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FP3 2021 June Q7

A Level / Edexcel / FP3

IAL 2021 June Paper · Question 7

题目

Problem

7.

Using calculus, find the exact values of

(i)

∫121x2−4x+5 dx\int_1^2 \frac{1}{x^2 - 4x + 5}\,\mathrm{d}x
(3)

(ii)

∫33x2−3x2 dx.\int_{\sqrt3}^3 \frac{\sqrt{x^2-3}}{x^2}\,\mathrm{d}x.
(5)
题目中文翻译

利用微积分求下列各积分的精确值：

(i)

∫121x2−4x+5 dx\int_1^2 \frac{1}{x^2 - 4x + 5}\,\mathrm{d}x
(3)

(ii)

∫33x2−3x2 dx\int_{\sqrt3}^3 \frac{\sqrt{x^2-3}}{x^2}\,\mathrm{d}x
(5)

解答

(i)

解法一

思路

展开

先对分母完成平方，化为 1+(x−2)21+(x-2)^2，再使用反正切的标准积分并代入上下限。

答题过程

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Complete the square:

x2−4x+5=(x−2)2+1.x^2-4x+5=(x-2)^2+1.

Therefore

∫121x2−4x+5 dx= [arctan⁡(x−2)]12= arctan⁡0−arctan⁡(−1)= π4.\begin{align*} \int_1^2\frac{1}{x^2-4x+5}\,\mathrm{d}x =&\,\big[\arctan(x-2)\big]_1^2\\[4mm] =&\,\arctan0-\arctan(-1)\\[4mm] =&\,\boxed{\frac\pi4}. \end{align*}

(ii)

解法一

思路

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选择分部积分，让 x−2x^{-2} 积分成 −x−1-x^{-1}。所得余项正好化为 ∫1/x2−3 dx\int1/\sqrt{x^2-3}\,\mathrm{d}x，可写成反双曲余弦或对数形式。

答题过程

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Let

u=x2−3,dv=1x2 dx.u=\sqrt{x^2-3}, \qquad \mathrm{d}v=\frac{1}{x^2}\,\mathrm{d}x.

Then

du=xx2−3 dx,v=−1x.\mathrm{d}u=\frac{x}{\sqrt{x^2-3}}\,\mathrm{d}x, \qquad v=-\frac1x.

Hence

∫x2−3x2 dx= −x2−3x +∫1x2−3 dx= −x2−3x+arcosh⁡ ⁣(x3).\begin{align*} \int\frac{\sqrt{x^2-3}}{x^2}\,\mathrm{d}x =&\,-\frac{\sqrt{x^2-3}}{x}\\[4mm] &\,\hspace{2pt}+\int\frac{1}{\sqrt{x^2-3}}\,\mathrm{d}x\\[4mm] =&\,-\frac{\sqrt{x^2-3}}{x} +\operatorname{arcosh}\!\big(\frac{x}{\sqrt3}\big). \end{align*}

Therefore

 ∫33x2−3x2 dx= [−x2−3x+arcosh⁡ ⁣(x3)]33= −63+arcosh⁡3= ln⁡(3+2)−63.\begin{align*} &\,\int_{\sqrt3}^{3} \frac{\sqrt{x^2-3}}{x^2}\,\mathrm{d}x\\[4mm] =&\,\big[ -\frac{\sqrt{x^2-3}}{x} +\operatorname{arcosh}\!\big(\frac{x}{\sqrt3}\big) \big]_{\sqrt3}^{3}\\[4mm] =&\,-\frac{\sqrt6}{3}+\operatorname{arcosh}\sqrt3\\[4mm] =&\,\boxed{\ln(\sqrt3+\sqrt2)-\frac{\sqrt6}{3}}. \end{align*}

解法二

思路

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令 x=3cosh⁡ux=\sqrt3\cosh u。根号、微分和分母约去后变成 tanh⁡2u\tanh^2u，再用 tanh⁡2u=1−sech⁡2u\tanh^2u=1-\operatorname{sech}^2u 积分。

答题过程

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Let

x=3cosh⁡u.x=\sqrt3\cosh u.

