题目

Problem

The curve $C_1$ has equation $y = \mathrm{f}(x)$ .

A table of values of $x$ and $y$ for $y = \mathrm{f}(x)$ is shown below, with the $y$ values rounded to 4 decimal places where appropriate.

$x$	0	0.5	1	1.5	2
$y$	3	2.6833	2.4	2.1466	1.92

(a) Use the trapezium rule with all the values of $y$ in the table to find an approximation for

$\int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x$

giving your answer to 3 decimal places.

(3)

Figure 1

Figure 2

The region $R$ , shown shaded in Figure 1, is bounded by

the curve $C_1$
the curve $C_2$ with equation $y = 2 - \dfrac{1}{4}x^2$
the line with equation $x = 2$
the $y$ -axis

The region $R$ forms part of the design for a logo shown in Figure 2.

The design consists of the shaded region $R$ inside a rectangle of width 2 and height 3

Using calculus and the answer to part (a),

(b) calculate an estimate for the percentage of the logo which is shaded.

(4)

解答

(a)

解法一

思路

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梯形法则（Trapezium Rule）用于估算曲线下方的面积。公式为：

\int_{a}^{b} y\,\mathrm{d}x \approx \frac{h}{2} \left[ y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1}) \right]

这里，积分区间为 $[0, 2]$ ，区间被分为 4 个等宽的条带（Strip）：

步长 $h = \frac{2 - 0}{4} = 0.5$ （或者直接通过 $x$ 表格中的间距 $0.5 - 0 = 0.5$ 观察得出）。
表格中的 $y$ $y$ 值对应为：
- $y_0 = 3$
- $y_1 = 2.6833$
- $y_2 = 2.4$
- $y_3 = 2.1466$
- $y_4 = 1.92$

代入公式计算并保留三位小数即可。

答题过程

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The width of each interval (strip) is:

h = 0.5

Applying the trapezium rule formula:

\int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x \approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right]

Substituting the $y$ values from the table:

\begin{align*} \int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x \approx&\,\, \frac{0.5}{2} \left[ 3 + 1.92 + 2(2.6833 + 2.4 + 2.1466) \right] \\[2mm] =&\,\, 0.25 \left[ 4.92 + 2(7.2299) \right] \\[2mm] =&\,\, 0.25 \left[ 4.92 + 14.4598 \right] \\[2mm] =&\,\, 0.25 \times 19.3798 \\[2mm] =&\,\, 4.84495 \\[2mm] \approx&\,\, 4.845 \quad (\text{to 3 decimal places}) \end{align*}

(b)

解法一

思路

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根据图 1 和题目描述，阴影区域 $R$ 夹在曲线 $C_1$ （上方边界）与曲线 $C_2$ （下方边界）之间，区间为 $x \in [0, 2]$ 。因此，阴影区域的面积可以用上方曲线下方的面积减去下方曲线下方的面积来计算：

\text{Area}(R) = \int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x - \int_{0}^{2} \left(2 - \frac{1}{4}x^2\right)\mathrm{d}x

计算 $C_2$ 下方的面积： $\int_{0}^{2} \left(2 - \frac{1}{4}x^2\right)\mathrm{d}x = \left[ 2x - \frac{1}{12}x^3 \right]_0^2 = \left(4 - \frac{8}{12}\right) - 0 = 4 - \frac{2}{3} = \frac{10}{3}$
计算阴影面积 $\text{Area}(R)$ ：使用 (a) 部分估算的 $4.845$ 代替第一项积分： $\text{Area}(R) \approx 4.845 - \frac{10}{3} \approx 1.5117$
计算阴影占比百分比：整个 Logo 的背景是一个宽为 2、高为 3 的矩形，其总面积为： $\text{Total Area} = 2 \times 3 = 6$ 阴影占比为： $\text{Percentage} = \frac{\text{Area}(R)}{\text{Total Area}} \times 100\% \approx \frac{1.5117}{6} \times 100\% \approx 25.2\%$

答题过程

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First, we find the exact area under the curve $C_2: y = 2 - \dfrac{1}{4}x^2$ from $x = 0$ to $x = 2$ using integration:

\begin{align*} \text{Area under } C_2 =&\,\, \int_{0}^{2} \left(2 - \frac{1}{4}x^2\right)\mathrm{d}x \\[2mm] =&\,\, \left[ 2x - \frac{x^3}{12} \right]_0^2 \\[2mm] =&\,\, \left( 2(2) - \frac{2^3}{12} \right) - 0 \\[2mm] =&\,\, 4 - \frac{8}{12} \\[2mm] =&\,\, 4 - \frac{2}{3} \\[2mm] =&\,\, \frac{10}{3} \end{align*}

The shaded region $R$ is the area between $C_1$ and $C_2$ . Using the approximation from part (a):

\begin{align*} \text{Area}(R) =&\,\, \int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x - \text{Area under } C_2 \\[2mm] \approx&\,\, 4.845 - \frac{10}{3} \\[2mm] =&\,\, 4.845 - 3.3333... \\[2mm] =&\,\, 1.51166... \end{align*}

The logo is a rectangle of width $2$ and height $3$ , so its total area is:

\text{Total Area} = 2 \times 3 = 6

The percentage of the logo which is shaded is:

\begin{align*} \text{Percentage} =&\,\, \frac{\text{Area}(R)}{\text{Total Area}} \times 100\% \\[2mm] \approx&\,\, \frac{1.51166...}{6} \times 100\% \\[2mm] =&\,\, 0.25194... \times 100\% \\[2mm] \approx&\,\, 25.2\% \quad (\text{to 3 significant figures}) \end{align>

IAL 2022 Oct Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程