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P4 2023 Oct Q3

A Level / Edexcel / P4

IAL 2023 Oct Paper · Question 3

题目

Problem

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

Solutions based on calculator technology are not acceptable.

(i) Use integration by parts to find the exact value of

∫04x2e2x dx\int_0^4 x^2e^{2x}\,\mathrm{d}x

giving your answer in simplest form.

(5)

(ii) Use integration by substitution to show that

 ∫32124x(2x−1)2 dx= a+ln⁡b\begin{align*} &\,\int_3^{\frac{21}{2}} \frac{4x}{(2x-1)^2}\,\mathrm{d}x\\ =&\,a+\ln b \end{align*}

where aa and bb are constants to be found.

(7)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

基于计算器技术的解法不被接受。

(i) 用分部积分法求

∫04x2e2x dx\int_0^4 x^2e^{2x}\,\mathrm{d}x

的精确值，并将答案化为最简形式。

(5)

(ii) 用换元积分法证明

 ∫32124x(2x−1)2 dx= a+ln⁡b\begin{align*} &\,\int_3^{\frac{21}{2}} \frac{4x}{(2x-1)^2}\,\mathrm{d}x\\ =&\,a+\ln b \end{align*}

其中 a,ba,b 为待求常数。

(7)

解答

(i)

解法一

思路

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按题目要求做两次分部积分：第一次把 x2x^2 降为 xx，第二次把 xx 降为常数。得到一个原函数后，分别求上、下限的值再相减。

答题过程

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For the first integration by parts, take

u=x2,dvdx=e2x.u=x^2,\qquad \frac{\mathrm{d}v}{\mathrm{d}x}=e^{2x}.

Then

dudx=2x,v=12e2x.\frac{\mathrm{d}u}{\mathrm{d}x}=2x,\qquad v=\frac12e^{2x}.

Let JJ denote an antiderivative of xe2xxe^{2x}. Thus an antiderivative FF of x2e2xx^2e^{2x} is

F(x)=12x2e2x−J(x).F(x)=\frac12x^2e^{2x}-J(x).

For JJ, integrate by parts using

u=x,dvdx=e2x.u=x,\qquad \frac{\mathrm{d}v}{\mathrm{d}x}=e^{2x}.

Then

dudx=1,v=12e2x.\frac{\mathrm{d}u}{\mathrm{d}x}=1,\qquad v=\frac12e^{2x}.

This gives

J(x)= 12xe2x −12∫e2x dx= 12xe2x−14e2x.\begin{align*} J(x)=&\,\frac12xe^{2x}\\ &\,\hspace{2pt}-\frac12\int e^{2x}\,\mathrm{d}x\\ =&\,\frac12xe^{2x}-\frac14e^{2x}. \end{align*}

Hence we can choose

F(x)= 12x2e2x −12xe2x+14e2x.\begin{align*} F(x)=&\,\frac12x^2e^{2x}\\ &\,\hspace{2pt}-\frac12xe^{2x} +\frac14e^{2x}. \end{align*}

At the limits,

F(4)= 8e8−2e8+14e8= 254e8,F(0)= 14.\begin{align*} F(4)=&\,8e^8-2e^8+\frac14e^8\\ =&\,\frac{25}{4}e^8,\\ F(0)=&\,\frac14. \end{align*}

Therefore the required integral is

F(4)−F(0)=25e8−14.F(4)-F(0)=\boxed{\frac{25e^8-1}{4}}.

解法二

思路

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D/I 表是把连续分部积分组织成表格，也可以使用这条路线，仍满足题目指定的方法。左列连续求导，右列连续积分；按正、负、正号取相邻对角乘积。

答题过程

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Use a differentiation/integration table.

SignDifferentiateIntegrate
++x2x^2e2xe^{2x}
−-2x2x12e2x\frac12e^{2x}
++2214e2x\frac14e^{2x}
0018e2x\frac18e^{2x}

The diagonal products give an antiderivative

F(x)= x2⋅12e2x −(2x)⋅14e2x +2⋅18e2x.\begin{align*} F(x)=&\,x^2\cdot\frac12e^{2x}\\ &\,\hspace{2pt}-(2x)\cdot\frac14e^{2x}\\ &\,\hspace{4pt}+2\cdot\frac18e^{2x}. \end{align*}

Factorising,

F(x)=e2x4(2x2−2x+1).F(x)=\frac{e^{2x}}4(2x^2-2x+1).

So

F(4)= e84(32−8+1)= 25e84,F(0)= 14.\begin{align*} F(4)=&\,\frac{e^8}{4}(32-8+1)\\ =&\,\frac{25e^8}{4},\\ F(0)=&\,\frac14. \end{align*}

The exact value is therefore

F(4)−F(0)=25e8−14.F(4)-F(0)=\boxed{\frac{25e^8-1}{4}}.

(ii)

解法一

思路

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取 u=2x−1u=2x-1，同时换掉变量、微分和上下限。新被积函数拆成 u−1+u−2u^{-1}+u^{-2}，分别积分后得到题目要求的形式。

答题过程

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Let the required integral be II.

