If n is a positive integer, then lim n→∞ 1/n[(1/n)^2+(2/n)^2++((n-1)/n)^2] can be expressed

This is a Riemann Sum.
First of all, Δx = 1/n.
So, now we have
lim (n→∞) Δx [1/(1+Δx) + 1/(1+2Δx)+ ... + 1/(1+nΔx)]
= lim (n→∞) Σ(k = 1 to n) [1/(1 + kΔx)] Δx.
Since x <---> a + kΔx
So, if we take a = 0, then b = 1, so that Δx = (b - a)/n = 1/n, etc.
(the choice of a is arbitrary, along as you choose b appropriately so that b - a = 1 in this case).
However, due to the (1 + kΔx) combination, I'll set a = 1 so that b = 2.
Then, f(1 + kΔx) <> f(x) ==> f(x) = 1/x.
Hence, this sum represents the integral
∫(x = 1 to 2) (1/x) dx, choice d.

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