If n is a positive integer, then lim (n--> infinity) (1/n) [1/(1+(1/n)) + 1/(1+(2/n))++1/(1+(n/n)] is?
the above can be expressed as
a.) integral from 0 to 1 of (1/x) dx
b.) integral from 1 to 2 of (1/(x+1))dx
c.) integral from 1 to 2 of (x)dx
d.) integral from 1 to 2 of (2/(x+1))
e.) integral from 1 to 2 of (1/x)

The correct answer to this question is d.) integral from 1 to 2 of (2/(x+1))
To solve this:

Since Δx = 1/n. 
lim (n→∞) Δx [1/(1+Δx) + 1/(1+2Δx)+ ... + 1/(1+nΔx)] 
= lim (n→∞) Σ(k = 1 to n) [1/(1 + kΔx)] Δx. 

x <---> a + kΔx 

a = 0, then b = 1, so that Δx = (b - a)/n = 1/n

Since (1 + kΔx) combination, a = 1 so that b = 2. 
Then, f(1 + kΔx) <> f(x) ==> f(x) = 1/x. 

This sum represents the integral 
∫(x = 1 to 2) (1/x) dx, so the correct answer is d.) integral from 1 to 2 of (2/(x+1))

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 

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