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IAL 2021 Oct FP3 Q6

A Level / Edexcel / FP3

IAL 2021 Oct Paper · Question 6

题目

Problem

Let

In=∫0π/2xncos⁡(x2) dxn≥1.I_n = \int_0^{\sqrt{\pi/2}} x^n \cos(x^2)\,dx \qquad n \ge 1.

(a) Prove that, for n≥5n \ge 5,

In=12(π2)(n−1)/2−(n−1)(n−3)4In−4.I_n = \frac{1}{2}\left(\frac{\pi}{2}\right)^{(n-1)/2} - \frac{(n-1)(n-3)}{4}I_{n-4}.

(b) Hence, determine the exact value of I5I_5, giving your answer in its simplest form.

(9)
题目中文翻译

设

In=∫0π/2xncos⁡(x2) dxn≥1I_n = \int_0^{\sqrt{\pi/2}} x^n \cos(x^2)\,dx \qquad n \ge 1

(a) 证明当 n≥5n \ge 5 时

In=12(π2)(n−1)/2−(n−1)(n−3)4In−4I_n = \frac{1}{2}\left(\frac{\pi}{2}\right)^{(n-1)/2} - \frac{(n-1)(n-3)}{4}I_{n-4}

(b) 因此求 I5I_5 的精确值，并写成最简形式。

解答

(a)

解法一：降低幂次

思路

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这是官方评分资料的第一条路线。先把 xnx^n 拆成 xn−1⋅xx^{n-1}\cdot x，对 xcos⁡(x2)x\cos(x^2) 积分；所得余项再把 xn−2x^{n-2} 拆成 xn−3⋅xx^{n-3}\cdot x，第二次分部积分后便出现 In−4I_{n-4}。条件 n≥5n\ge5 保证下限处的幂次项为零。

答题过程

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Let

L=π2.L=\sqrt{\frac{\pi}{2}}.

Writing xn=xn−1xx^n=x^{n-1}x and integrating by parts,

In= ∫0Lxn−1(xcos⁡(x2)) dx= [12xn−1sin⁡(x2)]0L −n−12∫0Lxn−2sin⁡(x2) dx.\begin{align*} I_n =&\,\int_0^L x^{n-1} \bigl(x\cos(x^2)\bigr)\,\mathrm{d}x \\ =&\,\left[ \frac12x^{n-1}\sin(x^2) \right]_0^L \\ &\,-\frac{n-1}{2} \int_0^L x^{n-2}\sin(x^2)\,\mathrm{d}x. \end{align*}

For the remaining integral, write xn−2=xn−3xx^{n-2}=x^{n-3}x and integrate by parts again:

 ∫0Lxn−2sin⁡(x2) dx= [−12xn−3cos⁡(x2)]0L +n−32∫0Lxn−4cos⁡(x2) dx= n−32In−4.\begin{align*} &\,\int_0^L x^{n-2}\sin(x^2)\,\mathrm{d}x \\ =&\,\left[ -\frac12x^{n-3}\cos(x^2) \right]_0^L \\ &\,+\frac{n-3}{2} \int_0^L x^{n-4}\cos(x^2)\,\mathrm{d}x \\ =&\,\frac{n-3}{2}I_{n-4}. \end{align*}

Here the boundary term is zero because L2=π2L^2=\frac{\pi}{2}, cos⁡(L2)=0\cos(L^2)=0, and n−3≥2n-3\ge2. Also,

[12xn−1sin⁡(x2)]0L=12Ln−1.\left[ \frac12x^{n-1}\sin(x^2) \right]_0^L =\frac12L^{n-1}.

