题目

Problem

Figure 4 shows a sketch of part of the curve with equation

y=\frac{3x-1}{x+2}\qquad x>-2

(a) Show that

\frac{3x-1}{x+2}=A+\frac{B}{x+2}

where $A$ and $B$ are constants to be found.

(2)

The finite region $R$ , shown shaded in Figure 4, is bounded by the curve, the line with equation $x=4$ , the $x$ -axis and the line with equation $x=1$ .

This region is rotated through $2\pi$ radians about the $x$ -axis to form a solid of revolution.

(b) Use the answer to part (a) and algebraic integration to find the exact volume of the solid generated, giving your answer in the form

\pi(p+q\ln 2)

where $p$ and $q$ are rational constants.

(6)

题目中文翻译

图 4 给出了曲线的一部分草图，其方程为

y=\frac{3x-1}{x+2}\qquad x>-2

(a) 证明

\frac{3x-1}{x+2}=A+\frac{B}{x+2}

其中 $A,B$ 为待求常数。

图 4 中阴影有限区域 $R$ 由该曲线、直线 $x=4$ 、 $x$ 轴以及直线 $x=1$ 围成。

将该区域绕 $x$ 轴旋转 $2\pi$ 弧度，得到一个旋转体。

(b) 利用 (a) 的结果和代数积分，求所生成立体的体积精确值，并将答案写成

\pi(p+q\ln 2)

的形式，其中 $p,q$ 为有理常数。

解答

(a)

We want

\frac{3x-1}{x+2}=A+\frac{B}{x+2}

Write the right hand side over the common denominator $x+2$ :

A+\frac{B}{x+2} =\frac{A(x+2)+B}{x+2}

3x-1=A(x+2)+B

Expanding,

3x-1=Ax+2A+B

Compare coefficients:

A=3

and

2A+B=-1

Substitute $A=3$ :

6+B=-1

B=-7

Therefore

\boxed{\frac{3x-1}{x+2}=3-\frac{7}{x+2}}

(b)

The volume of revolution about the $x$ -axis is

V=\pi\int_1^4 y^2\,\mathrm{d}x

Using part (a),

y=3-\frac{7}{x+2}

\begin{aligned} y^2 &=\left(3-\frac{7}{x+2}\right)^2 \\ &=9-\frac{42}{x+2}+\frac{49}{(x+2)^2} \end{aligned}

Therefore

V=\pi\int_1^4 \left(9-\frac{42}{x+2}+\frac{49}{(x+2)^2}\right)\,\mathrm{d}x

Integrate:

\begin{aligned} \int \left(9-\frac{42}{x+2}+\frac{49}{(x+2)^2}\right)\,\mathrm{d}x &=9x-42\ln(x+2)-\frac{49}{x+2} \end{aligned}

V=\pi\left[9x-42\ln(x+2)-\frac{49}{x+2}\right]_1^4

Apply the limits:

\begin{aligned} V &=\pi\left(\left(36-42\ln6-\frac{49}{6}\right) -\left(9-42\ln3-\frac{49}{3}\right)\right) \\ &=\pi\left(36-9-\frac{49}{6}+\frac{49}{3}-42\ln6+42\ln3\right) \\ &=\pi\left(27+\frac{49}{6}-42\ln2\right) \\ &=\pi\left(\frac{211}{6}-42\ln2\right) \end{aligned}

Hence the exact volume is

\boxed{\pi\left(\frac{211}{6}-42\ln2\right)}

IAL 2024 Oct Q7

题目

解答

(a)

(b)