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CIE 9231 2024 June Paper 21 Q5

A Level / CIE / FP2

CIE 9231 2024 June Paper 21 Paper · Question 5

题目

Problem

The diagram shows the curve with equation y=2x−x2y=2x-x^2 for 0≤x≤10\le x\le1, 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

∫01(2x−x2) dx<Un,\int_0^1(2x-x^2)\,\mathrm{d}x<U_n,

where

Un=(1+1n)(23−16n).U_n=\left(1+\frac{1}{n}\right)\left(\frac{2}{3}-\frac{1}{6n}\right).
[5]

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

∫01(2x−x2) dx.\int_0^1(2x-x^2)\,\mathrm{d}x.
[4]

(c) Show that

lim⁡n→∞(Un−Ln)=0.\lim_{n\to\infty}(U_n-L_n)=0.
[2]
题目中文翻译

图中显示曲线 y=2x−x2y=2x-x^2，其中 0≤x≤10\le x\le1，以及一组宽度均为 1n\frac{1}{n} 的 nn 个矩形。

(a) 通过考虑这些矩形的面积之和，证明

∫01(2x−x2) dx<Un\int_0^1(2x-x^2)\,\mathrm{d}x<U_n

其中

Un=(1+1n)(23−16n)U_n=\left(1+\frac{1}{n}\right)\left(\frac{2}{3}-\frac{1}{6n}\right)
(5)

(b) 使用类似方法，求关于 nn 的积分

∫01(2x−x2) dx\int_0^1(2x-x^2)\,\mathrm{d}x

的下界 LnL_n。

(4)

(c) 证明

lim⁡n→∞(Un−Ln)=0\lim_{n\to\infty}(U_n-L_n)=0
(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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曲线在 [0,1][0,1] 上严格递增，题图取右端点。将面积和拆成一次和二次两项，分别使用求和公式。

答题过程

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Let f(x)=2x−x2f(x)=2x-x^2. It is strictly increasing on [0,1][0,1], so the area sum of the right-endpoint rectangles is a strict upper bound. The areas are 1nf(1/n)\frac1n f(1/n), 1nf(2/n)\frac1n f(2/n), 1nf(3/n)\frac1n f(3/n), …\ldots, 1nf((n−1)/n)\frac1n f((n-1)/n), 1nf(1)\frac1n f(1). Their sum is

Un= 1n∑r=1nf(r/n)= 2n2∑r=1nr−1n3∑r=1nr2= 2n2⋅n(n+1)2 −n(n+1)(2n+1)6n3= n+1n−(n+1)(2n+1)6n2= (1+1n)(23−16n).\begin{align*} U_n=&\,\frac1n\sum_{r=1}^n f(r/n)\\ =&\,\frac2{n^2}\sum_{r=1}^n r-\frac1{n^3}\sum_{r=1}^n r^2\\ =&\,\frac{2}{n^2}\cdot\frac{n(n+1)}2\\ &\,\hspace{2pt}-\frac{n(n+1)(2n+1)}{6n^3}\\ =&\,\frac{n+1}{n}-\frac{(n+1)(2n+1)}{6n^2}\\ =&\,\bigg(1+\frac1n\bigg)\bigg(\frac23-\frac1{6n}\bigg). \end{align*}

Hence ∫01f(x) dx<Un\int_0^1 f(x)\,\mathrm{d}x<U_n.

(b)

解法一

思路

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改用左端点。由于 f(0)=0f(0)=0，一次项和二次项都只需从 11 加到 n−1n-1。

答题过程

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The left-endpoint rectangles give a lower bound because the curve is increasing. Their area sum is

Ln= 2n2∑r=1n−1r−1n3∑r=1n−1r2= 2n2⋅(n−1)n2 −(n−1)n(2n−1)6n3= n−1n−(n−1)(2n−1)6n2= (1−1n)(23+16n).\begin{align*} L_n=&\,\frac2{n^2}\sum_{r=1}^{n-1}r-\frac1{n^3}\sum_{r=1}^{n-1}r^2\\ =&\,\frac{2}{n^2}\cdot\frac{(n-1)n}{2}\\ &\,\hspace{2pt}-\frac{(n-1)n(2n-1)}{6n^3}\\ =&\,\frac{n-1}{n}-\frac{(n-1)(2n-1)}{6n^2}\\ =&\,\bigg(1-\frac1n\bigg)\bigg(\frac23+\frac1{6n}\bigg). \end{align*}

解法二

思路

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也可在右端点和中去掉 f(1)/nf(1)/n，再加上 f(0)/nf(0)/n。所有内部取样点都相同。

答题过程

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The left-endpoint sum differs from UnU_n only in replacing the final height f(1)=1f(1)=1 with the first height f(0)=0f(0)=0. Thus

Ln=Un−1n=(1−1n)(23+16n).L_n=U_n-\frac1n =\bigg(1-\frac1n\bigg)\bigg(\frac23+\frac1{6n}\bigg).

(c)

解法一

思路

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将上下界展开相减，常数项和二次小量抵消，只剩 1n\frac1n。

答题过程

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Subtracting the two bounds gives

Un−Ln= (23+12n−16n2) −(23−12n−16n2)= 1n⟶0.\begin{align*} U_n-L_n=&\,\bigg(\frac23+\frac1{2n}-\frac1{6n^2}\bigg)\\ &\,\hspace{2pt}-\bigg(\frac23-\frac1{2n}-\frac1{6n^2}\bigg)\\ =&\,\frac1n\longrightarrow0. \end{align*}