Therefore,

In= 12Ln−1−(n−1)(n−3)4In−4= 12(π2)(n−1)/2−(n−1)(n−3)4In−4.\begin{align*} I_n =&\,\frac12L^{n-1} -\frac{(n-1)(n-3)}{4}I_{n-4} \\ =&\,\boxed{ \frac12\bigg(\frac{\pi}{2}\bigg)^{(n-1)/2} -\frac{(n-1)(n-3)}{4}I_{n-4} }. \end{align*}

解法二：先建立 In+4I_{n+4} 的关系式

思路

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官方第二条路线先把 xnx^n 积分、把 cos⁡(x2)\cos(x^2) 求导，使余项的幂次上升；再做一次分部积分，得到 In+4I_{n+4} 关于 InI_n 的关系。最后把指标 nn 换成 n−4n-4，即可得到题设递推式。

答题过程

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Again let L=π2L=\sqrt{\frac{\pi}{2}}. Integrating by parts with u=cos⁡(x2)u=\cos(x^2) and dv=xn dx\mathrm{d}v=x^n\,\mathrm{d}x gives

In= [xn+1n+1cos⁡(x2)]0L +2n+1∫0Lxn+2sin⁡(x2) dx= 2n+1∫0Lxn+2sin⁡(x2) dx.\begin{align*} I_n =&\,\left[ \frac{x^{n+1}}{n+1}\cos(x^2) \right]_0^L \\ &\,+\frac{2}{n+1} \int_0^L x^{n+2}\sin(x^2)\,\mathrm{d}x \\ =&\,\frac{2}{n+1} \int_0^L x^{n+2}\sin(x^2)\,\mathrm{d}x. \end{align*}

Integrating the remaining integral by parts,

 ∫0Lxn+2sin⁡(x2) dx= [xn+3n+3sin⁡(x2)]0L −2n+3∫0Lxn+4cos⁡(x2) dx= Ln+3n+3−2n+3In+4.\begin{align*} &\,\int_0^L x^{n+2}\sin(x^2)\,\mathrm{d}x \\ =&\,\left[ \frac{x^{n+3}}{n+3}\sin(x^2) \right]_0^L \\ &\,-\frac{2}{n+3} \int_0^L x^{n+4}\cos(x^2)\,\mathrm{d}x \\ =&\,\frac{L^{n+3}}{n+3} -\frac{2}{n+3}I_{n+4}. \end{align*}

Hence

In= 2Ln+3(n+1)(n+3) −4(n+1)(n+3)In+4,\begin{align*} I_n =&\,\frac{2L^{n+3}}{(n+1)(n+3)} \\ &\,-\frac{4}{(n+1)(n+3)}I_{n+4}, \end{align*}

so

In+4=12Ln+3−(n+1)(n+3)4In.I_{n+4} =\frac12L^{n+3} -\frac{(n+1)(n+3)}{4}I_n.

Replacing nn by n−4n-4 and using L2=π2L^2=\frac{\pi}{2},

In=12(π2)(n−1)/2−(n−1)(n−3)4In−4.\boxed{ I_n =\frac12\bigg(\frac{\pi}{2}\bigg)^{(n-1)/2} -\frac{(n-1)(n-3)}{4}I_{n-4} }.

(b)

解法一

思路

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题目要求使用 (a)，所以先直接求出递推式所需的 I1I_1，再令 n=5n=5。此时 In−4=I1I_{n-4}=I_1，代入并化简即可。

答题过程

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First,

I1= ∫0Lxcos⁡(x2) dx= [12sin⁡(x2)]0L= 12.\begin{align*} I_1 =&\,\int_0^L x\cos(x^2)\,\mathrm{d}x \\ =&\,\left[\frac12\sin(x^2)\right]_0^L \\ =&\,\frac12. \end{align*}

Using the reduction formula from part (a) with n=5n=5,

I5= 12(π2)2−(5−1)(5−3)4I1= π28−2(12)= π28−1.\begin{align*} I_5 =&\,\frac12\bigg(\frac{\pi}{2}\bigg)^2 -\frac{(5-1)(5-3)}{4}I_1 \\ =&\,\frac{\pi^2}{8}-2\bigg(\frac12\bigg) \\ =&\,\boxed{\frac{\pi^2}{8}-1}. \end{align*}