I am calculating the value of the integral
$$
int_0^1int_1^2(r-u)r drdu
$$
I would like to understand how the following error is handled in R.

Looking around the forum, I realized that a vectorization of the first of the two integrals (in r) is required to compute the second one (in u). Therefore I assume that the following code is the correct way of solving the calculation (tell me please If I am wrong).

fI <- 
  function(u){
  
  integrand <- function(r, u){
    (r - u) * r
  }

  sapply(X = u,
         function(z){
           integrate(f = integrand, lower = 1, upper = 2, u = z)$value
         })

}

integrate(f = fI, lower = 0, upper = 1)

which returns

1.583333 with absolute error < 1.8e-14.

However, I could notice that even when incorrectly defining fI without vectorization, a small trick allows the integral to be calculated (I assume wrongly)

fI <- 
  function(u){
    
    integrand <- function(r, u){
      ((r - u) * r)
    }
    
    integrate(f = integrand, lower = 1, upper = 2, u = u)$value
  }

integrand2 <- 
  function(u){
    
    rep(1, length(u)) * fI(u = u)
  }

integrate(f = integrand2, lower = 0, upper = 1)

which delivers a different result

1.5 with absolute error < 1.7e-14.

Can anybody explain me what is going on in the second case and what is the value passed as u in fI(u=u) inside the integrand function? I understand that in this way u cannot take a range of values.
I need to understand this mechanism because my actual calculation is a bit more involved and I would need to understand what is going on in this simplified case.