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P4 2022 June Q8

A Level / Edexcel / P4

IAL 2022 June Paper · Question 8

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

Figure 2 shows the curve with equation

y= 10xe−12x, 0≤x≤10\begin{align*} y=&\,10xe^{-\frac12x},\\ &\,0\le x\le10 \end{align*}

The finite region RR, shown shaded in Figure 2, is bounded by the curve, the xx-axis and the line with equation x=10x=10

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

(a) Show that the volume, VV, of this solid is given by

V=k∫010x2e−x dxV=k\int_0^{10}x^2e^{-x}\,\mathrm{d}x

where kk is a constant to be found.

(2)

(b) Find ∫x2e−x dx\int x^2e^{-x}\,\mathrm{d}x

(3)

Figure 3 represents an exercise weight formed by joining two of these solids together.

The exercise weight has mass 55 kg and is 2020 cm long.

Given that

density=massvolume\text{density}=\frac{\text{mass}}{\text{volume}}

and using your answers to part (a) and part (b),

(c) find the density of this exercise weight. Give your answer in grams per cm3^3 to 3 significant figures.

(5)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

完全依赖计算器技术的解法不被接受。

图 2 给出了曲线

y= 10xe−12x, 0≤x≤10\begin{align*} y=&\,10xe^{-\frac12x},\\ &\,0\le x\le10 \end{align*}

图 2 中阴影所示的有限区域 RR，由该曲线、xx 轴和直线 x=10x=10 围成。

区域 RR 绕 xx 轴旋转 2π2\pi 弧度，形成一个旋转体。

(a) 证明该旋转体的体积 VV 可表示为

V=k∫010x2e−x dxV=k\int_0^{10}x^2e^{-x}\,\mathrm{d}x

其中 kk 是待求常数。

(2)

(b) 求 ∫x2e−x dx\int x^2e^{-x}\,\mathrm{d}x。

(3)

图 3 表示一个通过将两个这样的旋转体连接而成的哑铃。

该哑铃的质量为 55 kg，长度为 2020 cm。

已知

density=massvolume\text{density}=\frac{\text{mass}}{\text{volume}}

并利用 (a) 和 (b) 的答案，

(c) 求该哑铃的密度。答案以 g/cm3^3 为单位，精确到 3 位有效数字。

(5)

解答

(a)

解法一

思路

展开

区域绕 xx 轴旋转时，体积为 π∫y2 dx\pi\int y^2\,\mathrm{d}x。将曲线方程平方后，指数由 −12x-\frac12x 变成 −x-x，即可与题目给出的积分比较并确定 kk。

答题过程

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The volume of the solid of revolution is

V= π∫010y2 dx= π∫010(10xe−12x)2 dx= 100π∫010x2e−x dx.\begin{align*} V =&\,\pi\int_0^{10}y^2\,\mathrm{d}x\\ =&\,\pi\int_0^{10} \big(10xe^{-\frac12x}\big)^2\,\mathrm{d}x\\ =&\,100\pi\int_0^{10}x^2e^{-x}\,\mathrm{d}x. \end{align*}

Therefore,

k=100π.\boxed{k=100\pi}.

(b)

解法一

思路

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连续使用两次分部积分。第一次把 x2x^2 求导，留下 ∫xe−x dx\int xe^{-x}\,\mathrm{d}x；第二次再对该积分作同样处理，直至只剩下可直接积分的 e−xe^{-x}。

答题过程

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Using integration by parts with

u=x2,dvdx=e−x,u=x^2, \qquad \frac{\mathrm{d}v}{\mathrm{d}x}=e^{-x},

gives

 ∫x2e−x dx= −x2e−x +2∫xe−x dx.\begin{align*} &\,\int x^2e^{-x}\,\mathrm{d}x\\ =&\,-x^2e^{-x}\\ &\,\hspace{2pt}+2\int xe^{-x}\,\mathrm{d}x. \end{align*}

