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CIE 9231 2025 Nov Paper 23 Q3

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 23 Paper · Question 3

题目

Problem

The integral InI_n is defined by

In=∫01(1+x2)n dx.I_n = \int_0^1 \big(1 + x^2\big)^n \,\mathrm{d}x.

(a) By considering ddx(x(1+x2)n)\frac{\mathrm{d}}{\mathrm{d}x}\big(x(1 + x^2)^n\big), or otherwise, show that

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

(b) Find the exact value of I−2I_{-2}.

[4]
题目中文翻译

积分 InI_n 定义为

In=∫01(1+x2)n dx.I_n = \int_0^1 \big(1 + x^2\big)^n \,\mathrm{d}x.

(a) 通过考虑 ddx(x(1+x2)n)\frac{\mathrm{d}}{\mathrm{d}x}\big(x(1 + x^2)^n\big),或用其他方法,证明

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

(b) 求 I−2I_{-2} 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

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用乘积法则求导,把结果中的 x2x^2 写成 (1+x2)−1(1+x^2)-1,就能在积分后得到相邻的两个 II。

答题过程

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The product rule gives

ddx[x(1+x2)n]= (1+x2)n+2nx2(1+x2)n−1= (2n+1)(1+x2)n−2n(1+x2)n−1.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\big[x(1+x^2)^n\big] =&\,(1+x^2)^n+2nx^2(1+x^2)^{n-1}\\ =&\,(2n+1)(1+x^2)^n-2n(1+x^2)^{n-1}. \end{align*}

Integrating from 00 to 11 gives

[x(1+x2)n]01= (2n+1)In−2nIn−1,2n= (2n+1)In−2nIn−1.\begin{align*} \big[x(1+x^2)^n\big]_0^1 =&\,(2n+1)I_n-2nI_{n-1},\\ 2^n=&\,(2n+1)I_n-2nI_{n-1}. \end{align*}

Hence

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

(b)

解法一

思路

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先用反正切求 I−1I_{-1},再在递推关系中取 n=−1n=-1。

答题过程

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We have

I−1=∫0111+x2 dx=[arctan⁡x]01=π4.I_{-1}=\int_0^1\frac1{1+x^2}\,\mathrm{d}x =[\arctan x]_0^1=\frac\pi4.

With n=−1n=-1, the recurrence becomes

−π4=12−2I−2.-\frac{\pi}{4}=\frac12-2I_{-2}.

It follows that I−2=14+π8\boxed{I_{-2}=\frac14+\frac{\pi}{8}}.

解法二

思路

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也可直接作代换 x=tan⁡ux=\tan u,把被积函数和微分合并成 cos⁡2u du\cos^2u\,\mathrm{d}u。

答题过程

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With x=tan⁡ux=\tan u, we have 1+x2=sec⁡2u1+x^2=\sec^2u and dx=sec⁡2u du\mathrm{d}x=\sec^2u\,\mathrm{d}u. Thus

I−2= ∫0π/4cos⁡2u du= [u2+sin⁡2u4]0π/4= π8+14.\begin{align*} I_{-2}=&\,\int_0^{\pi/4}\cos^2u\,\mathrm{d}u\\ =&\,\bigg[\frac{u}{2}+\frac{\sin 2u}{4}\bigg]_0^{\pi/4}\\ =&\,\frac{\pi}{8}+\frac14. \end{align*}