CONTINUED To recap, we already know the answer , $\frac{b^{p+1}-a^{p+1}}{b-a}$, which we want for the area $A$ of the region under $x^p$ for $a \le x \le b$, we just want to prove it. (Assume for ease that $0 \lt a$.) A partition $P$ of $[a,b]$ is a sequence $a=x_0 \lt x_1 \lt \cdots \lt x_n=b$. The mesh $m(P)$ of $P$ is $\max(x_{i}-x_{i-1}).$ (There is rarely a reason to have unequal intervals, but Fermat gave one.) We use the sub-intervals, in two ways, as the bases of an assemblage of rectangles with heights determined by the endpoints. Since $x^p$ is monotonic, one is covered by the region and the other covers it. So the two areas provide a upper lower and an upper bound.
So for $p=5$, obviously $u^5$u^5 \lt \frac{v^5+v^4u+v^3u^2+v^2u^3+vu^4+u^5}{6} \lt v^5.$v^5.$$OK, so what? Why not use the average, the geometric mean or \left(\frac{b+a}{2}\right)^5? \left(\frac{u+v}{2}\right)^5? Well, (v-u)h(u,v)=\frac{v^6-u^6}{6} so \sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1}) collapses to \frac{b^6-a^6}{b-a}. It is slightly more fun to show that About as easily \sqrt{u} \lt \frac{2(v+\sqrt{vu}+u)}{3(\sqrt{v}+\sqrt{u})} \lt \sqrt{v}. It is slightly more fun to show that \sqrt{u} \lt \frac{2(v+\sqrt{vu}+u)}{3(\sqrt{v}+\sqrt{u})} \lt \sqrt{v}. To its credit I'll say that it does not show preference to any particular partition and uses nothing more complex than the two historic treatments above (although maybe it benefits from a modern frame of reference.) Also, rather than carefully converging to the correct answer as the partition evolves, it just starts there and stays unaffected. I don't immediately see that it can be applied to any other definite integrals. But the case of x^p has a certain primary importance. 2 added 2946 characters in body Take (k+1)^{p+1}-k^{p+1}=\sum_1^{p+1}\binom{p+1}{j}k^{p+1-j} and sum for k from 0 to n to get (n+1)^{p+1}-0^{p+1}=\sum_1^{p+1}\binom{p+1}{j}S_{p+1-j}(n). Since we know everything except S_p, the rest is algebra! This is quickly unpleasant and the final results are not as aesthetic as the first cases. HOWEVER, for the desired application we only need to establish that S_p(n)=\frac{n^{p+1}}{p+1}+\frac{n^p}{2}+O(n^{p-1}) That is not hard and shows that n subdivision subdivisions yield \frac{1}{p+1}-\frac{1}{2n} \lt \int_0^1x^pdx \lt \frac{1}{p+1}+\frac{1}{2n}. So that is the flavor of what I am asking about. I do not think this is a big list question unless there are a large number of approaches I have not seen. CONTINUED To recap, we already know the answer , \frac{b^{p+1}-a^{p+1}}{b-a}, which we want for the area A of the region under x^p for a \le x \le b, we just want to prove it. (Assume for ease that 0 \lt a.) A partition P of [a,b] is a sequence a=x_0 \lt x_1 \lt \cdots \lt x_n=b. The mesh m(P) of P is \max(x_{i}-x_{i-1}). (There is rarely a reason to have unequal intervals, but Fermat gave one.) We use the sub-intervals, in two ways, as the bases of an assemblage of rectangles with heights determined by the endpoints. Since x^p is monotonic, one is covered by the region and the other covers it. So the two areas provide a upper and an upper bound.$$ \sum_1^nx_{i-1}^p(x_i-x_{i-1}) \lt A \lt \sum_1^nx_{i}^p(x_i-x_{i-1})$$If we manage to compute or bound these bounds and show that, when the mesh goes to zero, they have a common limit (the one we expect), we are done. The actual bounds we compute are of value only for the interesting, but secondary, topic of speed of convergence. And anyway, if$m(P) \lt \epsilon$then the difference between the two bounds is less than$(b-a)(b^p-(b-\epsilon)^p),$which converges to zero. (For$p \lt 0$use$a-(a+\epsilon)^p$) So I propose to instead assign to each sub-interval$[u,v]$the height$h(u,v)=\frac{v^{p+1}-u^{p+1}}{(p+1)(v-u)}$and "compute"$\sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1})$which immediately collapses to, of course,$\frac{b^{p+1}-a^{p+1}}{p+1}.$Establishing that this has any relevance requires showing that the height$h(u,v)$is between$u^p$and$v^p$. This is easy in practice if one simplifies. If you simplify first, then the whole thing looks like magic until you see what was done. So for$p=5$, obviously$u^5 \lt \frac{v^5+v^4u+v^3u^2+v^2u^3+vu^4+u^5}{6} \lt v^5.$OK, so what? Why not use the average, the geometric mean or$\left(\frac{b+a}{2}\right)^5$? Well,$(v-u)h(u,v)=\frac{v^6-u^6}{6}$so$\sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1})$collapses to$\frac{b^6-a^6}{b-a}$. It is slightly more fun to show that$\frac{1}{\sqrt{v}} \lt \frac{2}{\sqrt{u}+\sqrt{v}} \lt \frac{1}{\sqrt{u}}\sqrt{u} \lt \frac{2(v+\sqrt{vu}+u)}{3(\sqrt{v}+\sqrt{u})} \lt \sqrt{v}.\frac{1}{v^2} \lt \frac{1}{uv} \lt \frac{1}{u^2}\frac{1}{v^4} \lt \left( \frac{1}{v^3u}+\frac{1}{v^2u^2}+\frac{1}{vu^3}\right)/3 \lt \frac{1}{u^4}$SO: Is this line of arguement valid? Is it interseting? Have you seen it before? To its credit I'll say that it does not show preference to any particular partition and uses nothing more complex than the two historic treatments above (although it benefits from a modern frame of reference.) I don't see that it can be applied to any other definite integrals. But the case of$x^p\$ has a certain primary importance.