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P3 2020 Jan Q8

A Level / Edexcel / P3

IAL 2020 Jan Paper · Question 8

题目

Problem

(i) Find, using algebraic integration, the exact value of

∫34223x−1 dx\int_3^{42}\frac{2}{3x-1}\,\mathrm{d}x

giving your answer in simplest form.

(4)

(ii)

h(x)=2x3−7x2+8x+1(x−1)2x>1h(x)=\frac{2x^3-7x^2+8x+1}{(x-1)^2}\qquad x>1

Given h(x)=Ax+B+C(x−1)2h(x)=Ax+B+\dfrac{C}{(x-1)^2} where AA, BB and CC are constants to be found, find

∫h(x) dx\int h(x)\,dx
(6)
题目中文翻译

(i) 利用代数积分法求

∫34223x−1 dx\int_3^{42}\frac{2}{3x-1}\,\mathrm{d}x

的精确值，并将答案化为最简形式。

(ii)

h(x)=2x3−7x2+8x+1(x−1)2x>1h(x)=\frac{2x^3-7x^2+8x+1}{(x-1)^2}\qquad x>1

已知

h(x)=Ax+B+C(x−1)2h(x)=Ax+B+\dfrac{C}{(x-1)^2}

其中 AA、BB 和 CC 为待求常数，求

∫h(x) dx\int h(x)\,dx

解答

(i)

解法一

思路

展开

令 u=3x−1u=3x-1，把被积函数化成 1/u1/u 的倍数，再用对数法则化简。

答题过程

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Let u=3x−1u=3x-1, so du=3 dx\mathrm{d}u=3\,\mathrm{d}x. The limits x=3x=3 and x=42x=42 give u=8u=8 and u=125u=125 respectively.

∫34223x−1 dx= 23∫81251u du= 23[ln⁡u]8125= 23ln⁡1258.\begin{align*} \int_3^{42}\frac{2}{3x-1}\,\mathrm{d}x =&\,\frac23\int_8^{125}\frac{1}{u}\,\mathrm{d}u\\ =&\,\frac23\big[\ln u\big]_8^{125}\\ =&\,\frac23\ln\frac{125}{8}. \end{align*}

Since 125/8=(5/2)3125/8=(5/2)^3, this simplifies to

ln⁡254.\boxed{\ln\frac{25}{4}}.

(ii)

解法一

思路

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乘去分母并比较系数，求出 A,B,CA,B,C 后逐项积分。

答题过程

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We are given

h(x)=2x3−7x2+8x+1(x−1)2h(x)=\frac{2x^3-7x^2+8x+1}{(x-1)^2}

and

h(x)=Ax+B+C(x−1)2h(x)=Ax+B+\frac{C}{(x-1)^2}

Multiply by (x−1)2(x-1)^2:

2x3−7x2+8x+1=(Ax+B)(x−1)2+C2x^3-7x^2+8x+1=(Ax+B)(x-1)^2+C

Expand:

(Ax+B)(x2−2x+1)+C(Ax+B)(x^2-2x+1)+C

So

2x3−7x2+8x+1=Ax3+(B−2A)x2+(A−2B)x+(B+C)2x^3-7x^2+8x+1 =Ax^3+(B-2A)x^2+(A-2B)x+(B+C)

Compare coefficients:

A=2A=2

Then

B−2A=−7B-2A=-7

so

B−4=−7B-4=-7

and therefore

B=−3B=-3

Now check the coefficient of xx:

A−2B=2−2(−3)=8A-2B=2-2(-3)=8

which agrees with the numerator.

For the constant term:

B+C=1B+C=1

so

−3+C=1-3+C=1

and therefore

C=4C=4

Thus

h(x)=2x−3+4(x−1)2h(x)=2x-3+\frac{4}{(x-1)^2}

Now integrate:

∫h(x) dx=∫(2x−3+4(x−1)−2) dx=x2−3x+4⋅(x−1)−1−1=x2−3x−4x−1+c\begin{aligned} \int h(x)\,\mathrm{d}x &=\int \left(2x-3+4(x-1)^{-2}\right)\,\mathrm{d}x\\ &=x^2-3x+4\cdot \frac{(x-1)^{-1}}{-1}\\ &=x^2-3x-\frac{4}{x-1}+c \end{aligned}

Therefore

∫h(x) dx=x2−3x−4x−1+c\boxed{\int h(x)\,\mathrm{d}x=x^2-3x-\frac{4}{x-1}+c}

解法二

思路

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将分子除以 (x−1)2(x-1)^2，利用一次商与常数余数直接读出系数。

答题过程

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Divide 2x3−7x2+8x+12x^3-7x^2+8x+1 by (x−1)2=x2−2x+1(x-1)^2=x^2-2x+1. Subtracting 2x(x−1)22x(x-1)^2 leaves −3x2+6x+1-3x^2+6x+1. Subtracting −3(x−1)2-3(x-1)^2 then leaves 44.

Thus the quotient is 2x−32x-3 and the remainder is 44, so A=2A=2, B=−3B=-3 and C=4C=4. Integrating term by term gives

∫h(x) dx= ∫(2x−3+4(x−1)−2) dx= x2−3x−4x−1+c.\begin{align*} \int h(x)\,\mathrm{d}x =&\,\int \Bigl(2x-3+4(x-1)^{-2}\Bigr)\,\mathrm{d}x\\ =&\,\boxed{x^2-3x-\frac{4}{x-1}+c}. \end{align*}