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IAL 2024 June FP3 Q8

A Level / Edexcel / FP3

IAL 2024 June Paper · Question 8

题目

Problem

In=∫0kxn(k−x)1/2 dxn≥0I_n=\int_0^k x^n(k-x)^{1/2}\,dx \qquad n\ge 0

where kk is a positive constant.

(a) Show that

In=2kn3+2nIn−1n≥1I_n=\frac{2kn}{3+2n}I_{n-1}\qquad n\ge 1

Given that

∫0kx2(k−x)1/2 dx=93280\int_0^k x^2(k-x)^{1/2}\,dx=\frac{9\sqrt3}{280}

(b) use the result in part (a) to determine the exact value of kk.

(9)
题目中文翻译

设

In=∫0kxn(k−x)1/2 dxn≥0I_n=\int_0^k x^n(k-x)^{1/2}\,dx \qquad n\ge 0

其中 kk 为正实数。

(a) 证明

In=2kn3+2nIn−1n≥1I_n=\frac{2kn}{3+2n}I_{n-1}\qquad n\ge 1

已知

∫0kx2(k−x)1/2 dx=93280\int_0^k x^2(k-x)^{1/2}\,dx=\frac{9\sqrt3}{280}

(b) 利用 (a) 的结果求 kk 的精确值。

解答

(a)

解法一：直接分部积分

思路

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令 u=xnu=x^n，把 (k−x)1/2(k-x)^{1/2} 作为被积部分作分部积分。边界项在 n⩾1n\geqslant1 时为零；再把 (k−x)3/2(k-x)^{3/2} 拆成 (k−x)(k−x)1/2(k-x)(k-x)^{1/2}，便能将剩余积分写成 In−1I_{n-1} 与 InI_n。

答题过程

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For n⩾1n\geqslant1, integrate by parts with

u=xn,dvdx=(k−x)1/2.u=x^n, \qquad \frac{\mathrm{d}v}{\mathrm{d}x}=(k-x)^{1/2}.

Then

dudx=nxn−1,v=−23(k−x)3/2.\frac{\mathrm{d}u}{\mathrm{d}x}=nx^{n-1}, \qquad v=-\frac23(k-x)^{3/2}.

Hence

In= [−23xn(k−x)3/2]0k +2n3∫0kxn−1(k−x)3/2 dx.\begin{align*} I_n =&\,\biggl[ -\frac23x^n(k-x)^{3/2} \biggr]_0^k \\ &\,+\frac{2n}{3}\int_0^k x^{n-1}(k-x)^{3/2} \,\mathrm{d}x. \end{align*}

The boundary term is zero: at x=kx=k, (k−x)3/2=0(k-x)^{3/2}=0, while at x=0x=0, xn=0x^n=0 because n⩾1n\geqslant1. Therefore,

In= 2n3∫0kxn−1(k−x)(k−x)1/2 dx= 2n3[k∫0kxn−1(k−x)1/2 dx −∫0kxn(k−x)1/2 dx]= 2n3(kIn−1−In).\begin{align*} I_n =&\,\frac{2n}{3}\int_0^k x^{n-1}(k-x)(k-x)^{1/2} \,\mathrm{d}x \\ =&\,\frac{2n}{3}\biggl[ k\int_0^k x^{n-1}(k-x)^{1/2}\,\mathrm{d}x \\ &\,\hspace{54pt} -\int_0^k x^n(k-x)^{1/2}\,\mathrm{d}x \biggr] \\ =&\,\frac{2n}{3}\bigl(kI_{n-1}-I_n\bigr). \end{align*}

It follows that

(1+2n3)In= 2kn3In−1,(3+2n)In= 2knIn−1.\begin{align*} \biggl(1+\frac{2n}{3}\biggr)I_n =&\,\frac{2kn}{3}I_{n-1}, \\ (3+2n)I_n =&\,2knI_{n-1}. \end{align*}

Thus

In=2kn3+2nIn−1(n⩾1).\boxed{ I_n=\frac{2kn}{3+2n}I_{n-1} } \qquad(n\geqslant1).

