A calculus question....Approximating finite sums with integrals.?

Question

I am confused on how to go about this....

1. Evaluate: limit as n ---> infinite: (1^5 + 2^5 +3^5+...+n^5) / n^6 by showing that the limit is: integral symbol -with a 1 on top, 0 on bottom, x^5 dx ........and evaluating the integral.

2. Then we are supposed to evaluate from the previous problem the limit as n---> infinite of 1/n^4 (1^3 + 2^3 + 3^3 +...+n^3).

Answer 1

Consider a partition of the interval [0, 1] into n segments of equal length. The endpoints of the segments are x₀=0, x₁=1/n, x₂=2/n... x_k=k/n, x_n=1. The kth segment is from x_(k-1) to x_k, so evaluating each segment at the right endpoint gives a Riemann sum for a function f on the interval [0, 1] of:

[k=1, n]∑f(k/n)*1/n

Letting f(x)=x⁵, we have:

[k=1, n]∑k⁵/n⁶

Which is the function whose limit we are evaluating as n→∞. Per the definition of the integral, [0, 1]∫x⁵ dx is the limit of these Riemann sums as the mesh of the partitions approaches 0, which happens as n→∞. This means that:

[n→∞]lim [k=1, n]∑k⁵/n⁶ = [0, 1]∫x⁵ dx

And by the fundamental theorem of calculus, [0, 1]∫x⁵ dx = x⁶/6 |[0, 1] = 1/6.

#2: As mentioned the right Riemann sum for a function on the interval [0, 1] on the partition of the interval into n segments of equal length is:

[k=1, n]∑f(k/n)*1/n

Letting f(x)=x³, we have:

[k=1, n]∑k³/n⁴

Obviously, then:

[n→∞]lim [k=1, n]∑k³/n⁴ = [0, 1]∫x³ dx = x⁴/4 |[0, 1] = 1/4