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P2 2022 Oct Q6

A Level / Edexcel / P2

IAL 2022 Oct Paper · Question 6

题目

Problem

The curve C1C_1 has equation y=f(x)y = \mathrm{f}(x).

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

xx00.511.52
yy32.68332.42.14661.92

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

∫02f(x) dx\int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x

giving your answer to 3 decimal places.

(3)

The region RR, shown shaded in Figure 1, is bounded by

  • the curve C1C_1
  • the curve C2C_2 with equation y=2−14x2y = 2 - \dfrac{1}{4}x^2
  • the line with equation x=2x = 2
  • the yy-axis

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

The design consists of the shaded region RR 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)
题目中文翻译

曲线 C1C_1 的方程为 y=f(x)y=\mathrm{f}(x)。下表给出部分 x,yx,y 值，必要时 yy 值已舍入到四位小数。

xx00.511.52
yy32.68332.42.14661.92

(a) 用梯形法则及表中全部 yy 值估算

∫02f(x) dx,\int_0^2\mathrm{f}(x)\,\mathrm{d}x,

答案保留三位小数。（3 分）

图 1 的阴影区域 RR 由曲线 C1C_1、方程为 y=2−14x2y=2-\frac14x^2 的曲线 C2C_2、直线 x=2x=2 和 yy 轴围成。

区域 RR 是图 2 所示标志设计的一部分。该设计由宽为 2、高为 3 的矩形及其内部的阴影区域 RR 组成。

使用微积分及 (a) 的答案，

(b) 估算该标志中阴影部分所占的百分比。（4 分）

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：

∫aby dx≈h2[y0+yn+2(y1+y2+⋯+yn−1)]\int_{a}^{b} y\,\mathrm{d}x \approx \frac{h}{2} \Big[ y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1}) \Big]

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

  • 步长 h=2−04=0.5h = \frac{2 - 0}{4} = 0.5（或者直接通过 xx 表格中的间距 0.5−0=0.50.5 - 0 = 0.5 观察得出）。
  • 表格中的 yy 值对应为：
    • y0=3y_0 = 3
    • y1=2.6833y_1 = 2.6833
    • y2=2.4y_2 = 2.4
    • y3=2.1466y_3 = 2.1466
    • y4=1.92y_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=0.5h = 0.5

Applying the trapezium rule formula:

∫02f(x) dx≈h2[y0+y4+2(y1+y2+y3)]\int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x \approx \frac{h}{2} \Big[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \Big]

Substituting the yy values from the table:

∫02f(x) dx≈ 0.52[3+1.92 +2(2.6833+2.4+2.1466)]= 0.25[4.92+2(7.2299)]= 0.25[4.92+14.4598]= 0.25×19.3798= 4.84495≈ 4.845(to 3 decimal places)\begin{align*} \int_0^2\mathrm{f}(x)\,\mathrm{d}x \approx&\,\frac{0.5}{2}\Big[3+1.92\\ &\,\hspace{2pt}+2(2.6833+2.4+2.1466)\Big]\\ =&\, 0.25 \Big[ 4.92 + 2(7.2299) \Big] \\ =&\, 0.25 \Big[ 4.92 + 14.4598 \Big] \\ =&\, 0.25 \times 19.3798 \\ =&\, 4.84495 \\ \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 RR 夹在曲线 C1C_1（上方边界）与曲线 C2C_2（下方边界）之间，区间为 x∈[0,2]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：

Area(R)=∫02f(x) dx−∫02(2−14x2) dx\text{Area}(R) = \int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x - \int_{0}^{2} \Big(2 - \frac{1}{4}x^2\Big)\,\mathrm{d}x

计算 C2C_2 下方的面积：

∫02(2−14x2) dx= [2x−x312]02= 4−23=103.\begin{align*} \int_0^2\Big(2-\frac14x^2\Big)\,\mathrm{d}x =&\,\Big[2x-\frac{x^3}{12}\Big]_0^2\\ =&\,4-\frac23=\frac{10}{3}. \end{align*}

计算阴影面积 Area(R)\text{Area}(R)： 使用 (a) 部分估算的 4.8454.845 代替第一项积分：

Area(R)≈4.845−103≈1.5117\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：

Total Area= 2×3= 6\begin{align*} \text{Total Area}=&\,2 \times 3\\ =&\,6 \end{align*}

阴影占比为：

Percentage= Area(R)6×100%≈ 1.51176×100%≈ 25.2%.\begin{align*} \text{Percentage} =&\,\frac{\text{Area}(R)}6\times100\%\\ \approx&\,\frac{1.5117}{6}\times100\%\\ \approx&\,25.2\%. \end{align*}

答题过程

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First, we find the exact area under the curve C2:y=2−14x2C_2: y = 2 - \dfrac{1}{4}x^2 from x=0x = 0 to x=2x = 2 using integration:

Area under C2= ∫02(2−14x2) dx= [2x−x312]02= (2(2)−2312)−0= 4−812= 4−23= 103\begin{align*} \text{Area under } C_2 =&\, \int_{0}^{2} \Big(2 - \frac{1}{4}x^2\Big)\,\mathrm{d}x \\ =&\, \Big[ 2x - \frac{x^3}{12} \Big]_0^2 \\ =&\, \Big( 2(2) - \frac{2^3}{12} \Big) - 0 \\ =&\, 4 - \frac{8}{12} \\ =&\, 4 - \frac{2}{3} \\ =&\, \frac{10}{3} \end{align*}

The shaded region RR is the area between C1C_1 and C2C_2. Using the approximation from part (a):

Area(R)= ∫02f(x) dx−Area under C2≈ 4.845−103= 4.845−3.3333...= 1.51166...\begin{align*} \text{Area}(R) =&\, \int_{0}^{2} \mathrm{f}(x)\,\mathrm{d}x - \text{Area under } C_2 \\ \approx&\, 4.845 - \frac{10}{3} \\ =&\, 4.845 - 3.3333... \\ =&\, 1.51166... \end{align*}

The logo is a rectangle of width 22 and height 33, so its total area is:

Total Area= 2×3= 6\begin{align*} \text{Total Area}=&\,2 \times 3\\ =&\,6 \end{align*}

The percentage of the logo which is shaded is:

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