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CIE 9231 2025 Nov Paper 21 Q5

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 21 Paper · Question 5

题目

Problem

The diagram shows the curve with equation y=13x3+xy = \frac{1}{3}x^3 + x for 0≤x≤10 \leq x \leq 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

∫01(13x3+x) dx<Un,\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\mathrm{d}x < U_n,

where

Un=112(1+1n)(7+1n).U_n = \frac{1}{12}\bigg(1 + \frac{1}{n}\bigg) \bigg(7 + \frac{1}{n}\bigg).
[5]

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

∫01(13x3+x) dx.\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\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

题图显示曲线 y=13x3+xy = \frac{1}{3}x^3 + x，其中 0≤x≤10 \leq x \leq 1，以及一组宽为 1n\frac{1}{n} 的 nn 个矩形。

(a) 通过考虑这些矩形面积之和，证明

∫01(13x3+x) dx<Un,\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\mathrm{d}x < U_n,

其中

Un=112(1+1n)(7+1n).U_n = \frac{1}{12}\bigg(1 + \frac{1}{n}\bigg) \bigg(7 + \frac{1}{n}\bigg).
(5)

(b) 用类似方法，求出关于 nn 的

∫01(13x3+x) dx\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\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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函数严格递增。题图中每个矩形都取区间右端点的高度，因此矩形面积和严格大于积分。把高度中的一次项和三次项分别求和。

答题过程

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Let f(x)=13x3+xf(x)=\frac13x^3+x. Since f′(x)=x2+1>0f'(x)=x^2+1>0, the right-endpoint rectangles lie above the curve on the interior of each interval. Their total area can be written as

Un= 1n[f(1n)+f(2n)+f(3n) +⋯+f(n−1n)+f(1)].\begin{align*} U_n=&\,\frac1n\Big[f\Big(\frac1n\Big)+f\Big(\frac2n\Big)+f\Big(\frac3n\Big)\\ &\,\hspace{2pt}+\cdots+f\Big(\frac{n-1}{n}\Big)+f(1)\Big]. \end{align*}

Using ∑r=1nr=n(n+1)/2\sum_{r=1}^n r=n(n+1)/2 and ∑r=1nr3=n2(n+1)2/4\sum_{r=1}^n r^3=n^2(n+1)^2/4, we obtain

Un= 1n∑r=1nf(rn)=13n4∑r=1nr3+1n2∑r=1nr= (n+1)212n2+n+12n=112(1+1n)(7+1n).\begin{align*} U_n=&\,\frac1n\sum_{r=1}^n f\bigg(\frac rn\bigg) =\frac{1}{3n^4}\sum_{r=1}^n r^3+\frac1{n^2}\sum_{r=1}^n r\\ =&\,\frac{(n+1)^2}{12n^2}+\frac{n+1}{2n} =\frac1{12}\bigg(1+\frac1n\bigg)\bigg(7+\frac1n\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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改用左端点的高度。因为 f(0)=0f(0)=0，可以从 r=1r=1 开始求和，到 n−1n-1 为止。

答题过程

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The left-endpoint rectangles lie below the increasing curve, so their total area is a lower bound. Using f(0)=0f(0)=0,

Ln= 1n∑r=0n−1f(rn)= 13n4∑r=1n−1r3+1n2∑r=1n−1r= (n−1)212n2+n−12n=112(1−1n)(7−1n).\begin{align*} L_n=&\,\frac1n\sum_{r=0}^{n-1} f\bigg(\frac rn\bigg)\\ =&\,\frac1{3n^4}\sum_{r=1}^{n-1}r^3+\frac1{n^2}\sum_{r=1}^{n-1}r\\ =&\,\frac{(n-1)^2}{12n^2}+\frac{n-1}{2n} =\frac1{12}\bigg(1-\frac1n\bigg)\bigg(7-\frac1n\bigg). \end{align*}

解法二

思路

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也可直接比较两组取样点。它们只有最左和最右的高度没有抵消。

答题过程

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The left-endpoint area sum LnL_n and the right-endpoint area sum UnU_n use the same heights at all interior division points. Thus

Ln= Un−f(1)−f(0)n= (n+1)(7n+1)−16n12n2= (n−1)(7n−1)12n2.\begin{align*} L_n=&\,U_n-\frac{f(1)-f(0)}n\\ =&\,\frac{(n+1)(7n+1)-16n}{12n^2}\\ =&\,\frac{(n-1)(7n-1)}{12n^2}. \end{align*}

(c)

解法一

思路

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将两个界的表达式相减，常数项和 1n2\frac1{n^2} 项抵消。

答题过程

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Subtracting gives

Un−Ln= (n+1)(7n+1)−(n−1)(7n−1)12n2= 43n⟶0(n→∞).\begin{align*} U_n-L_n=&\,\frac{(n+1)(7n+1)-(n-1)(7n-1)}{12n^2}\\ =&\,\frac{4}{3n}\longrightarrow 0\qquad (n\to\infty). \end{align*}