Set

u=2x−1.u=2x-1.

Then

x=u+12,dx=12 du.x=\frac{u+1}{2},\qquad \mathrm{d}x=\frac12\,\mathrm{d}u.

The new limits are

x=3 ⟹u=5,x=212 ⟹u=20.\begin{align*} x=3&\,\Longrightarrow u=5,\\ x=\frac{21}{2}&\,\Longrightarrow u=20. \end{align*}

Substitution gives

I=∫520u+1u2 du.I=\int_5^{20}\frac{u+1}{u^2}\,\mathrm{d}u.

Since

u+1u2=u−1+u−2,\frac{u+1}{u^2}=u^{-1}+u^{-2},

an antiderivative is

F(u)=ln⁡u−1u.F(u)=\ln u-\frac1u.

Therefore

I= F(20)−F(5)= ln⁡20−120 −ln⁡5+15= ln⁡4+4−120= 320+ln⁡4.\begin{align*} I=&\,F(20)-F(5)\\ =&\,\ln20-\frac1{20}\\ &\,\hspace{2pt}-\ln5+\frac15\\ =&\,\ln4+\frac{4-1}{20}\\ =&\,\frac3{20}+\ln4. \end{align*}

Thus

a=320,b=4.\boxed{a=\frac3{20}},\qquad \boxed{b=4}.

换元后也可以用分部积分处理 (u+1)u−2(u+1)u^{-2}。取 U=u+1U=u+1、dW/du=u−2\mathrm{d}W/\mathrm{d}u=u^{-2}，则

 ∫(u+1)u−2 du= −u+1u +∫1u du= −1−1u+ln⁡u+C.\begin{align*} &\,\int (u+1)u^{-2}\,\mathrm{d}u\\ =&\,-\frac{u+1}{u}\\ &\,\hspace{2pt}+\int\frac1u\,\mathrm{d}u\\ =&\,-1-\frac1u+\ln u+C. \end{align*}

这里的原函数只比 F(u)F(u) 多常数 −1-1，在上下限相减时消去，结果相同。

解法二

思路

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也可以令 u=(2x−1)2u=(2x-1)^2。本题积分区间内 2x−1>02x-1>0，所以 2x−1=u2x-1=\sqrt u；新上下限为 2525 和 400400，用幂函数与对数积分即可。

答题过程

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Let

u=(2x−1)2.u=(2x-1)^2.

On the given interval, 2x−1>02x-1>0, so

x=u+12.x=\frac{\sqrt u+1}{2}.

Differentiating,

dudx=4(2x−1)=4u.\frac{\mathrm{d}u}{\mathrm{d}x} =4(2x-1)=4\sqrt u.

Hence

dx=14u du.\mathrm{d}x=\frac1{4\sqrt u}\,\mathrm{d}u.

In particular,

4x dx=u+12u du.4x\,\mathrm{d}x =\frac{\sqrt u+1}{2\sqrt u}\,\mathrm{d}u.

The new limits are

x=3 ⟹u=25,x=212 ⟹u=400.\begin{align*} x=3&\,\Longrightarrow u=25,\\ x=\frac{21}{2}&\,\Longrightarrow u=400. \end{align*}

Therefore the required integral is

I=12∫25400(u−1+u−3/2) du.I=\frac12\int_{25}^{400} \big(u^{-1}+u^{-3/2}\big)\,\mathrm{d}u.

An antiderivative of the integrand including the factor 1/21/2 is

G(u)=12ln⁡u−u−1/2.G(u)=\frac12\ln u-u^{-1/2}.

Evaluating,

I= G(400)−G(25)= 12ln⁡400−120 −12ln⁡25+15= 12ln⁡16+320= ln⁡4+320.\begin{align*} I=&\,G(400)-G(25)\\ =&\,\frac12\ln400-\frac1{20}\\ &\,\hspace{2pt}-\frac12\ln25+\frac15\\ =&\,\frac12\ln16+\frac3{20}\\ =&\,\ln4+\frac3{20}. \end{align*}

Thus

a=320,b=4.\boxed{a=\frac3{20}},\qquad \boxed{b=4}.

解法三

思路

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也可以令 u=2xu=2x。换元后得到 u/(u−1)2u/(u-1)^2，使用分部积分；新的上下限为 66 和 2121。

答题过程

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Let

u=2x.u=2x.

Then

dx=12 du,\mathrm{d}x=\frac12\,\mathrm{d}u,

and the limits change as follows:

x=3 ⟹u=6,x=212 ⟹u=21.\begin{align*} x=3&\,\Longrightarrow u=6,\\ x=\frac{21}{2}&\,\Longrightarrow u=21. \end{align*}

The required integral becomes

I=∫621u(u−1)2 du.I=\int_6^{21}\frac{u}{(u-1)^2}\,\mathrm{d}u.

To integrate by parts, choose

U=u,dWdu=(u−1)−2.U=u,\qquad \frac{\mathrm{d}W}{\mathrm{d}u}=(u-1)^{-2}.

Then

dUdu=1,W=−1u−1.\frac{\mathrm{d}U}{\mathrm{d}u}=1,\qquad W=-\frac1{u-1}.

