How to compute the Riemann sum of $x^n$ for any positive integer $n$?

Question

I'm trying to compute the integral $$ \int_{0}^{1}{x^n}~dx $$ for any given positive integer $n$ using ONLY the Riemann sums. I'd really appreciate it if you could help me out. Thanks.

Answer 1

One Riemann sum for the integral is $$\frac{1}{m} \sum_{i = 1}^m \left(\frac{i}{m}\right)^n = \frac{1}{m^{n + 1}} \sum_{i = 1}^m i^n.$$ This sum is given by Faulhaber's Formula, and is an polynomial of degree $n + 1$ in $m$: $$ \sum_{i = 1}^m i^n = \frac{1}{n + 1} \sum_{j = 0}^n (-1)^j {{n + 1}\choose{j}} B_j m^{n + 1 - j} = \frac{1}{n + 1} m^{n + 1} + \cdots, $$ where $B_j$ is the $j$th Bernoulli number (but it turns out that all we need to know here is that $B_0 = 1$, which gives the first term in the rightmost expressions).

When taking the limit as $m \to \infty$ to evaluate the integral, we see that only the leading term of the polynomial contributes anything: By the above, $$\sum_{i = 1}^m i^n = \frac{1}{n + 1} m^{n + 1} + r(m)$$ for some remainder $r(m)$ of degree $\leq n$, and so $$\int_0^1 x^n \,dx = \lim_{m \to \infty} \frac{1}{m^{n + 1}} \sum_{i = 1}^m i^n = \lim_{m \to \infty} \frac{1}{m^{n + 1}} \left[\frac{1}{n + 1} m^{n + 1} + r(m)\right] = \frac{1}{n + 1}.$$

Answer 2

Take a partition of $[0,1]$ comprising of $N$ equal segments: $P = [0, 1/N, 2/N, ..., (N-1)/N, 1]$. Then the Riemann upper sum for $f(x) = x^n$ is

$$U(f,P) = \sum_{j=1}^N \frac{1}{N} \left(\frac{j}{N}\right)^n = \frac{1}{N^{n+1}} \sum_{j=1}^N j^n $$

You can write a similar expression for the lower Riemann sum $L(f,P)$. But as $f$ is continuous, we know already the upper and lower sums will be equal in the limit, so let's just stick with the upper sum.

Now that sum is a bit tricky: $$ \sum_{j=1}^N j^n = \frac{1}{1+n} N^{n+1} + \frac{1}{2} N^n + \frac{1}{n+1} \sum_{k=2}^n {p+1 \choose k} B_k N^{n+1-k}$$

where the $B_k$ are Bernoulli numbers. Taking that as read, you can see that all terms of the Riemann sum except the first one will result in reciprocal powers of $N$. Hence

$$\lim_{N\to\infty} U(f,P) = \lim_{N\to\infty} \left( \frac{1}{n+1} + \frac{1}{2} \frac{1}{N} + \frac{1}{n+1} \sum_{k=2}^n {p+1 \choose k} B_k \frac{1}{N^k} \right) = \frac{1}{n+1}$$