Compare a left-endpoint sum with the integral · source reference

Printed question

The graph $y=2^x$, $0\le x\le1$, is shown. Rectangles, each width $1/n$, where $n$ is an integer, are drawn under the curve. Show their total area is $1/[n(2^{1/n}-1)]$. [2]

By considering the area bounded by $y=2^x$, $x=1$ and the axes, briefly explain why $\ln2<n(2^{1/n}-1)$. [3]

[2008/NJC/Prelim/I/3]

Printed working

$$\begin{aligned}\text{total area}&=\frac1n[2^0+2^{1/n}+2^{2/n}+\cdots+2^{(n-1)/n}]\\&=\frac1n\left[\frac{1((2^{1/n})^n-1)}{2^{1/n}-1}\right]=\frac{2^1-1}{n(2^{1/n}-1)}=\frac1{n(2^{1/n}-1)}\quad\text{(shown)}.\end{aligned}$$

All rectangles lie below the curve. Total area of rectangles $<$ area under curve on $[0,1]$:

$$\frac1{n(2^{1/n}-1)}<\int_0^12^x dx=\frac1{\ln2}[2^x]_0^1=\frac{2-1}{\ln2}.$$

Hence $\ln2<n(2^{1/n}-1)$ (shown).

Teacher notes

The source diagram labels O, $y=2^x$, $1/n,2/n,3/n,(n-3)/n,(n-2)/n,(n-1)/n,1$. Lower rectangles use left-end heights, with a middle gap indicating continuation. Native rendering retains all symbolic labels.

Printed answer key

No separate short-key entry is printed for Q14; its two “shown” results are supplied in the detailed solution.

Source diagrams in native form