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CIE 9231 2024 November Paper 22 Q6

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 22 Paper · Question 6

题目

Problem

The diagram shows the curve with equation y=e1xy = e^{1 - x} for 0x10 \le x \le 1, together with a set of nn rectangles of width 1n\frac{1}{n}.

(a) By considering the sum of the areas of these rectangles, show that

01e1xdx<Un,\int_0^1 e^{1 - x}\,\mathrm{d}x < U_n,

where

Un=e1n(1e1n).U_n = \frac{e - 1}{n\left(1 - e^{-\frac{1}{n}}\right)}.
[4]

(b) Use a similar method to find, in terms of nn, a lower bound LnL_n for

01e1xdx.\int_0^1 e^{1 - x}\,\mathrm{d}x.
[4]

(c) Show that

limn(UnLn)=0.\lim_{n\to\infty}(U_n - L_n) = 0.
[2]

(d) Use the Maclaurin’s series for exe^x given in the list of formulae (MF19) to find the first three terms of the series expansion of z(1e1z)z\left(1 - e^{-\frac{1}{z}}\right), in ascending powers of 1z\frac{1}{z}, and deduce the value of

limn(Un).\lim_{n\to\infty}(U_n).
[3]
题目中文翻译

图中显示曲线 y=e1xy = e^{1 - x},其中 0x10 \le x \le 1,以及一组宽度均为 1n\frac{1}{n}nn 个矩形。

(a) 通过考虑这些矩形面积之和,证明

01e1xdx<Un\int_0^1 e^{1 - x}\,\mathrm{d}x < U_n

其中

Un=e1n(1e1n)U_n = \frac{e - 1}{n\left(1 - e^{-\frac{1}{n}}\right)}

(b) 使用类似方法,求关于 nn 的积分

01e1xdx\int_0^1 e^{1 - x}\,\mathrm{d}x

的下界 LnL_n

(c) 证明

limn(UnLn)=0\lim_{n\to\infty}(U_n - L_n) = 0

(d) 使用公式表(MF19)中给出的 exe^x 的 Maclaurin 级数,求 z(1e1z)z\left(1 - e^{-\frac{1}{z}}\right) 的级数展开的前三项,按 1z\frac{1}{z} 的升幂排列,并据此推出

limn(Un)\lim_{n\to\infty}(U_n)

的值。

解答