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P4 2020 Oct Q7

A Level / Edexcel / P4

IAL 2020 Oct Paper · Question 7

题目

Problem

(i) Using a suitable substitution, find, using calculus, the value of

∫153x2x−1 dx\int_1^5 \frac{3x}{\sqrt{2x-1}}\,\mathrm{d}x

(Solutions relying entirely on calculator technology are not acceptable.)

(6)

(ii) Find

∫6x2−16(x+1)(2x−3) dx\int \frac{6x^2-16}{(x+1)(2x-3)}\,\mathrm{d}x
(6)
题目中文翻译

(i) 采用适当的代换，使用微积分求

∫153x2x−1 dx\int_1^5 \frac{3x}{\sqrt{2x-1}}\,\mathrm{d}x

（完全依赖计算器技术的解法不接受。）

(6)

(ii) 求

∫6x2−16(x+1)(2x−3) dx\int \frac{6x^2-16}{(x+1)(2x-3)}\,\mathrm{d}x
(6)

解答

(i)

解法一

思路

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令 u=2x−1u=\sqrt{2x-1}，这样根式本身直接变为 uu，而 xx 与 dx\mathrm{d}x 也都能改写成 uu。定积分换元时同时更换上下限，所得被积函数只是关于 uu 的多项式。

答题过程

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Let

u=2x−1.u=\sqrt{2x-1}.

Then

u2= 2x−1,x= u2+12,dx= u du.\begin{align*} u^2 =&\,2x-1,\\ x =&\,\frac{u^2+1}{2},\\ \mathrm{d}x =&\,u\,\mathrm{d}u. \end{align*}

The limits become

x=1⇒ u=1,x=5⇒ u=3.\begin{align*} x=1\Rightarrow&\,u=1,\\ x=5\Rightarrow&\,u=3. \end{align*}

Therefore,

I=∫153x2x−1 dx.I=\int_1^5\frac{3x}{\sqrt{2x-1}}\,\mathrm{d}x.

Then

I= ∫133(u2+1)2u(u du)= 32∫13(u2+1) du= [12u3+32u]13= (18)−(2)= 16.\begin{align*} I =&\,\int_1^3 \frac{3(u^2+1)}{2u} \big(u\,\mathrm{d}u\big)\\ =&\,\frac32\int_1^3(u^2+1)\,\mathrm{d}u\\ =&\,\bigg[ \frac12u^3+\frac32u \bigg]_1^3\\ =&\,(18)-(2)\\ =&\,\boxed{16}. \end{align*}

解法二

思路

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也可直接令 u=2x−1u=2x-1。此时根式变为 u1/2u^{1/2}，而 x=u+12x=\frac{u+1}{2}；展开后得到两个幂函数，逐项积分即可。

答题过程

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Let

u=2x−1.u=2x-1.

Then

x= u+12,dx= 12 du.\begin{align*} x =&\,\frac{u+1}{2},\\ \mathrm{d}x =&\,\frac12\,\mathrm{d}u. \end{align*}

The limits are u=1u=1 when x=1x=1, and u=9u=9 when x=5x=5. Hence,

I=∫153x2x−1 dx.I=\int_1^5\frac{3x}{\sqrt{2x-1}}\,\mathrm{d}x.

Therefore,

 I= 34∫19(u1/2+u−1/2) du= [12u3/2+32u1/2]19= (18)−(2)= 16.\begin{align*} &\,I\\ =&\,\frac34\int_1^9 \big(u^{1/2}+u^{-1/2}\big)\,\mathrm{d}u\\ =&\,\bigg[ \frac12u^{3/2}+\frac32u^{1/2} \bigg]_1^9\\ =&\,(18)-(2)\\ =&\,\boxed{16}. \end{align*}

(ii)

解法一

思路

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分子与分母同为二次式，可以先作多项式除法，再把余下的真分式作部分分式分解。积分时注意 2x−32x-3 的导数是 22，因此对应的对数项会带有系数 12\frac12。

答题过程

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Since

 (x+1)(2x−3)= 2x2−x−3,\begin{align*} &\,(x+1)(2x-3)\\ =&\,2x^2-x-3, \end{align*}

polynomial division gives

 F(x)= 6x2−16(x+1)(2x−3).\begin{align*} &\,F(x)\\ =&\,\frac{6x^2-16}{(x+1)(2x-3)}. \end{align*}

