression above is equivalent to (Δx/3)*[f(x0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+⋯+4f(xn−1)+f(xn)].
x(3:2:end-2) %j:i:k creates a regularly-spaced vector using i as the increment between elements.
So here indexing into the elements of x at odd places. We left out the end as f(xn) is multiplied with 1 and not 2. Similarly, access elements of x at even positions using
x(2:2:end)
Since we are passing a vector to f, the ouput is a vector and sum is used to find the sum of the output vector.
sum(f(x(3:2:end-2)))
