Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2025 Nov Paper 24 Q1

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 24 Paper · Question 1

题目

Problem

It is given that In=∫01xnsinh⁡x dxI_n = \int_0^1 x^n\sinh x\,\mathrm{d}x, where nn is a non-negative integer.

(a) Show that, for n≥2n \ge 2,

In=cosh⁡1−nsinh⁡1+n(n−1)In−2.I_n = \cosh 1 - n\sinh 1 + n(n - 1)I_{n - 2}.
[3]

(b) Find the exact value of I2I_2.

[2]
题目中文翻译

已知 In=∫01xnsinh⁡x dxI_n = \int_0^1 x^n\sinh x\,\mathrm{d}x,其中 nn 是非负整数。

(a) 证明当 n≥2n \ge 2 时,

In=cosh⁡1−nsinh⁡1+n(n−1)In−2.I_n = \cosh 1 - n\sinh 1 + n(n - 1)I_{n - 2}.
(3)

(b) 求 I2I_2 的精确值。

(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

展开

连续做两次分部积分,每次把 xx 的幂降低一次。因为 n≥2n\ge2,两次得到的边界项在 x=0x=0 都为零。

答题过程

展开

First use u=xnu=x^n, dv=sinh⁡x dx\mathrm{d}v=\sinh x\,\mathrm{d}x, so v=cosh⁡xv=\cosh x. For the remaining integral, use u=xn−1u=x^{n-1} and dv=cosh⁡x dx\mathrm{d}v=\cosh x\,\mathrm{d}x, giving

∫01xn−1cosh⁡x dx=[xn−1sinh⁡x]01−(n−1)In−2.\int_0^1x^{n-1}\cosh x\,\mathrm{d}x =[x^{n-1}\sinh x]_0^1-(n-1)I_{n-2}.

Thus the two integrations by parts give

In= [xncosh⁡x]01−n∫01xn−1cosh⁡x dx= [xncosh⁡x]01−n[xn−1sinh⁡x]01 +n(n−1)∫01xn−2sinh⁡x dx= cosh⁡1−nsinh⁡1+n(n−1)In−2.\begin{align*} I_n =&\,[x^n\cosh x]_0^1-n\int_0^1 x^{n-1}\cosh x\,\mathrm{d}x\\ =&\,[x^n\cosh x]_0^1-n[x^{n-1}\sinh x]_0^1\\ &\,\hspace{2pt}+n(n-1)\int_0^1x^{n-2}\sinh x\,\mathrm{d}x\\ =&\,\cosh1-n\sinh1+n(n-1)I_{n-2}. \end{align*}

(b)

解法一

思路

展开

在递推式中取 n=2n=2,再直接积分求 I0I_0。

答题过程

展开

Since I0=[cosh⁡x]01=cosh⁡1−1I_0=[\cosh x]_0^1=\cosh1-1, we have

I2=cosh⁡1−2sinh⁡1+2I0=3cosh⁡1−2sinh⁡1−2.I_2=\cosh1-2\sinh1+2I_0 =\boxed{3\cosh1-2\sinh1-2}.