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P4 2021 Oct Q8

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 8

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) Find

∫x2ln⁡x dx\int x^2\ln x\,\mathrm{d}x
(3)

Figure 3 shows a sketch of part of the curve with equation

y=xln⁡xx>0y=x\ln x \qquad x>0

The region RR, shown shaded in Figure 3, lies entirely above the xx-axis and is bounded by the curve, the xx-axis and the line with equation x=ex=e.

This region is rotated through 2π2\pi radians about the xx-axis to form a solid of revolution.

(b) Find the exact volume of the solid formed, giving your answer in simplest form.

(4)
题目中文翻译

本题你必须写出所有计算步骤。

不接受依赖计算器技术的解法。

(a) 求

∫x2ln⁡x dx\int x^2\ln x\,\mathrm{d}x
(3)

图 3 给出了曲线一部分的草图，其方程为

y=xln⁡xx>0y=x\ln x \qquad x>0

图 3 中阴影部分所示区域 RR 完全位于 xx 轴上方，并由该曲线、xx 轴和直线 x=ex=e 围成。

将该区域绕 xx 轴旋转 2π2\pi 弧度，形成一个旋转体。

(b) 求所形成立体的精确体积，并将答案写成最简形式。

(4)

解答

(a)

解法一

思路

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使用分部积分，把 ln⁡x\ln x 作为求导部分、x2x^2 作为积分部分。这样求导后会出现 1/x1/x，使剩余积分化为简单的幂函数积分。

答题过程

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Using integration by parts, take

u=ln⁡x,dv=x2 dx.u=\ln x, \qquad \mathrm{d}v=x^2\,\mathrm{d}x.

Then

du=1x dx,v=x33.\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x, \qquad v=\frac{x^3}{3}.

Therefore,

 ∫x2ln⁡x dx= x33ln⁡x −∫x33⋅1x dx= x33ln⁡x −13∫x2 dx= x33ln⁡x−x39+C.\begin{align*} &\,\int x^2\ln x\,\mathrm{d}x\\ =&\,\frac{x^3}{3}\ln x\\ &\,\hspace{2pt}-\int\frac{x^3}{3}\cdot\frac1x\,\mathrm{d}x\\ =&\,\frac{x^3}{3}\ln x \\ &\,\hspace{2pt}-\frac13\int x^2\,\mathrm{d}x\\ =&\,\boxed{ \frac{x^3}{3}\ln x-\frac{x^3}{9}+C }. \end{align*}

(b)

解法一

思路

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曲线与 xx 轴相交于 x=1x=1，所以旋转体体积为 π∫1e(xln⁡x)2 dx\pi\int_1^e(x\ln x)^2\,\mathrm{d}x。对 ∫x2(ln⁡x)2 dx\int x^2(\ln x)^2\,\mathrm{d}x 作分部积分，并用 (a) 的结果处理余下积分。

答题过程

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The curve meets the xx-axis when

xln⁡x=0.x\ln x=0.

Since x>0x>0, this gives ln⁡x=0\ln x=0, so x=1x=1. Hence the volume is

V=π∫1ex2(ln⁡x)2 dx.V=\pi\int_1^e x^2(\ln x)^2\,\mathrm{d}x.

Integrating by parts with

u= (ln⁡x)2,dv= x2 dx,\begin{align*} u=&\,(\ln x)^2,\\ \mathrm{d}v=&\,x^2\,\mathrm{d}x, \end{align*}

gives

 ∫x2(ln⁡x)2 dx= x33(ln⁡x)2 −23∫x2ln⁡x dx.\begin{align*} &\,\int x^2(\ln x)^2\,\mathrm{d}x\\ =&\,\frac{x^3}{3}(\ln x)^2\\ &\,-\frac23\int x^2\ln x\,\mathrm{d}x. \end{align*}

Using the result from part (a),

 ∫x2(ln⁡x)2 dx= x33(ln⁡x)2 −23( x33ln⁡x −x39)+C= x33(ln⁡x)2 −2x39ln⁡x +2x327+C.\begin{align*} &\,\int x^2(\ln x)^2\,\mathrm{d}x\\ =&\,\frac{x^3}{3}(\ln x)^2\\ &\,-\frac23\left( \begin{aligned} &\,\frac{x^3}{3}\ln x\\ &\,-\frac{x^3}{9} \end{aligned} \right)+C\\ =&\,\frac{x^3}{3}(\ln x)^2 \\ &\,\hspace{2pt}-\frac{2x^3}{9}\ln x\\ &\,\hspace{4pt}+\frac{2x^3}{27}+C. \end{align*}

