Express the quantity as a finite integral

Question

lim (1/n)Σ(k/n)³ as n tends to infinity and where k from 1 to n.

Answer 1

This is a pattern matching problem.

lim n→∞ ((b-a)/n)∑_k=1ⁿ f(a+k/n) = ∫_a^b f(x) dx

For this case, f(a+k/n) = (k/n)³, so a = 0

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

The integral equivalent to the limit summation expression is:

∫₀¹x³dx

Answer 2

It is easiest to compare it to

∫f(x) dx

where the integral is from 0 to 1.

Then the task is to find the right f(x).

That integral can be approximated by dividing the interval [0,1] into n segments,

for some n.

Segment number k (where k = 1, ..., n) goes from x = (k-1)/n to x = k/n.

For the segment we can approximate the area under the curve of f(x) with a rectangle

of width 1/n and height f(k/n).

The area of such a rectangle is (1/n)×f(k/n).

The sum of these rectangles will approximate the integral.

The larger the n the better is the approximation.

So

∫f(x) dx = lim (n->∞) ∑1/n f(k/n) = lim (n->∞)1/n ∑f(k/n)

where the sum goes from k=1 to n.

Comparing it with the given expression f(x) = x³