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P4 2022 Jan Q7

A Level / Edexcel / P4

IAL 2022 Jan Paper · Question 7

题目

Problem

Figure 3 shows the design of a doorknob.

The shape of the doorknob is formed by rotating the curve shown in Figure 4 through 360∘360^\circ about the xx-axis, where the units are centimetres.

The equation of the curve is given by

 f(x)=14(4−x)ex, 0≤x≤4.\begin{align*} &\,\mathrm{f}(x)=\frac{1}{4}(4-x)e^x,\\ &\,0\leq x\leq 4. \end{align*}

(a) Show that the volume, VV cm3^3, of the doorknob is given by

 V= K∫04(x2−8x+16) ×e2x dx\begin{align*} &\,V\\ =&\,K\int_0^4 (x^2-8x+16)\\ &\,\hspace{2pt}\times e^{2x}\,\mathrm{d}x \end{align*}

where KK is a constant to be found.

(3)

(b) Hence, find the exact value of the volume of the doorknob.

Give your answer in the form pπ(eq+r)p\pi(e^q+r) cm3^3 where pp, qq and rr are simplified rational numbers to be found.

(5)
题目中文翻译

图 3 给出了一个门把手的设计。

该门把手的形状由图 4 所示曲线绕 xx 轴旋转 360∘360^\circ 形成，其中单位为厘米。

该曲线的方程为

 f(x)=14(4−x)ex, 0≤x≤4.\begin{align*} &\,\mathrm{f}(x)=\frac{1}{4}(4-x)e^x,\\ &\,0\leq x\leq 4. \end{align*}

(a) 证明该门把手的体积 VV cm3^3 可表示为

 V= K∫04(x2−8x+16) ×e2x dx\begin{align*} &\,V\\ =&\,K\int_0^4 (x^2-8x+16)\\ &\,\hspace{2pt}\times e^{2x}\,\mathrm{d}x \end{align*}

其中 KK 为待求常数。

(3)

(b) 进而求该门把手体积的精确值。

将答案写成 pπ(eq+r)p\pi(e^q+r) cm3^3 的形式，其中 p,q,rp,q,r 为待求的最简有理数。

(5)

解答

(a)

解法一

思路

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曲线绕 xx 轴旋转所得体积为 π∫y2 dx\pi\int y^2\,\mathrm{d}x。将 f(x)\mathrm{f}(x) 平方，展开 (4−x)2(4-x)^2，再与题目给出的积分比较即可确定 KK。

答题过程

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The volume of the solid of revolution is

 V= π∫04(f(x))2 dx= π∫04(14(4−x)ex)2 dx= π16∫04(4−x)2e2x dx= π16∫04(x2−8x+16)e2x dx.\begin{align*} &\,V\\ =&\,\pi\int_0^4\big(\mathrm{f}(x)\big)^2 \,\mathrm{d}x\\ =&\,\pi\int_0^4 \bigg(\frac14(4-x)e^x\bigg)^2 \,\mathrm{d}x\\ =&\,\frac{\pi}{16}\int_0^4 (4-x)^2e^{2x}\,\mathrm{d}x\\ =&\,\frac{\pi}{16}\int_0^4 (x^2-8x+16)e^{2x}\\ &\,\hspace{2pt}\mathrm{d}x. \end{align*}

Therefore,

K=π16.\boxed{K=\frac{\pi}{16}}.

(b)

解法一

思路

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承接 (a)，直接对二次多项式乘指数函数使用分部积分。第一次求导后降为一次式，第二次分部积分即可完成；最后代入 0,40,4 并乘上 K=π16K=\frac{\pi}{16}。

也可以把积分拆成 x2e2xx^2e^{2x}、−8xe2x-8xe^{2x} 和 16e2x16e^{2x} 三项。分别求原函数后再合并，得到与整体分部积分相同的结果。

答题过程

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Let

 I= ∫(x2−8x+16)e2x dx.\begin{align*} &\,I\\ =&\,\int(x^2-8x+16)e^{2x}\\ &\,\hspace{2pt}\mathrm{d}x. \end{align*}

Integrating by parts gives

 I= 12(x2−8x+16)e2x −12∫(2x−8)e2x dx.\begin{align*} &\,I\\ =&\,\frac12(x^2-8x+16)e^{2x}\\ &\,\hspace{2pt} -\frac12\int(2x-8)e^{2x}\,\mathrm{d}x. \end{align*}

For the remaining integral,

 ∫(2x−8)e2x dx= 12(2x−8)e2x −∫e2x dx= (x−4)e2x−12e2x.\begin{align*} &\,\int(2x-8)e^{2x}\,\mathrm{d}x\\ =&\,\frac12(2x-8)e^{2x}\\ &\,\hspace{2pt}-\int e^{2x}\,\mathrm{d}x\\ =&\,(x-4)e^{2x}-\frac12e^{2x}. \end{align*}

