题目

Problem

The constant $a$ is such that $\displaystyle\int_0^a x\mathrm{e}^{-2x}\,\mathrm{d}x=\frac18$ .

(a) Show that $a=\frac12\ln(4a+2)$ .

(5)

(b) Verify by calculation that $a$ lies between $0.5$ and $1$ .

(2)

(c) Use an iterative formula based on the equation in (a) to determine $a$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places.

(3)

题目中文翻译

常数 $a$ 满足 $\displaystyle\int_0^a x\mathrm{e}^{-2x}\,\mathrm{d}x=\frac18$ 。

(a) 证明 $a=\frac12\ln(4a+2)$ 。

(5)

(b) 通过计算验证 $a$ 位于 $0.5$ 与 $1$ 之间。

(2)

(3)

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

用分部积分求定积分，代入 $1/8$ 后整理出指数等式，再取对数。这个推导本身不必假定 $a$ 为正。

答题过程

展开

Integrating by parts with $u=x$ and $\mathrm{d}v=e^{-2x}\,\mathrm{d}x$ ,

\begin{align*} \int xe^{-2x}\,\mathrm{d}x =&\,-\frac x2e^{-2x}+\frac12\int e^{-2x}\,\mathrm{d}x\\[4mm] =&\,-\frac x2e^{-2x}-\frac14e^{-2x}+C. \end{align*}

Using the given definite integral,

-\frac a2e^{-2a}-\frac14e^{-2a}+\frac14=\frac18.

Hence

\begin{align*} (4a+2)e^{-2a}=&\,1,\\[4mm] e^{2a}=&\,4a+2. \end{align*}

In particular $4a+2>0$ , so taking logarithms is valid:

2a=\ln(4a+2).

Therefore

\boxed{a=\frac12\ln(4a+2)}.

CIE 9709 2023 June Paper 31 Q9

题目

解答

(a)

解法一

思路

答题过程