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P2 2023 May Q1

A Level / Edexcel / P2

IAL 2023 May Paper · Question 1

题目

Problem

The continuous curve CC has equation y=f(x)y = f(x).

A table of values of xx and yy for y=f(x)y = f(x) is shown below.

xx4.04.04.24.24.44.44.64.64.84.85.05.0
yy9.29.28.45568.45563.85123.85125.03425.03427.82977.82978.68.6

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

∫45f(x) dx\int_4^5 f(x)\,\mathrm{d}x

giving your answer to 3 decimal places.

(3)
题目中文翻译

连续曲线 CC 的方程为 y=f(x)y = f(x)。

下表给出了 y=f(x)y = f(x) 的 xx 和 yy 值。

xx4.04.04.24.24.44.44.64.64.84.85.05.0
yy9.29.28.45568.45563.85123.85125.03425.03427.82977.82978.68.6

用梯形法则,使用表中所有 yy 值,求 ∫45f(x) dx\displaystyle\int_4^5 f(x)\,\mathrm{d}x 的近似值,答案保留3位小数。

(3分)

解答

解法一

思路

展开

表中有六个纵坐标,对应五个区间;步长为 h=(5−4)/5=0.2h=(5-4)/5=0.2。梯形法则将首末纵坐标各计一次,四个内点各计两次。计算结果最后才保留三位小数。

答题过程

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There are six ordinates and five intervals, so

h=5−45=0.2.h=\frac{5-4}{5}=0.2.

Using all six ordinates,

∫45f(x) dx≈ h2[y0+y5 +2(y1+y2+y3+y4)]= 0.1[9.2+8.6 +2(8.4556+3.8512 +5.0342+7.8297)]= 0.1[17.8+2(25.1707)]= 0.1(68.1414)= 6.81414.\begin{align*} \int_4^5 f(x)\,\mathrm{d}x \approx&\,\frac h2\big[y_0+y_5\\ &\,\hspace{2pt}+2(y_1+y_2+y_3+y_4)\big]\\ =&\,0.1\big[9.2+8.6\\ &\,\hspace{2pt}+2(8.4556+3.8512\\ &\,\hspace{4pt}+5.0342+7.8297)\big]\\ =&\,0.1\big[17.8+2(25.1707)\big]\\ =&\,0.1(68.1414)\\ =&\,6.81414. \end{align*}

Therefore the approximation, to 3 decimal places, is

6.814.\boxed{6.814}.

也可以分别求五个小梯形的面积后相加:

I≈ 0.1[(9.2+8.4556) +(8.4556+3.8512) +(3.8512+5.0342) +(5.0342+7.8297) +(7.8297+8.6)]= 6.81414.\begin{align*} I\approx&\,0.1\big[(9.2+8.4556)\\ &\,\hspace{2pt}+(8.4556+3.8512)\\ &\,\hspace{4pt}+(3.8512+5.0342)\\ &\,\hspace{6pt}+(5.0342+7.8297)\\ &\,\hspace{8pt}+(7.8297+8.6)\big]\\ =&\,6.81414. \end{align*}