解法二：先拆开因子再分部积分

思路

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先把 (k−x)1/2(k-x)^{1/2} 写成 (k−x)(k−x)−1/2(k-x)(k-x)^{-1/2}，从而将 InI_n 拆成两个积分。分别对这两个积分作分部积分后，边界项都为零，剩余部分直接变成 In−1I_{n-1} 与 InI_n。

答题过程

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First,

In= ∫0kxn(k−x)(k−x)−1/2 dx= k∫0kxn(k−x)−1/2 dx −∫0kxn+1(k−x)−1/2 dx.\begin{align*} I_n =&\,\int_0^k x^n(k-x)(k-x)^{-1/2} \,\mathrm{d}x \\ =&\,k\int_0^k x^n(k-x)^{-1/2} \,\mathrm{d}x \\ &\,-\int_0^k x^{n+1}(k-x)^{-1/2} \,\mathrm{d}x. \end{align*}

For the first integral, integration by parts gives

k∫0kxn(k−x)−1/2 dx= [−2kxn(k−x)1/2]0k +2kn∫0kxn−1(k−x)1/2 dx= 2knIn−1.\begin{align*} k\int_0^k x^n(k-x)^{-1/2}\,\mathrm{d}x =&\,\bigl[-2kx^n(k-x)^{1/2}\bigr]_0^k \\ &\,+2kn\int_0^k x^{n-1}(k-x)^{1/2} \,\mathrm{d}x \\ =&\,2knI_{n-1}. \end{align*}

Similarly,

∫0kxn+1(k−x)−1/2 dx= [−2xn+1(k−x)1/2]0k +2(n+1)∫0kxn(k−x)1/2 dx= 2(n+1)In.\begin{align*} \int_0^k x^{n+1}(k-x)^{-1/2}\,\mathrm{d}x =&\,\bigl[-2x^{n+1}(k-x)^{1/2}\bigr]_0^k \\ &\,+2(n+1)\int_0^k x^n(k-x)^{1/2} \,\mathrm{d}x \\ =&\,2(n+1)I_n. \end{align*}

Thus

In= 2knIn−1−2(n+1)In,(3+2n)In= 2knIn−1.\begin{align*} I_n =&\,2knI_{n-1}-2(n+1)I_n, \\ (3+2n)I_n =&\,2knI_{n-1}. \end{align*}

Therefore,

In=2kn3+2nIn−1(n⩾1).\boxed{ I_n=\frac{2kn}{3+2n}I_{n-1} } \qquad(n\geqslant1).

(b)

解法一

思路

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必须使用 (a) 的递推式：先依次把 I2I_2 化为 I1I_1、再化为 I0I_0，然后直接计算 I0I_0。与题给 I2I_2 比较后得到关于正数 kk 的方程，并识别出同为 7/27/2 次幂的形式。

答题过程

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Using the reduction formula from part (a),

I2=4k7I1,I1=2k5I0.I_2=\frac{4k}{7}I_1, \qquad I_1=\frac{2k}{5}I_0.

Also,

I0= ∫0k(k−x)1/2 dx= [−23(k−x)3/2]0k= 23k3/2.\begin{align*} I_0 =&\,\int_0^k(k-x)^{1/2}\,\mathrm{d}x \\ =&\,\biggl[-\frac23(k-x)^{3/2}\biggr]_0^k \\ =&\,\frac23k^{3/2}. \end{align*}

Therefore,

I2= 4k7⋅2k5⋅23k3/2= 16105k7/2.\begin{align*} I_2 =&\,\frac{4k}{7}\cdot\frac{2k}{5} \cdot\frac23k^{3/2} \\ =&\,\frac{16}{105}k^{7/2}. \end{align*}

Using the given value of I2I_2,

16105k7/2= 93280,k7/2= 273128= (34)7/2.\begin{align*} \frac{16}{105}k^{7/2} =&\,\frac{9\sqrt3}{280}, \\ k^{7/2} =&\,\frac{27\sqrt3}{128} \\ =&\,\biggl(\frac34\biggr)^{7/2}. \end{align*}

Since kk is positive,

k=34.\boxed{k=\frac34}.