Applying integration by parts again,

 ∫xe−x dx= −xe−x+∫e−x dx= −xe−x−e−x.\begin{align*} &\,\int xe^{-x}\,\mathrm{d}x\\ =&\,-xe^{-x}+\int e^{-x}\,\mathrm{d}x\\ =&\,-xe^{-x}-e^{-x}. \end{align*}

Hence

 ∫x2e−x dx= −x2e−x−2xe−x −2e−x+C= −(x2+2x+2)e−x +C.\begin{align*} &\,\int x^2e^{-x}\,\mathrm{d}x\\ =&\,-x^2e^{-x}-2xe^{-x}\\ &\,\hspace{2pt}-2e^{-x}+C\\ =&\,-(x^2+2x+2)e^{-x}\\ &\,\hspace{2pt}+C. \end{align*}

解法二

思路

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把连续分部积分整理成 DI 表格：左列不断求导，右列不断积分。按交替的正负号，将左列每项与右列下一行相乘；当左列变成 00 时停止。

答题过程

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Differentiate x2x^2 repeatedly and integrate e−xe^{-x} repeatedly:

DI+x2e−x−2x−e−x+2e−x0−e−x\begin{array}{c|c|c} & D & I\\ \hline + & x^2 & e^{-x}\\ - & 2x & -e^{-x}\\ + & 2 & e^{-x}\\ & 0 & -e^{-x} \end{array}

Multiply each entry in the DD column by the entry one row below it in the II column, using the indicated sign. Since the final derivative is zero, no integral remains:

 ∫x2e−x dx= x2(−e−x)−2x(e−x) +2(−e−x)+C= −(x2+2x+2)e−x +C.\begin{align*} &\,\int x^2e^{-x}\,\mathrm{d}x\\ =&\,x^2(-e^{-x})-2x(e^{-x})\\ &\,\hspace{2pt}+2(-e^{-x})+C\\ =&\,-(x^2+2x+2)e^{-x}\\ &\,\hspace{2pt}+C. \end{align*}

(c)

解法一

思路

展开

先用 (b) 的原函数计算定积分，再用 (a) 的常数求一个旋转体的体积。哑铃由两个相同旋转体连接而成，因此总体积还要乘以 22；最后把 55 kg 化为 50005000 g 后计算密度。

答题过程

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Using the result from part (b), let

 F(x)= −(x2+2x+2)e−x.\begin{align*} &\,F(x)\\ =&\,-(x^2+2x+2)e^{-x}. \end{align*}

Then

 F(10)= −(100+20+2)e−10= −122e−10, F(0)=−2.\begin{align*} &\,F(10)\\ =&\,-(100+20+2)e^{-10}\\ =&\,-122e^{-10},\\ &\,F(0)=-2. \end{align*}

Therefore,

 ∫010x2e−x dx= F(10)−F(0)= 2−122e−10.\begin{align*} &\,\int_0^{10}x^2e^{-x}\,\mathrm{d}x\\ =&\,F(10)-F(0)\\ =&\,2-122e^{-10}. \end{align*}

The exercise weight consists of two identical solids, so its total volume is

 Vtotal= 2(100π)(2−122e−10)= 200π(2−122e−10) cm3.\begin{align*} &\,V_{\text{total}}\\ =&\,2(100\pi)\big(2-122e^{-10}\big)\\ =&\,200\pi\big(2-122e^{-10}\big)\text{ cm}^3. \end{align*}

Also, 5 kg=5000 g5\text{ kg}=5000\text{ g}. Writing the density as ρ\rho,

ρ= 5000200π(2−122e−10)= 3.989… g/cm3.\begin{align*} \rho =&\,\frac{5000} {200\pi\big(2-122e^{-10}\big)}\\ =&\,3.989\ldots\text{ g/cm}^3. \end{align*}

To three significant figures,

ρ=3.99 g/cm3.\boxed{\rho=3.99\text{ g/cm}^3}.