Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2025 March Paper 32 Q11

A Level / CIE / P3

CIE 9709 2025 March Paper 32 Paper · Question 11

题目

Problem

Find the exact value of ∫0πx2cos⁡13x dx\displaystyle\int_0^\pi x^2\cos\frac13x\,\mathrm{d}x.

(6)
题目中文翻译

求 ∫0πx2cos⁡13x dx\displaystyle\int_0^\pi x^2\cos\frac13x\,\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

展开

两次分部积分依次降低多项式次数。三角函数内部的三分之一会使积分系数乘三,务必逐步保留。

答题过程

展开

Let

I=∫0πx2cos⁡(x/3) dx.I=\int_0^\pi x^2\cos(x/3)\,\mathrm{d}x.

For the first integration by parts, take u=x2u=x^2 and dv=cos⁡(x/3) dx\mathrm{d}v=\cos(x/3)\,\mathrm{d}x. Then du=2x dx\mathrm{d}u=2x\,\mathrm{d}x and v=3sin⁡(x/3)v=3\sin(x/3), giving

I= [3x2sin⁡(x/3)]0π −6∫0πxsin⁡(x/3) dx.\begin{align*} I=&\,\big[3x^2\sin(x/3)\big]_0^\pi\\[4mm] &\,\hspace{2pt}-6\int_0^\pi x\sin(x/3)\,\mathrm{d}x. \end{align*}

For the remaining integral, take u=xu=x and dv=sin⁡(x/3) dx\mathrm{d}v=\sin(x/3)\,\mathrm{d}x, so v=−3cos⁡(x/3)v=-3\cos(x/3). Hence

∫0πxsin⁡(x/3) dx= [−3xcos⁡(x/3)]0π +3∫0πcos⁡(x/3) dx= [−3xcos⁡(x/3)+9sin⁡(x/3)]0π.\begin{align*} \int_0^\pi x\sin(x/3)\,\mathrm{d}x =&\,\big[-3x\cos(x/3)\big]_0^\pi\\[4mm] &\,\hspace{2pt}+3\int_0^\pi\cos(x/3)\,\mathrm{d}x\\[4mm] =&\,\big[-3x\cos(x/3)+9\sin(x/3)\big]_0^\pi. \end{align*}

Substituting this result into II,

I= [3x2sin⁡(x/3)+18xcos⁡(x/3) −54sin⁡(x/3)]0π.\begin{align*} I=&\,\big[3x^2\sin(x/3)+18x\cos(x/3)\\[4mm] &\,\hspace{2pt}-54\sin(x/3)\big]_0^\pi. \end{align*}

At x=0x=0, all terms vanish. At x=πx=\pi, use sin⁡(π/3)=3/2\sin(\pi/3)=\sqrt3/2 and cos⁡(π/3)=1/2\cos(\pi/3)=1/2:

I=332π2+9π−273.\boxed{I=\frac{3\sqrt3}{2}\pi^2+9\pi-27\sqrt3}.