Shouldn't these two results be the same? Why are they not?
integrate(\(x) {x * min(-x+10, 10)},lower = 0, upper = 10)$value
> [1] 1.085709
integrate(\(x) {x * (-x+10)},lower = 0, upper = 10)$value
> [1] 166.6667
Keep in mind that from x values 0 to 10 we should never expect to get a value of y = -x + 10 that is higher than 10, so the min(-x+10, 10) will always return (-x+10) as long as we are between 0 and 10. So the two integrals should be identical.
Why are they not?
The problem is that min returns one value only, it's not vectorized.
f <- \(x) {x * min(-x+10, 10)}
g <- \(x) {x * (-x + 10)}
f <- Vectorize(f, "x")
g <- Vectorize(g, "x")
# curve(f, from = 0, to = 10)
# # this overplots perfectly
# curve(g, from = 0, to = 10, add = TRUE, col = "red")
integrate(f, lower = 0, upper = 10)$value
#> [1] 166.6667
integrate(g, lower = 0, upper = 10)$value
#> [1] 166.6667
Created on 2023-08-21 with reprex v2.0.2
If you don't want to assign twice to the same names, the first to create the function, the second to vectorize it, use the new pipe operator.
f <- (\(x) {x * min(-x+10, 10)}) |> Vectorize()
g <- (\(x) {x * (-x + 10)}) |> Vectorize()
Created on 2023-08-21 with reprex v2.0.2
@RuiBarradas' answer works, but I found a way to mitigate the issue using pmin:
integrate(\(x) {x * pmin(-x+10, 10)},lower = 0, upper = 10)$value
> [1] 166.6667
integrate(\(x) {x * (-x+10)},lower = 0, upper = 10)$value
> [1] 166.6667
If anyone for some reason experiences other problems with min() not being vectorized, this might solve it.