CalcGospel 國際數學圖譜
返回

P4 2022 June Q5

A Level / Edexcel / P4

IAL 2022 June Paper · Question 5

题目

Problem

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

Solutions relying on calculator technology are not acceptable.

(a) Use the substitution x=2sin⁡ux=2\sin u to show that

 ∫013x+2(4−x2)32 dx= ∫0p(32sec⁡utan⁡u +12sec⁡2u) du\begin{align*} &\,\int_0^1\frac{3x+2}{(4-x^2)^{\frac32}}\,\mathrm{d}x\\ =&\,\int_0^p\bigg(\frac32\sec u\tan u\\ &\,\hspace{2pt}+\frac12\sec^2u\bigg)\,\mathrm{d}u \end{align*}

where pp is a constant to be found.

(4)

(b) Hence find the exact value of

∫013x+2(4−x2)32 dx\int_0^1\frac{3x+2}{(4-x^2)^{\frac32}}\,\mathrm{d}x
(4)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

依赖计算器技术的解法不被接受。

(a) 令 x=2sin⁡ux=2\sin u，证明

 ∫013x+2(4−x2)32 dx= ∫0p(32sec⁡utan⁡u +12sec⁡2u) du\begin{align*} &\,\int_0^1\frac{3x+2}{(4-x^2)^{\frac32}}\,\mathrm{d}x\\ =&\,\int_0^p\bigg(\frac32\sec u\tan u\\ &\,\hspace{2pt}+\frac12\sec^2u\bigg)\,\mathrm{d}u \end{align*}

其中 pp 是待求常数。

(4)

(b) 进而求

∫013x+2(4−x2)32 dx\int_0^1\frac{3x+2}{(4-x^2)^{\frac32}}\,\mathrm{d}x

的精确值。

(4)

解答

(a)

解法一

思路

展开

由 x=2sin⁡ux=2\sin u 同时替换 xx、dx\mathrm{d}x 和积分上下限。利用 1−sin⁡2u=cos⁡2u1-\sin^2u=\cos^2u 化简分母；在新的积分区间内 cos⁡u>0\cos u>0，因此 (4cos⁡2u)3/2=8cos⁡3u(4\cos^2u)^{3/2}=8\cos^3u。

答题过程

展开

Let

x=2sin⁡u.x=2\sin u.

Then

dx=2cos⁡u du.\mathrm{d}x=2\cos u\,\mathrm{d}u.

The limits transform as follows:

xu001sin⁡−1(12)=π6\begin{array}{c|c} x&u\\ \hline 0&0\\ 1&\sin^{-1}\big(\frac12\big)=\frac{\pi}{6} \end{array}

Hence p=π6p=\frac{\pi}{6}. Since 0≤u≤π60\le u\le\frac{\pi}{6}, we have cos⁡u>0\cos u>0, and therefore

(4−x2)32= (4−4sin⁡2u)32= (4cos⁡2u)32= 8cos⁡3u.\begin{align*} (4-x^2)^{\frac32} =&\,(4-4\sin^2u)^{\frac32}\\ =&\,(4\cos^2u)^{\frac32}\\ =&\,8\cos^3u. \end{align*}

Thus

 ∫013x+2(4−x2)32 dx= ∫0π66sin⁡u+24cos⁡2u du= ∫0π6(3sin⁡u2cos⁡2u +12cos⁡2u) du= ∫0π6(32sec⁡utan⁡u +12sec⁡2u) du,\begin{align*} &\,\int_0^1 \frac{3x+2}{(4-x^2)^{\frac32}}\,\mathrm{d}x\\ =&\,\int_0^{\frac{\pi}{6}} \frac{6\sin u+2}{4\cos^2u}\,\mathrm{d}u\\ =&\,\int_0^{\frac{\pi}{6}} \bigg( \frac{3\sin u}{2\cos^2u}\\ &\,\hspace{2pt}+\frac{1}{2\cos^2u} \bigg)\,\mathrm{d}u\\ =&\,\int_0^{\frac{\pi}{6}} \bigg( \frac32\sec u\tan u\\ &\,\hspace{2pt}+\frac12\sec^2u \bigg)\,\mathrm{d}u, \end{align*}

as required, with

p=π6.\boxed{p=\frac{\pi}{6}}.

(b)

解法一

思路

展开

承接 (a) 的换元结果，分别使用 ∫sec⁡utan⁡u du=sec⁡u\int\sec u\tan u\,\mathrm{d}u=\sec u 和 ∫sec⁡2u du=tan⁡u\int\sec^2u\,\mathrm{d}u=\tan u，再代入新的上下限并化简精确值。

答题过程

展开

Using the result from part (a),

 ∫013x+2(4−x2)32 dx= [32sec⁡u+12tan⁡u]0π6= 32(23) +12(13)−32= 3+36−32= 73−96.\begin{align*} &\,\int_0^1 \frac{3x+2}{(4-x^2)^{\frac32}}\,\mathrm{d}x\\ =&\,\Bigg[ \frac32\sec u+\frac12\tan u \Bigg]_0^{\frac{\pi}{6}}\\ =&\,\frac32\bigg(\frac{2}{\sqrt3}\bigg) \\ &\,\hspace{2pt}+\frac12\bigg(\frac{1}{\sqrt3}\bigg)-\frac32\\ =&\,\sqrt3+\frac{\sqrt3}{6}-\frac32\\ =&\,\boxed{\frac{7\sqrt3-9}{6}}. \end{align*}