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CIE 9231 2023 June Paper 21 Q4

A Level / CIE / FP2

CIE 9231 2023 June Paper 21 Paper · Question 4

题目

Problem

The integral InI_n is defined by In=∫01(1+x5)n dxI_n=\displaystyle\int_0^1\big(1+x^5\big)^n\,\mathrm{d}x.

(a) By considering ddx(x(1+x5)n)\dfrac{\mathrm{d}}{\mathrm{d}x}\big(x\big(1+x^5\big)^n\big), or otherwise, show that

(5n+1)In=2n+5nIn−1.(5n+1)I_n=2^n+5nI_{n-1}.
(5)

(b) Find the exact value of I3I_3.

(4)
题目中文翻译

积分 InI_n 定义为 In=∫01(1+x5)n dxI_n=\displaystyle\int_0^1\big(1+x^5\big)^n\,\mathrm{d}x。

(a) 通过考虑 ddx(x(1+x5)n)\dfrac{\mathrm{d}}{\mathrm{d}x}\big(x\big(1+x^5\big)^n\big),或用其他方法,证明

(5n+1)In=2n+5nIn−1.(5n+1)I_n=2^n+5nI_{n-1}.
(5)

(b) 求 I3I_3 的精确值。

(4)
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

展开

求乘积的导数,再把其中的 x5x^5 写成 (1+x5)−1(1+x^5)-1。这样积分后只含相邻两个 II。

答题过程

展开

The product rule gives

ddx[x(1+x5)n]= (1+x5)n+5nx5(1+x5)n−1= (5n+1)(1+x5)n−5n(1+x5)n−1.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\big[x(1+x^5)^n\big] =&\,(1+x^5)^n+5nx^5(1+x^5)^{n-1}\\ =&\,(5n+1)(1+x^5)^n-5n(1+x^5)^{n-1}. \end{align*}

Integrating on [0,1][0,1] gives 2n=(5n+1)In−5nIn−12^n=(5n+1)I_n-5nI_{n-1}. This rearranges to (5n+1)In=2n+5nIn−1(5n+1)I_n=2^n+5nI_{n-1}.

解法二

思路

展开

把被积函数看作 (1+x5)n⋅1(1+x^5)^n\cdot1,分部积分后用 x5=(1+x5)−1x^5=(1+x^5)-1,便出现 InI_n 和 In−1I_{n-1}。

答题过程

展开

Take u=(1+x5)nu=(1+x^5)^n and dv=dx\mathrm{d}v=\mathrm{d}x, so du=5nx4(1+x5)n−1 dx\mathrm{d}u=5nx^4(1+x^5)^{n-1}\,\mathrm{d}x and v=xv=x. Integration by parts gives

In= [x(1+x5)n]01 −5n∫01x5(1+x5)n−1 dx= 2n−5n∫01[(1+x5)−1](1+x5)n−1 dx= 2n−5nIn+5nIn−1.\begin{align*} I_n=&\,\big[x(1+x^5)^n\big]_0^1\\ &\,\hspace{2pt}-5n\int_0^1x^5(1+x^5)^{n-1}\,\mathrm{d}x\\ =&\,2^n-5n\int_0^1\big[(1+x^5)-1\big](1+x^5)^{n-1}\,\mathrm{d}x\\ =&\,2^n-5nI_n+5nI_{n-1}. \end{align*}

Hence (5n+1)In=2n+5nIn−1(5n+1)I_n=2^n+5nI_{n-1}.

(b)

解法一

思路

展开

由 I0=1I_0=1 开始,依次取 n=1,2,3n=1,2,3。每一步保留分数。

答题过程

展开

Starting from I0=1I_0=1, the recurrence gives

I1= 2+56=76,I2= 4+10(7/6)11=4733.\begin{align*} I_1=&\,\frac{2+5}{6}=\frac76,\\ I_2=&\,\frac{4+10(7/6)}{11}=\frac{47}{33}. \end{align*}

Then

I3=8+15(47/33)16=323176.I_3=\frac{8+15(47/33)}{16}=\boxed{\frac{323}{176}}.

解法二

思路

展开

n=3n=3 时也可直接展开被积函数。各项都是整数次幂,逐项积分即可。

答题过程

展开

Since (1+x5)3=1+3x5+3x10+x15(1+x^5)^3=1+3x^5+3x^{10}+x^{15},

I3=1+36+311+116=323176.I_3=1+\frac36+\frac3{11}+\frac1{16}=\frac{323}{176}.