Hence

 I= 12(x2−8x+16)e2x −12(x−4)e2x+14e2x.\begin{align*} &\,I\\ =&\,\frac12(x^2-8x+16)e^{2x}\\ &\,\hspace{2pt} -\frac12(x-4)e^{2x}+\frac14e^{2x}. \end{align*}

Alternatively, split II into three terms. By integration by parts,

 ∫xe2x dx= x2e2x−12∫e2x dx= e2x(x2−14).\begin{align*} &\,\int xe^{2x}\,\mathrm{d}x\\ =&\,\frac{x}{2}e^{2x}-\frac12\int e^{2x}\,\mathrm{d}x\\ =&\,e^{2x}\bigg(\frac{x}{2}-\frac14\bigg). \end{align*}

Applying integration by parts once more,

 ∫x2e2x dx= x22e2x−∫xe2x dx= e2x(x22−x2+14).\begin{align*} &\,\int x^2e^{2x}\,\mathrm{d}x\\ =&\,\frac{x^2}{2}e^{2x}-\int xe^{2x}\,\mathrm{d}x\\ =&\,e^{2x}\bigg(\frac{x^2}{2}-\frac{x}{2}+\frac14\bigg). \end{align*}

Also, ∫16e2x dx=8e2x\int16e^{2x}\,\mathrm{d}x=8e^{2x}. Combining the three terms gives

 I= e2x(x22−x2+14) −8e2x(x2−14) +8e2x= e2x(x22−9x2+414).\begin{align*} &\,I\\ =&\,e^{2x}\bigg(\frac{x^2}{2}-\frac{x}{2}+\frac14\bigg)\\ &\,\hspace{2pt}-8e^{2x}\bigg(\frac{x}{2}-\frac14\bigg)\\ &\,\hspace{4pt}+8e^{2x}\\ =&\,e^{2x}\bigg(\frac{x^2}{2}-\frac{9x}{2}+\frac{41}{4}\bigg). \end{align*}

Using the result from part (a),

 V= π16[e2x ×(12(x2−8x+16) −12(x−4)+14)]04= π16(14e8−414)= π64(e8−41) cm3.\begin{align*} &\,V\\ =&\,\frac{\pi}{16}\Bigg[e^{2x}\\ &\,\hspace{2pt}\times\bigg(\frac12(x^2-8x+16)\\ &\,\hspace{4pt}-\frac12(x-4)+\frac14\bigg) \Bigg]_0^4\\ =&\,\frac{\pi}{16} \bigg(\frac14e^8-\frac{41}{4}\bigg)\\ =&\,\boxed{\frac{\pi}{64}(e^8-41)\text{ cm}^3}. \end{align*}

Thus p=164p=\frac1{64}, q=8q=8 and r=−41r=-41.

解法二

思路

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注意到积分中的二次式就是 (4−x)2(4-x)^2。保留这个结构作两次分部积分，求导会依次得到 −2(4−x)-2(4-x) 和常数，因此运算更短。

答题过程

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Since

x2−8x+16=(4−x)2,x^2-8x+16=(4-x)^2,

let

J=∫(4−x)2e2x dx.J=\int(4-x)^2e^{2x}\,\mathrm{d}x.

Integrating by parts,

 J= 12(4−x)2e2x +∫(4−x)e2x dx.\begin{align*} &\,J\\ =&\,\frac12(4-x)^2e^{2x}\\ &\,\hspace{2pt}+\int(4-x)e^{2x}\,\mathrm{d}x. \end{align*}

A second integration by parts gives

 ∫(4−x)e2x dx= 12(4−x)e2x +12∫e2x dx= 12(4−x)e2x+14e2x.\begin{align*} &\,\int(4-x)e^{2x}\,\mathrm{d}x\\ =&\,\frac12(4-x)e^{2x}\\ &\,\hspace{2pt}+\frac12\int e^{2x}\,\mathrm{d}x\\ =&\,\frac12(4-x)e^{2x}+\frac14e^{2x}. \end{align*}

Therefore,

 V= π16[e2x ×(12(4−x)2 +12(4−x)+14)]04= π16(14e8−414)= π64(e8−41) cm3.\begin{align*} &\,V\\ =&\,\frac{\pi}{16}\Bigg[e^{2x}\\ &\,\hspace{2pt}\times\bigg(\frac12(4-x)^2\\ &\,\hspace{4pt}+\frac12(4-x)+\frac14\bigg) \Bigg]_0^4\\ =&\,\frac{\pi}{16} \bigg(\frac14e^8-\frac{41}{4}\bigg)\\ =&\,\boxed{\frac{\pi}{64}(e^8-41)\text{ cm}^3}. \end{align*}