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Find the area $A$ A_p(a,b)$of the region under$x^p$for$a \le x \le b.$We know would write this with an integral sign and use the answer we wantFundamental theorem of Calculus but they did not have those tools (as is the case with modern day Calculus students for a lecture or two). Even without these tools they (and modern day Calculus students) show for$p=0,1,2,3$that $$A_p(a,b)=\frac{b^{p+1}-a^{p+1}}{b-a}.$$ It is easy to get as a corollary the cases$p=1/2,1/3$and the conjecture is and was obvious: The equation should hold for "all"$\frac{b^{p+1}-a^{p+1}}{b-a},$but we insist on establishing a,b,p$ except when it using doesn't (make sense). There is much further supporting evidence, but that is not a proof. Useful identities for $p=1,2,3$ are $S_2(n)=\sum_1^nk^2=\frac{n(n+1)(2n+1)}{6}$ along with $S_1(n)=\frac{n(n+1)}{2}$ and $S_3(n)=S_1(n)^2.$ The desired method is a familiar kind of geometric/algebraic argument using the area under step functions over $[a,b].$
$$\sum_1^nx_{i-1}^p(x_i-x_{i-1}) \lt A A_p(a,b) \lt \sum_1^nx_{i}^p(x_i-x_{i-1}).$$ 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 $(b^p-(b-\epsilon)^p)(b-a) \lt p \epsilon b^p (b-a)$ which converges to zero. (For $p \lt 0,$ use $a-(a+\epsilon)^p$.)