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P2 2025 May Q5

A Level / Edexcel / P2

IAL 2025 May Paper · Question 5

题目

Problem

xy−212−0.54.24311.52.50.53040.1885.50.06670.023\begin{array}{c|r} x&y\\ \hline -2&12\\ -0.5&4.243\\ 1&1.5\\ 2.5&0.530\\ 4&0.188\\ 5.5&0.066\\ 7&0.023 \end{array}

The table above shows corresponding values of xx and yy for

y=3(12)xy=3\bigg(\frac12\bigg)^x

The values of yy are given to 33 decimal places as appropriate.

(a) Using the trapezium rule with all the values of yy in the given table, obtain an estimate for

∫−273(12)x dx\int_{-2}^{7}3\bigg(\frac12\bigg)^x\,\mathrm{d}x

giving the answer to one decimal place.

(3)

Using the answer to part (a) and making your method clear, estimate

(b) (i)

∫−273(12)x+2 dx\int_{-2}^{7}3\bigg(\frac12\bigg)^{x+2}\,\mathrm{d}x

(ii)

∫−27(2−x+2x) dx\int_{-2}^{7}\bigg(2^{-x}+2x\bigg)\,\mathrm{d}x
(3)
题目中文翻译

下表给出了 y=3(12)xy=3\big(\frac12\big)^x 的对应 xx、yy 值。yy 值在适当情况下保留至小数点后三位。

xy−212−0.54.24311.52.50.53040.1885.50.06670.023\begin{array}{c|r} x&y\\ \hline -2&12\\ -0.5&4.243\\ 1&1.5\\ 2.5&0.530\\ 4&0.188\\ 5.5&0.066\\ 7&0.023 \end{array}

(a) 使用表中所有 yy 值，用梯形法则估计

∫−273(12)x dx,\int_{-2}^{7}3\bigg(\frac12\bigg)^x\,\mathrm{d}x,

答案保留至小数点后一位。（3分）

利用第 (a) 问的答案，并清楚说明方法，估计

(b) (i)

∫−273(12)x+2 dx,\int_{-2}^{7}3\bigg(\frac12\bigg)^{x+2}\,\mathrm{d}x,

(ii)

∫−27(2−x+2x) dx.\int_{-2}^{7}(2^{-x}+2x)\,\mathrm{d}x.

（第 (b) 问共3分）

解答

(a)

解法一

思路

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表格中的 xx 每次增加 1.51.5，所以梯形宽度 h=1.5h=1.5。套 trapezium rule 时，首尾项各算一次，中间项乘 22。

答题过程

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The interval width is

h=1.5.\begin{align*} h=1.5. \end{align*}

The sum of the interior ordinates is

T= 4.243+1.5+0.530 +0.188+0.066= 6.527.\begin{align*} T=&\,4.243+1.5+0.530\\ &\,\hspace{2pt}+0.188+0.066\\ =&\,6.527. \end{align*}

Using the trapezium rule,

estimate= 1.52[12+0.023+2T]= 0.75(12.023+13.054)= 18.80775.\begin{align*} \text{estimate} =&\,\frac{1.5}{2}\big[12+0.023+2T\big]\\ =&\,0.75(12.023+13.054)\\ =&\,18.80775. \end{align*}

Therefore, to one decimal place,

∫−273(12)x dx≈18.8.\begin{align*} \int_{-2}^{7}3\bigg(\frac12\bigg)^x\,\mathrm{d}x\approx18.8. \end{align*}

(b)(i)

解法一

思路

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(12)x+2=(12)x(12)2\bigg(\frac12\bigg)^{x+2}=\bigg(\frac12\bigg)^x\bigg(\frac12\bigg)^2，所以整个 integrand 是原来 integrand 的 14\frac14。

答题过程

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Since

(12)x+2=14(12)x,\bigg(\frac12\bigg)^{x+2} =\frac14\bigg(\frac12\bigg)^x,

we have

3(12)x+2=14⋅3(12)x.3\bigg(\frac12\bigg)^{x+2} =\frac14\cdot3\bigg(\frac12\bigg)^x.

Let

I=∫−273(12)x dx.I=\int_{-2}^{7}3\bigg(\frac12\bigg)^x\,\mathrm{d}x.

From part (a), I≈18.8I\approx18.8. The required integral is exactly 14I\frac14I.

Hence the estimate is

14(18.8)=4.7.\frac14(18.8)=4.7.

(b)(ii)

解法一

思路

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因为 3(12)x=3⋅2−x3\bigg(\frac12\bigg)^x=3\cdot2^{-x}，所以 2−x2^{-x} 的积分估计值是第 (a) 题结果的 13\frac13。另外 ∫2x dx\int 2x\,\mathrm{d}x 可以精确计算。

答题过程

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Since

3(12)x=3⋅2−x,\begin{align*} 3\bigg(\frac12\bigg)^x=3\cdot2^{-x}, \end{align*}

we get

2−x=13⋅3(12)x.\begin{align*} 2^{-x}=\frac13\cdot3\bigg(\frac12\bigg)^x. \end{align*}

Using II as defined above,

 ∫−27(2−x+2x) dx= 13I+∫−272x dx= 13I+[x2]−27= 13I+45.\begin{align*} &\,\int_{-2}^{7}(2^{-x}+2x)\,\mathrm{d}x\\ =&\,\frac13I+\int_{-2}^{7}2x\,\mathrm{d}x\\ =&\,\frac13I+\big[x^2\big]_{-2}^{7}\\ =&\,\frac13I+45. \end{align*}

Since I≈18.8I\approx18.8, the estimate is

18.83+45= 51.266….\begin{align*} \frac{18.8}{3}+45=&\,51.266\ldots. \end{align*}

Therefore, the estimate is

51.3\begin{align*} 51.3 \end{align*}

to one decimal place.