Therefore,

 V=π[ x33(ln⁡x)2 −2x39ln⁡x +2x327]1e= π[ ( e33−2e39 +2e327) −227]= π27(5e3−2).\begin{align*} &\,V=\pi\left[ \begin{aligned} &\,\frac{x^3}{3}(\ln x)^2\\ &\,\hspace{2pt}-\frac{2x^3}{9}\ln x\\ &\,\hspace{4pt}+\frac{2x^3}{27} \end{aligned} \right]_1^e\\ =&\,\pi\left[ \begin{aligned} &\,\left(\begin{aligned} &\,\frac{e^3}{3}-\frac{2e^3}{9}\\ &\,\hspace{2pt}+\frac{2e^3}{27} \end{aligned}\right)\\ &\,\hspace{2pt}-\frac{2}{27} \end{aligned} \right]\\ =&\,\boxed{\frac{\pi}{27}\big(5e^3-2\big)}. \end{align*}

解法二

思路

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把被积函数写成 ln⁡x⋅(x2ln⁡x)\ln x\cdot(x^2\ln x)，并直接把 (a) 的原函数当作分部积分中的 vv。这样同样能得到 ∫x2(ln⁡x)2 dx\int x^2(\ln x)^2\,\mathrm{d}x 的原函数。

答题过程

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From part (a), let

F(x)=x33ln⁡x−x39,F(x)=\frac{x^3}{3}\ln x-\frac{x^3}{9},

so that

F′(x)=x2ln⁡x.F'(x)=x^2\ln x.

Then, by integration by parts,

 ∫x2(ln⁡x)2 dx= ∫ln⁡x F′(x) dx= F(x)ln⁡x −∫F(x)x dx= (x33ln⁡x−x39)ln⁡x −13∫x2ln⁡x dx +19∫x2 dx.\begin{align*} &\,\int x^2(\ln x)^2\,\mathrm{d}x\\ =&\,\int \ln x\,F'(x)\,\mathrm{d}x\\ =&\,F(x)\ln x \\ &\,\hspace{2pt}-\int\frac{F(x)}{x}\,\mathrm{d}x\\ =&\,\bigg( \frac{x^3}{3}\ln x-\frac{x^3}{9} \bigg)\ln x\\ &\,-\frac13\int x^2\ln x\,\mathrm{d}x \\ &\,\hspace{2pt}+\frac19\int x^2\,\mathrm{d}x. \end{align*}

Using part (a) again,

 ∫x2(ln⁡x)2 dx= (x33ln⁡x−x39)ln⁡x −13(x33ln⁡x−x39) +x327+C= x33(ln⁡x)2 −2x39ln⁡x +2x327+C.\begin{align*} &\,\int x^2(\ln x)^2\,\mathrm{d}x\\ =&\,\bigg( \frac{x^3}{3}\ln x-\frac{x^3}{9} \bigg)\ln x\\ &\,-\frac13\bigg( \frac{x^3}{3}\ln x-\frac{x^3}{9} \bigg)\\ &\,\hspace{2pt}+\frac{x^3}{27}+C\\ =&\,\frac{x^3}{3}(\ln x)^2 \\ &\,\hspace{2pt}-\frac{2x^3}{9}\ln x\\ &\,\hspace{4pt}+\frac{2x^3}{27}+C. \end{align*}

Therefore,

 V=π[ x33(ln⁡x)2 −2x39ln⁡x +2x327]1e= π27(5e3−2).\begin{align*} &\,V=\pi\left[ \begin{aligned} &\,\frac{x^3}{3}(\ln x)^2\\ &\,\hspace{2pt}-\frac{2x^3}{9}\ln x\\ &\,\hspace{4pt}+\frac{2x^3}{27} \end{aligned} \right]_1^e\\ =&\,\boxed{\frac{\pi}{27}\big(5e^3-2\big)}. \end{align*}