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}\equiv 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}\equiv A+\frac{B}{x+2}

其中 $A,B$ 为待求常数。

(2)

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。

(6)

解答

(a)

解法一

思路

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通分后比较分子系数，求出 $A,B$ ；最后必须写出完整的分式恒等式。

答题过程

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We want

\frac{3x-1}{x+2}\equiv 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)

解法一

思路

展开

旋转体体积用 $\pi\int y^2\,\mathrm{d}x$ 。利用 (a) 先展开平方，分别积分并计算端点，最后合并对数。

答题过程

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The volume of revolution is

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

Using part (a), $y=3-7/(x+2)$ . Expand the square:

\begin{align*} y^2=&\,\bigg(3-\frac7{x+2}\bigg)^2\\ =&\,9-\frac{42}{x+2}+\frac{49}{(x+2)^2}. \end{align*}

Integrating these terms gives an antiderivative

F(x)=9x-42\ln(x+2)-\frac{49}{x+2}.

Here $x+2>0$ throughout the interval. Hence

V=\pi\big(F(4)-F(1)\big).

Compute each endpoint separately:

\begin{align*} F(4)=&\,36-42\ln6-\frac{49}{6}\\ =&\,\frac{167}{6}-42\ln6 \end{align*}

and

\begin{align*} F(1)=&\,9-42\ln3-\frac{49}{3}\\ =&\,-\frac{22}{3}-42\ln3. \end{align*}

Subtract the rational and logarithmic terms:

\frac{167}{6}-\bigg(-\frac{22}{3}\bigg)=\frac{211}{6},

-42\ln6+42\ln3=-42\ln2.

Therefore

\boxed{V=\pi\bigg(\frac{211}{6}-42\ln2\bigg)}.

若令 $u=x+2$ ，上下限变为 $3$ 与 $6$ ，则 $y^2=9-42/u+49/u^2$ ，原函数为 $9u-42\ln u-49/u$ 。它与上面的直接积分只是变量平移，代入上下限得到同一结果。

解法二

思路

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利用 (a) 令 $u=7/(x+2)$ ，把曲线写成 $y=3-u$ 。换元同时改变微分和上下限，保留反向积分的负号。

答题过程

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Use the result of part (a), $y=3-7/(x+2)$ , and set

u=\frac7{x+2}.

Then $y=3-u$ and

x=\frac7u-2, \qquad \mathrm{d}x=-\frac7{u^2}\,\mathrm{d}u.

The limits become

x=1:\quad u=\frac73=a,

x=4:\quad u=\frac76=b.

Retaining this direction of integration,

V=\pi\int_a^b(3-u)^2\bigg(-\frac7{u^2}\bigg)\,\mathrm{d}u.

The transformed integrand is

-\frac{63}{u^2}+\frac{42}{u}-7.

Since $u>0$ , its antiderivative is

G(u)=\frac{63}{u}+42\ln u-7u.

Evaluate the endpoints:

\begin{align*} G(b)=&\,54+42\ln\frac76-\frac{49}{6}\\ =&\,\frac{275}{6}+42\ln\frac76 \end{align*}

and

\begin{align*} G(a)=&\,27+42\ln\frac73-\frac{49}{3}\\ =&\,\frac{32}{3}+42\ln\frac73. \end{align*}

Thus $V=\pi(G(b)-G(a))$ . The rational terms give $275/6-32/3=211/6$ , while the logarithms give

42\ln\bigg(\frac{7/6}{7/3}\bigg)=-42\ln2.

Therefore

\boxed{V=\pi\bigg(\frac{211}{6}-42\ln2\bigg)}.

P4 2024 Oct Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程