Then

x2−3=3sinh⁡u,dx=3sinh⁡u du.\sqrt{x^2-3}=\sqrt3\sinh u, \qquad \mathrm{d}x=\sqrt3\sinh u\,\mathrm{d}u.

The limits become

x=3⇒u=0,x=3⇒u=arcosh⁡3.x=\sqrt3\Rightarrow u=0, \qquad x=3\Rightarrow u=\operatorname{arcosh}\sqrt3.

Thus

 ∫33x2−3x2 dx= ∫0arcosh⁡3(3sinh⁡u)(3sinh⁡u)3cosh⁡2u du= ∫0arcosh⁡3tanh⁡2u du= ∫0arcosh⁡3(1−sech⁡2u) du= [u−tanh⁡u]0arcosh⁡3= arcosh⁡3−63= ln⁡(3+2)−63.\begin{align*} &\,\int_{\sqrt3}^{3} \frac{\sqrt{x^2-3}}{x^2}\,\mathrm{d}x\\[4mm] =&\,\int_0^{\operatorname{arcosh}\sqrt3} \frac{(\sqrt3\sinh u)(\sqrt3\sinh u)}{3\cosh^2u}\,\mathrm{d}u\\ =&\,\int_0^{\operatorname{arcosh}\sqrt3} \tanh^2u\,\mathrm{d}u\\ =&\,\int_0^{\operatorname{arcosh}\sqrt3} (1-\operatorname{sech}^2u)\,\mathrm{d}u\\[4mm] =&\,\big[u-\tanh u\big]_0^{\operatorname{arcosh}\sqrt3}\\[4mm] =&\,\operatorname{arcosh}\sqrt3-\frac{\sqrt6}{3}\\[4mm] =&\,\boxed{\ln(\sqrt3+\sqrt2)-\frac{\sqrt6}{3}}. \end{align*}

解法三

思路

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令 x=3sec⁡ux=\sqrt3\sec u，使根号化为 3tan⁡u\sqrt3\tan u。换元后用 tan⁡2u=sec⁡2u−1\tan^2u=\sec^2u-1，把被积式拆成 sec⁡u−cos⁡u\sec u-\cos u，即可使用标准积分。

答题过程

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Let x=3sec⁡ux=\sqrt3\sec u, with 0≤u<π/20\le u<\pi/2. Then

x2−3=3tan⁡u,dx=3sec⁡utan⁡u du.\sqrt{x^2-3}=\sqrt3\tan u, \qquad \mathrm{d}x=\sqrt3\sec u\tan u\,\mathrm{d}u.

When x=3x=\sqrt3, u=0u=0. When x=3x=3, u=αu=\alpha, where

sec⁡α=3,tan⁡α=2,sin⁡α=63.\sec\alpha=\sqrt3,\qquad \tan\alpha=\sqrt2,\qquad \sin\alpha=\frac{\sqrt6}{3}.

Therefore

 ∫33x2−3x2 dx= ∫0α(3tan⁡u)(3sec⁡utan⁡u)3sec⁡2u du= ∫0αtan⁡2usec⁡u du= ∫0α(sec⁡u−cos⁡u) du= [ln⁡(sec⁡u+tan⁡u)−sin⁡u]0α= ln⁡(3+2)−63−(ln⁡1−0)= ln⁡(3+2)−63.\begin{align*} &\,\int_{\sqrt3}^3\frac{\sqrt{x^2-3}}{x^2}\,\mathrm{d}x\\ =&\,\int_0^\alpha \frac{(\sqrt3\tan u)(\sqrt3\sec u\tan u)}{3\sec^2u}\,\mathrm{d}u\\ =&\,\int_0^\alpha\frac{\tan^2u}{\sec u}\,\mathrm{d}u\\ =&\,\int_0^\alpha(\sec u-\cos u)\,\mathrm{d}u\\ =&\,\big[\ln(\sec u+\tan u)-\sin u\big]_0^\alpha\\ =&\,\ln(\sqrt3+\sqrt2)-\frac{\sqrt6}{3}-(\ln1-0)\\ =&\,\boxed{\ln(\sqrt3+\sqrt2)-\frac{\sqrt6}{3}}. \end{align*}