题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 1

Figure 1 shows a sketch of part of the curve $C_1$ with equation

y=x^2+3,\qquad x>0

and part of the curve $C_2$ with equation

y=13-\frac9{x^2},\qquad x>0

The curves $C_1$ and $C_2$ intersect at the points $P$ and $Q$ as shown in Figure 1.

(a) Use algebra to find the $x$ coordinate of $P$ and the $x$ coordinate of $Q$ .

(4)

The finite region $R$ , shown shaded in Figure 1, is bounded by $C_1$ and $C_2$

(b) Use algebraic integration to find the exact area of $R$ .

(4)

解答

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交点处两个 $y$ 相等。因为有 $\frac9{x^2}$ ，先乘以 $x^2$ ，把它化成关于 $x^2$ 的二次方程。

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At the intersections,

\begin{align*} x^2+3=&\,13-\frac9{x^2}. \end{align*}

Multiply by $x^2$ :

\begin{align*} x^4+3x^2=&\,13x^2-9\\ x^4-10x^2+9=&\,0. \end{align*}

Factorise as a quadratic in $x^2$ :

\begin{align*} (x^2-1)(x^2-9)=&\,0. \end{align*}

\begin{align*} x^2=1\quad\text{or}\quad x^2=9. \end{align*}

Since $x>0$ ,

\begin{align*} x=1\quad\text{or}\quad x=3. \end{align*}

Therefore, the $x$ coordinates of $P$ and $Q$ are $1$ and $3$ .

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在 $1<x<3$ 上，上方曲线是 $C_2$ ，下方曲线是 $C_1$ 。面积是上减下的积分。

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The area is

\begin{align*} \int_1^3\left(13-\frac9{x^2}-(x^2+3)\right)\,\mathrm{d}x. \end{align*}

Simplify:

\begin{align*} 13-\frac9{x^2}-(x^2+3) =&\,10-9x^{-2}-x^2. \end{align*}

\begin{align*} \text{Area} =&\,\int_1^3(10-9x^{-2}-x^2)\,\mathrm{d}x\\ =&\,\left[10x+9x^{-1}-\frac{x^3}{3}\right]_1^3\\ =&\,\left(30+3-9\right)-\left(10+9-\frac13\right)\\ =&\,24-\frac{56}{3}\\ =&\,\frac{16}{3}. \end{align*}

Therefore, the exact area of $R$ is

\begin{align*} \frac{16}{3}. \end{align*}