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IAL 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\int_3^{\frac{21}{2}} \frac{4x}{(2x-1)^2}\,\mathrm{d}x =a+\ln b

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。

(ii) 用换元积分法证明

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

其中 a,ba,b 为待求常数。

解答

(i)

解法一

思路

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按照题目要求使用分部积分。第一次令 u=x2u=x^2，把问题化为 ∫xe2x dx\int xe^{2x}\,\mathrm{d}x；再对这个积分做第二次分部积分，最后统一代入上下限。

答题过程

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Using integration by parts with

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

gives

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

Therefore,

∫x2e2x dx=12x2e2x−∫xe2x dx.\int x^2e^{2x}\,\mathrm{d}x =\frac12x^2e^{2x} -\int xe^{2x}\,\mathrm{d}x.

Applying integration by parts again,

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

Hence

∫x2e2x dx=12x2e2x−12xe2x+14e2x.\int x^2e^{2x}\,\mathrm{d}x =\frac12x^2e^{2x} -\frac12xe^{2x} +\frac14e^{2x}.

Applying the limits,

∫04x2e2x dx= [12x2e2x−12xe2x+14e2x]04= (8e8−2e8+14e8)−14= 25e8−14.\begin{align*} \int_0^4 x^2e^{2x}\,\mathrm{d}x =&\,\bigg[ \frac12x^2e^{2x} -\frac12xe^{2x} +\frac14e^{2x} \bigg]_0^4\\ =&\,\bigg(8e^8-2e^8+\frac14e^8\bigg) -\frac14\\ =&\,\frac{25e^8-1}{4}. \end{align*}

Therefore,

∫04x2e2x dx=25e8−14.\boxed{\int_0^4 x^2e^{2x}\,\mathrm{d}x =\frac{25e^8-1}{4}}.

(ii)

解法一

思路

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按照题目要求换元，取 u=2x−1u=2x-1，因为它正好是分母的基本结构。把 xx、dx\mathrm{d}x 和上下限全部改写成 uu 后，被积函数会拆成 1u+1u2\dfrac1u+\dfrac1{u^2}，可以直接积分。

答题过程

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Let

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 limits become

x=3⟹u=5,x=212⟹u=20.x=3\Longrightarrow u=5, \qquad x=\frac{21}{2}\Longrightarrow u=20.

Therefore,

 ∫32124x(2x−1)2 dx= ∫5202(u+1)u2⋅12 du= ∫520(1u+1u2) du= [ln⁡u−1u]520= ln⁡20−120−ln⁡5+15= ln⁡4+320.\begin{align*} &\,\int_3^{\frac{21}{2}} \frac{4x}{(2x-1)^2}\,\mathrm{d}x\\ =&\,\int_5^{20} \frac{2(u+1)}{u^2}\cdot\frac12\,\mathrm{d}u\\ =&\,\int_5^{20} \bigg(\frac1u+\frac1{u^2}\bigg)\,\mathrm{d}u\\ =&\,\bigg[\ln u-\frac1u\bigg]_5^{20}\\ =&\,\ln20-\frac1{20} -\ln5+\frac15\\ =&\,\ln4+\frac3{20}. \end{align*}

Thus the result has the required form

320+ln⁡4,\boxed{\frac3{20}+\ln4},

so

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