Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2024 November Paper 22 Q1

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 22 Paper · Question 1

题目

Problem

Find the value of

∫671(x−5)2−1 dx,\int_6^7 \frac{1}{\sqrt{(x - 5)^2 - 1}}\,\mathrm{d}x,

giving your answer in the form ln⁡(a+b)\ln(a + \sqrt{b}), where aa and bb are integers to be determined.

[4]
题目中文翻译

求

∫671(x−5)2−1 dx\int_6^7 \frac{1}{\sqrt{(x - 5)^2 - 1}}\,\mathrm{d}x

的值,并将答案写成 ln⁡(a+b)\ln(a + \sqrt{b}) 的形式,其中 aa 和 bb 为待确定的整数。

(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

解法一

思路

展开

识别反双曲余弦的积分形式。下限处分母为零,先从 66 的右侧取极限,再将答案改写成对数。

答题过程

展开

For x>6x>6, an antiderivative is arcosh⁡(x−5)\operatorname{arcosh}(x-5). Thus

 ∫671(x−5)2−1 dx= lim⁡ε→0+[arcosh⁡(x−5)]6+ε7= arcosh⁡2−arcosh⁡1= ln⁡(2+3).\begin{align*} &\,\int_6^7\frac1{\sqrt{(x-5)^2-1}}\,\mathrm{d}x\\[4mm] =&\,\lim_{\varepsilon\to0^+}[\operatorname{arcosh}(x-5)]_{6+\varepsilon}^7\\[4mm] =&\,\operatorname{arcosh}2-\operatorname{arcosh}1\\[4mm] =&\,\boxed{\ln(2+\sqrt3)}. \end{align*}

Here arcosh⁡u=ln⁡(u+u2−1)\operatorname{arcosh}u=\ln(u+\sqrt{u^2-1}), so arcosh⁡2=ln⁡(2+3)\operatorname{arcosh}2=\ln(2+\sqrt3) and arcosh⁡1=ln⁡1=0\operatorname{arcosh}1=\ln1=0. Hence a=2a=2 and b=3b=3.