Hence an antiderivative is

H(u)= −uu−1 +∫1u−1 du= −uu−1+ln⁡(u−1).\begin{align*} H(u)=&\,-\frac{u}{u-1}\\ &\,\hspace{2pt}+\int\frac1{u-1}\,\mathrm{d}u\\ =&\,-\frac{u}{u-1}+\ln(u-1). \end{align*}

So

H(21)= −2120+ln⁡20,H(6)= −65+ln⁡5.\begin{align*} H(21)=&\,-\frac{21}{20}+\ln20,\\ H(6)=&\,-\frac65+\ln5. \end{align*}

Therefore

I= H(21)−H(6)= −2120+2420+ln⁡4= 320+ln⁡4.\begin{align*} I=&\,H(21)-H(6)\\ =&\,-\frac{21}{20}+\frac{24}{20}+\ln4\\ =&\,\frac3{20}+\ln4. \end{align*}

Thus

a=320,b=4.\boxed{a=\frac3{20}},\qquad \boxed{b=4}.

解法四

思路

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可以结合部分分式与换元：先作部分分式，再对其中的平方分母项明确代换。不能把代换步骤删掉；若全过程没有代换，虽可算对答案，却不满足本题指定方法，不能获得全部过程分。

答题过程

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Write

 4x(2x−1)2= A2x−1 +B(2x−1)2.\begin{align*} &\,\frac{4x}{(2x-1)^2}\\ =&\,\frac{A}{2x-1}\\ &\,\hspace{2pt}+\frac{B}{(2x-1)^2}. \end{align*}

Multiplying by (2x−1)2(2x-1)^2,

4x=A(2x−1)+B.4x=A(2x-1)+B.

Comparing coefficients,

2A=4,−A+B=0.2A=4,\qquad -A+B=0.

Hence A=B=2A=B=2.

Split the integral as I=I1+I2I=I_1+I_2, where

I1=∫321/222x−1 dx,I_1=\int_3^{21/2}\frac2{2x-1}\,\mathrm{d}x,

and

I2=∫321/22(2x−1)2 dx.I_2=\int_3^{21/2}\frac2{(2x-1)^2}\,\mathrm{d}x.

The first is

I1= [ln⁡(2x−1)]321/2= ln⁡20−ln⁡5= ln⁡4.\begin{align*} I_1=&\,\big[\ln(2x-1)\big]_3^{21/2}\\ =&\,\ln20-\ln5\\ =&\,\ln4. \end{align*}

For the second, use the substitution

u=2x−1,du=2 dx.u=2x-1,\qquad \mathrm{d}u=2\,\mathrm{d}x.

The limits become 55 and 2020, so

I2= ∫520u−2 du= [−u−1]520= −120+15= 320.\begin{align*} I_2=&\,\int_5^{20}u^{-2}\,\mathrm{d}u\\ =&\,\big[-u^{-1}\big]_5^{20}\\ =&\,-\frac1{20}+\frac15\\ =&\,\frac3{20}. \end{align*}

Therefore

I=I1+I2=320+ln⁡4,I=I_1+I_2=\frac3{20}+\ln4,

and

a=320,b=4.\boxed{a=\frac3{20}},\qquad \boxed{b=4}.

这种混合方法包含局部代换，符合题目要求；如果完全不做任何代换，最多只能获得本小题 4/74/7 分。

解法五

思路

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另一种方法是先分部积分，再对剩下的简单分式积分明确代换。保留该代换及上下限，才能符合题目要求。

答题过程

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For integration by parts, take

U=4x,dWdx=(2x−1)−2.U=4x,\qquad \frac{\mathrm{d}W}{\mathrm{d}x}=(2x-1)^{-2}.

Then

dUdx=4,W=−12(2x−1).\frac{\mathrm{d}U}{\mathrm{d}x}=4,\qquad W=-\frac1{2(2x-1)}.

Hence

I=[−2x2x−1]321/2+J,I=\Big[-\frac{2x}{2x-1}\Big]_3^{21/2}+J,

where

J=∫321/222x−1 dx.J=\int_3^{21/2}\frac2{2x-1}\,\mathrm{d}x.

For JJ, set

u=2x−1,du=2 dx.u=2x-1,\qquad \mathrm{d}u=2\,\mathrm{d}x.

Changing the limits to 55 and 2020 gives

J= ∫5201u du= ln⁡20−ln⁡5= ln⁡4.\begin{align*} J=&\,\int_5^{20}\frac1u\,\mathrm{d}u\\ =&\,\ln20-\ln5\\ =&\,\ln4. \end{align*}

The boundary contribution is

[−2x2x−1]321/2= −2120+65= 320.\begin{align*} \Big[-\frac{2x}{2x-1}\Big]_3^{21/2} =&\,-\frac{21}{20}+\frac65\\ =&\,\frac3{20}. \end{align*}

Therefore

I=320+ln⁡4,I=\frac3{20}+\ln4,

so

a=320,b=4.\boxed{a=\frac3{20}},\qquad \boxed{b=4}.