For n independent uniforms on [0,1], the k-th smallest has expectation k/(n+1). So the maximum averages n/(n+1) and the minimum 1/(n+1).
The intuition is pure symmetry: n points cut the interval into n+1 exchangeable gaps summing to 1, so each gap averages 1/(n+1). No integration needed, and saying it this way is much stronger than deriving it.
The exponential fact worth memorising: the minimum of independent exponentials is exponential with rate equal to the sum of the rates. That answers "expected time until the first of several machines fails" immediately.