$\int_0^1 x^{m} = 1/\iota(m+1)$, computed by the fundamental theorem and checked against the definition

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

$\int_0^1 x^{m}\,\mathrm{d}x \;=\; \frac{1}{\iota(m+1)} ,$

where $\iota(m+1)$ is the canonical natural of $m+1$ in $\mathbb{R}$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field) and is positive because $m+1 \ge 1$ (Canonical naturals are positive and strictly increasing).

The $\iota$ is not decoration. A natural number is a von Neumann natural, that is a set, so $m+1$ is not an element of $\mathbb{R}$ and $1/(m+1)$ is not an expression of the field; what the display says is $1/\iota(m+1)$, and that is why the reader meets $\iota$ here at all.

Two independent checks are carried out below: the value at $m = 0$ against the published formula for the integral of a constant (If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$), and the monotonicity of the answer in $m$ against the pointwise inequality $x^{m+1} \le x^{m}$ on $[0,1]$.

• The primitive is the only input, and the index range inside it is the trap. For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term states its power rule for $n \ge 1$ only, because at $n = 0$ the formula $\iota(0)c^{-1}$ mentions $c^{-1}$, which is undefined at $c = 0$. The exponent used above is $m+1$, which is $\ge 1$ for every $m \in \mathbb{N}$ including $m = 0$, so claim 2 of that lemma applies with no case split and the example is correct at $m = 0$ as well.

• The same computation from the definition is a different exercise. The previous page's companion evaluates $\int_0^1 x^{2}$ from Darboux sums on uniform partitions and a closed form for $\sum_{k<n}\iota(k)^{2}$; that is the value $1/3$, which is $1/\iota(3)$ and agrees with the formula above at $m = 2$. The point of computing by the fundamental theorem is that no closed form for a power sum is needed at any $m$.

• Linearity extends this to every polynomial. For $P(x) = \sum_{j<n}a_jx^{j}$ one gets $\int_0^1 P = \sum_{j<n}a_j/\iota(j+1)$ from Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$; that is a routine consequence and is not stated as a separate claim.