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FP3 2025 June Q5

A Level / Edexcel / FP3

IAL 2025 June Paper · Question 5

题目

Problem

5.

In= ∫16xn(3x−2)−1/2 dxn≥0\begin{align*} I_n =&\, \int_1^6 x^n(3x-2)^{-1/2}\,\mathrm{d}x \qquad n\ge 0 \end{align*}

(a) Show that, for n≥1n \ge 1

(3+6n)In= 4nIn−1+8⋅6n−2\begin{align*} (3+6n)I_n =&\, 4nI_{n-1} + 8\cdot 6^n-2 \end{align*}
(5)

(b) Use the reduction formula in part (a) to determine the exact value of

∫16x3(3x−2)−1/2 dx\begin{align*} \int_1^6 x^3(3x-2)^{-1/2}\,\mathrm{d}x \end{align*}
(4)
题目中文翻译 In= ∫16xn(3x−2)−1/2 dxn≥0\begin{align*} I_n =&\, \int_1^6 x^n(3x-2)^{-1/2}\,\mathrm{d}x \qquad n\ge 0 \end{align*}

(a) 证明，当 n≥1n \ge 1 时

(3+6n)In= 4nIn−1+8⋅6n−2\begin{align*} (3+6n)I_n =&\, 4nI_{n-1} + 8\cdot 6^n-2 \end{align*}
(5)

(b) 利用 (a) 的递推公式求

∫16x3(3x−2)−1/2 dx\begin{align*} \int_1^6 x^3(3x-2)^{-1/2}\,\mathrm{d}x \end{align*}

的准确值。

(4)

解答

(a)

解法一

思路

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对 InI_n 分部积分，令 u=xnu=x^n，并把 (3x−2)−1/2(3x-2)^{-1/2} 积分为 23(3x−2)1/2\frac23(3x-2)^{1/2}。余下的根式改写成 (3x−2)(3x−2)−1/2(3x-2)(3x-2)^{-1/2}，便能拆出 3In−2In−13I_n-2I_{n-1}；最后计算边界项并整理。

答题过程

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Using integration by parts, let

u=xnu=x^n

and

dv=(3x−2)−1/2 dx.\mathrm{d}v=(3x-2)^{-1/2}\,\mathrm{d}x.

Then

du=nxn−1 dx\mathrm{d}u=nx^{n-1}\,\mathrm{d}x

and

v=23(3x−2)1/2.v=\frac23(3x-2)^{1/2}.

Therefore,

In= [23xn(3x−2)1/2]16 −2n3∫16xn−1(3x−2)1/2 dx.\begin{align*} I_n =&\,\Big[\frac23x^n(3x-2)^{1/2}\Big]_1^6\\ &\,-\frac{2n}{3}\int_1^6 x^{n-1}(3x-2)^{1/2}\,\mathrm{d}x. \end{align*}

Now

(3x−2)1/2=(3x−2)(3x−2)−1/2,(3x-2)^{1/2} =(3x-2)(3x-2)^{-1/2},

so

 ∫16xn−1(3x−2)1/2 dx= ∫16xn−1(3x−2) ×(3x−2)−1/2 dx= 3In−2In−1.\begin{align*} &\,\int_1^6x^{n-1}(3x-2)^{1/2}\,\mathrm{d}x\\ =&\,\int_1^6x^{n-1}(3x-2)\\ &\,\hspace{2pt}\times(3x-2)^{-1/2}\,\mathrm{d}x\\ =&\,3I_n-2I_{n-1}. \end{align*}

Also,

[23xn(3x−2)1/2]16= 23(4⋅6n−1)= 83 6n−23.\begin{align*} \Big[\frac23x^n(3x-2)^{1/2}\Big]_1^6 =&\,\frac23\big(4\cdot6^n-1\big)\\ =&\,\frac83\,6^n-\frac23. \end{align*}

Hence

In= 83 6n−23 −2n3(3In−2In−1),3In= 8⋅6n−2−6nIn+4nIn−1.\begin{align*} I_n =&\,\frac83\,6^n-\frac23\\ &\,-\frac{2n}{3}\big(3I_n-2I_{n-1}\big),\\ 3I_n =&\,8\cdot6^n-2-6nI_n+4nI_{n-1}. \end{align*}

Therefore,

(3+6n)In=4nIn−1+8⋅6n−2,\boxed{ (3+6n)I_n =4nI_{n-1}+8\cdot6^n-2},

as required.

(b)

解法一

思路

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所求积分就是 I3I_3。先直接求 I0I_0，再在 (a) 的递推式中依次令 n=1,2,3n=1,2,3，逐级算出 I1,I2,I3I_1,I_2,I_3；全程保留精确分数。

答题过程

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First,

I0= ∫16(3x−2)−1/2 dx= [23(3x−2)1/2]16= 23(4−1)= 2.\begin{align*} I_0 =&\,\int_1^6(3x-2)^{-1/2}\,\mathrm{d}x\\ =&\,\Big[\frac23(3x-2)^{1/2}\Big]_1^6\\ =&\,\frac23(4-1)\\ =&\,2. \end{align*}

Using the reduction formula with n=1n=1,

9I1= 4I0+8⋅6−2= 4(2)+48−2= 54,\begin{align*} 9I_1 =&\,4I_0+8\cdot6-2\\ =&\,4(2)+48-2\\ =&\,54, \end{align*}

so I1=6I_1=6.

With n=2n=2,

15I2= 8I1+8⋅62−2= 8(6)+288−2= 334,\begin{align*} 15I_2 =&\,8I_1+8\cdot6^2-2\\ =&\,8(6)+288-2\\ =&\,334, \end{align*}

so

I2=33415.I_2=\frac{334}{15}.

Finally, with n=3n=3,

21I3= 12I2+8⋅63−2= 12(33415)+1728−2= 99665.\begin{align*} 21I_3 =&\,12I_2+8\cdot6^3-2\\ =&\,12\bigg(\frac{334}{15}\bigg)+1728-2\\ =&\,\frac{9966}{5}. \end{align*}

Therefore,

I3=332235.\boxed{I_3=\frac{3322}{35}}.