$\int_0^1 x^{2} = 1/3$, computed from the Darboux definition with uniform partitions and the closed form $\sum_{k<n} k^{2} = n(n-1)(2n-1)/6$

Let $f : [0,1] \to \mathbb{R}$ be $f(x) = x^{2}$ (Integer powers $a^m$). Then $f$ is Riemann integrable on $[0,1]$ and

$\int_0^1 x^{2}\,\mathrm{d}x \;=\; \frac{1}{3} .$

Everything is computed from the definition. For the uniform partition $U_N$ of $[0,1]$ into $N \ge 1$ parts (Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions) the two Darboux sums (For bounded $f$ on $[a,b]$ and a partition $P$: the infimum $m_i$ and supremum $M_i$ of $f$ on the $i$-th subinterval, and the lower and upper Darboux sums $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$) are

$L(f,U_N) \;=\; \frac{(N-1)(2N-1)}{6N^{2}}, \qquad U(f,U_N) \;=\; \frac{(N+1)(2N+1)}{6N^{2}}, \qquad U(f,U_N) - L(f,U_N) \;=\; \frac{1}{N},$

where a natural $N$ multiplying or dividing a real stands for its canonical natural $\iota(N)$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), as in clause 2 of Laws of finite sums and finite products. Both expressions converge to $1/3$ as $N$ grows, the lower sums from below and the upper sums from above, and the gap $1/N$ is what Archimedes' property drives to $0$.

The arithmetic rests on one closed form, valid for every $n \in \mathbb{N}$:

$6\sum_{k<n} \iota(k)^{2} \;=\; \iota(n)\bigl(\iota(n)-1\bigr)\bigl(2\iota(n)-1\bigr) ,$

which at $n = 0$ reads $0 = 0$ and is the familiar $n(n-1)(2n-1)/6$ written so that the empty sum is the case $n = 0$ rather than an exception. Note the indexing: the sum runs over $k < n$, so its last term is $\iota(n-1)^{2}$ and not $\iota(n)^2$.

• The first index is $0$ and the last term of the lower sum is not $1$. The lower sum uses $m_i = t_i^{2}$ with $i$ running from $0$, so its first term is $0$ and its last is $\bigl((N-1)/N\bigr)^{2}\cdot(1/N)$; the upper sum uses $M_i = t_{i+1}^{2}$ and so ends at $1 \cdot (1/N)$. Reading the closed form with the other convention, $\sum_{k=1}^{n}k^{2} = n(n+1)(2n+1)/6$, and attaching it to the wrong sum is the standard way to lose the factor and land on $1/2$ or $1/6$ instead of $1/3$.

• The gap is exactly $1/N$, not merely $O(1/N)$. Step 2.3 computes it by telescoping without evaluating either sum, which is also the cheapest route to integrability through Riemann's criterion: a bounded $f$ on $[a,b]$ is Darboux integrable if and only if for every real $\varepsilon > 0$ there is a partition $P$ with $U(f,P) - L(f,P) < \varepsilon$: the continuity of $x^2$ is not needed for that, only for the shortcut taken in step 1.1.

• What this example does not do. No antiderivative and no fundamental theorem of calculus is used or available at this point in the reading order; the value $1/3$ is extracted from the two sums and the Archimedean property alone. The same computation with $x^{3}$ needs $\sum_{k<n}k^{3}$ and is no harder, but it is not carried out here.