Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 18 / a

Partition the finite interval as $a=x_0<x_1<\cdots <x_N=b$, with $\Delta x_i=x_{i+1}-x_i$. For a Riemann integrable function, the left-endpoint Riemann sums satisfyThe maximum subinterval width, rather than merely the number of subintervals, must tend to zero. Continuity of $f$ on the closed interval is a sufficient integrability hypothesis. More generally, any chosen sample point in each subinterval gives the same definite integral under these assumptions.

Index the stated geometric progression by $x_i=(1+ϵ{)}^i$, $i=0,1,\dots$. Then $\Delta x_i=ϵ(1+ϵ{)}^i$, and its left-endpoint contribution isHence the approximation is a geometric series withSince $ϵ>0$ and $\alpha >0$, the ratio satisfies $0<q<1$ and the series converges. The integrand is decreasing, so the left-endpoint sum is an upper bound for the improper integral.

The binomial series gives $(1+ϵ{)}^{-\alpha }=1-\alpha ϵ+O({ϵ}^2)$. ThereforeTo justify the limiting approximation over the infinite interval, also use the right-endpoint sum. It is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$, and monotonicity gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds have the same limit. The geometric partition bounds for a power-law improper integral and the squeeze theorem thus proveThis bounding argument matters because the geometric partition has unbounded widths far along the half-line; a finite-interval maximum-mesh argument cannot simply be applied to the entire infinite partition at fixed $ϵ$.