Then

 F(x)= 3+3x−7(x+1)(2x−3).\begin{align*} &\,F(x)\\ =&\,3+\frac{3x-7}{(x+1)(2x-3)}. \end{align*}

For the remaining fraction, write

 3x−7(x+1)(2x−3)= Ax+1+ B2x−3.\begin{align*} &\,\frac{3x-7}{(x+1)(2x-3)}\\ =&\,\frac{A}{x+1}\\ +&\,\frac{B}{2x-3}. \end{align*}

Thus

 3x−7= A(2x−3) +B(x+1).\begin{align*} &\,3x-7\\ =&\,A(2x-3)\\ &\,\hspace{2pt}+B(x+1). \end{align*}

Setting x=−1x=-1 gives A=2A=2, and setting x=32x=\frac32 gives B=−1B=-1. Therefore,

F(x)= 3+2x+1− 12x−3.\begin{align*} F(x) =&\,3+\frac{2}{x+1}\\ -&\,\frac{1}{2x-3}. \end{align*}

Hence,

 ∫F(x) dx= ∫3 dx+ 2∫1x+1 dx− ∫12x−3 dx=  3x+2ln⁡∣x+1∣ −12ln⁡∣2x−3∣+c.\begin{align*} &\,\int F(x)\,\mathrm{d}x\\ =&\,\int3\,\mathrm{d}x\\ +&\,2\int\frac{1}{x+1}\,\mathrm{d}x\\ -&\,\int\frac{1}{2x-3}\,\mathrm{d}x\\ =&\,\boxed{ \begin{aligned} &\,3x+2\ln|x+1|\\ &\,\hspace{2pt}-\frac12\ln|2x-3|+c \end{aligned} }. \end{align*}

解法二

思路

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也可以直接把原分式设为一个常数加两个真分式，不必先做除法。通分后比较二次项系数，求出常数项；再代入使某个因子为零的 xx 值，分别求出另外两个系数。最后逐项积分，注意线性因子 2x−32x-3 带来的系数 12\frac12，并保留积分常数。

答题过程

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Write

 6x2−16(x+1)(2x−3)= A+Bx+1 +C2x−3.\begin{align*} &\,\frac{6x^2-16}{(x+1)(2x-3)}\\ =&\,A+\frac{B}{x+1}\\ &\,\hspace{2pt}+\frac{C}{2x-3}. \end{align*}

Multiplying by (x+1)(2x−3)(x+1)(2x-3) gives the identity

 6x2−16= A(x+1)(2x−3) +B(2x−3) +C(x+1).\begin{align*} &\,6x^2-16\\ =&\,A(x+1)(2x-3)\\ &\,\hspace{2pt}+B(2x-3)\\ &\,\hspace{4pt}+C(x+1). \end{align*}

Comparing the coefficients of x2x^2,

6= 2A,A= 3.\begin{align*} 6=&\,2A,\\ A=&\,3. \end{align*}

Setting x=−1x=-1 in the identity,

−10= −5B,B= 2.\begin{align*} -10=&\,-5B,\\ B=&\,2. \end{align*}

Setting x=32x=\frac32,

−52= 52C,C= −1.\begin{align*} -\frac52=&\,\frac52C,\\ C=&\,-1. \end{align*}

Therefore,

 6x2−16(x+1)(2x−3)= 3+2x+1 −12x−3.\begin{align*} &\,\frac{6x^2-16}{(x+1)(2x-3)}\\ =&\,3+\frac{2}{x+1}\\ &\,\hspace{2pt}-\frac{1}{2x-3}. \end{align*}

Integrating term by term,

 ∫6x2−16(x+1)(2x−3) dx= 3∫ dx +2∫1x+1 dx −∫12x−3 dx=  3x+2ln⁡∣x+1∣ −12ln⁡∣2x−3∣+c.\begin{align*} &\,\int\frac{6x^2-16}{(x+1)(2x-3)}\,\mathrm{d}x\\ =&\,3\int\,\mathrm{d}x\\ &\,\hspace{2pt}+2\int\frac{1}{x+1}\,\mathrm{d}x\\ &\,\hspace{4pt}-\int\frac{1}{2x-3}\,\mathrm{d}x\\ =&\,\boxed{\begin{aligned} &\,3x+2\ln|x+1|\\ &\,\hspace{2pt}-\frac12\ln|2x-3|+c \end{aligned}}. \end{align*}