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IAL 2021 June FP3 Q7

A Level / Edexcel / FP3

IAL 2021 June Paper · Question 7

题目

Problem

Using calculus, find the exact values of

(i) ∫12dxx2−4x+5\int_1^2 \frac{\mathrm{d}x}{x^2 - 4x + 5}

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

(8)
题目中文翻译

利用微积分求下列各积分的精确值：

(i) ∫12dxx2−4x+5\int_1^2 \frac{\mathrm{d}x}{x^2 - 4x + 5}

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

解答

(i)

解法一

思路

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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 =&\,\left[\arctan(x-2)\right]_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}\!\left(\frac{x}{\sqrt3}\right). \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] =&\,\left[ -\frac{\sqrt{x^2-3}}{x} +\operatorname{arcosh}\!\left(\frac{x}{\sqrt3}\right) \right]_{\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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官方替代解法令 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⁡3tanh⁡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} \tanh^2u\,\mathrm{d}u\\[4mm] =&\,\left[u-\tanh u\right]_0^{\operatorname{arcosh}\sqrt3}\\[4mm] =&\,\operatorname{arcosh}\sqrt3-\frac{\sqrt6}{3}\\[4mm] =&\,\boxed{\ln(\sqrt3+\sqrt2)-\frac{\sqrt6}{3}}. \end{align*}