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IAL 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−12x0≤x≤10y=10xe^{-\frac12x}\qquad 0\le x\le 10

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}\,dx

where kk is a constant to be found.

(2)

(b) Find ∫x2e−x dx\int x^2e^{-x}\,dx

(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−12x0≤x≤10y=10xe^{-\frac12x}\qquad 0\le x\le 10

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}\,dx

其中 kk 是待求常数。

(b) 求 ∫x2e−x dx\int x^2e^{-x}\,dx。

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。

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

展开

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} +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}+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

展开

Using the result from part (b),

∫010x2e−x dx= [−x2e−x−2xe−x−2e−x]010= −100e−10−20e−10 −2e−10+2= 2−122e−10.\begin{align*} \int_0^{10}x^2e^{-x}\,\mathrm{d}x =&\,\Bigg[ -x^2e^{-x}-2xe^{-x}-2e^{-x} \Bigg]_0^{10}\\ =&\,-100e^{-10}-20e^{-10}\\ &\,\hspace{2pt}-2e^{-10}+2\\ =&\,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}. Therefore,

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

To three significant figures,

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