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CIE 9709 2026 June Paper 31 Q8

A Level / CIE / P3

CIE 9709 2026 June Paper 31 Paper · Question 8

题目

Problem

Find the exact value of ∫013x2e3x dx\displaystyle\int_0^1 3x^2\mathrm{e}^{3x}\,\mathrm{d}x.

(6)
题目中文翻译

求 ∫013x2e3x dx\displaystyle\int_0^1 3x^2\mathrm{e}^{3x}\,\mathrm{d}x 的精确值。

(6)
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

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先对指数函数积分,连续使用两次分部积分,使多项式的次数逐次降低。

答题过程

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Let I=∫013x2e3x dxI=\int_0^1 3x^2e^{3x}\,\mathrm{d}x. Using integration by parts gives

I=[x2e3x]01−∫012xe3x dx.I=\big[x^2e^{3x}\big]_0^1-\int_0^1 2xe^{3x}\,\mathrm{d}x.

For the remaining integral, integrate by parts again:

∫012xe3x dx= [23xe3x]01−23∫01e3x dx= [23xe3x−29e3x]01.\begin{align*} \int_0^1 2xe^{3x}\,\mathrm{d}x =&\,\bigg[\frac23xe^{3x}\bigg]_0^1 -\frac23\int_0^1e^{3x}\,\mathrm{d}x\\[4mm] =&\,\bigg[\frac23xe^{3x}-\frac29e^{3x}\bigg]_0^1. \end{align*}

Therefore

I= [e3x(x2−23x+29)]01= e3(1−23+29)−29= 59e3−29.\begin{align*} I=&\,\bigg[e^{3x}\bigg(x^2-\frac23x+\frac29\bigg)\bigg]_0^1\\[4mm] =&\,e^3\bigg(1-\frac23+\frac29\bigg)-\frac29\\[4mm] =&\,\frac59e^3-\frac29. \end{align*}

解法二

思路

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也可设原函数为指数函数乘一个二次多项式。求导并比较系数,可以直接确定这个多项式。

答题过程

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Let an antiderivative be e3x(ax2+bx+c)e^{3x}(ax^2+bx+c). Differentiating gives

e3x(3ax2+(2a+3b)x+b+3c).e^{3x}\bigl(3ax^2+(2a+3b)x+b+3c\bigr).

Matching this to 3x2e3x3x^2e^{3x} gives

3a= 3,2a+3b= 0,b+3c= 0.\begin{align*} 3a=&\,3,\\[4mm] 2a+3b=&\,0,\\[4mm] b+3c=&\,0. \end{align*}

Thus a=1a=1, b=−2/3b=-2/3 and c=2/9c=2/9. Evaluating at 11 and 00 gives

e3(1−23+29)−29=59e3−29.e^3\bigg(1-\frac23+\frac29\bigg)-\frac29 =\frac59e^3-\frac29.