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IAL 2025 May Q5

A Level / Edexcel / P2

IAL 2025 May Paper · Question 5

题目

Problem

x−2−0.512.545.57y124.2431.50.5300.1880.0660.023\begin{array}{c|rrrrrrr} x&-2&-0.5&1&2.5&4&5.5&7\\ \hline y&12&4.243&1.5&0.530&0.188&0.066&0.023 \end{array}

The table above shows corresponding values of xx and yy for

y=3(12)xy=3\left(\frac12\right)^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\left(\frac12\right)^x\,dx

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\left(\frac12\right)^{x+2}\,dx

(ii)

∫−27(2−x+2x) dx\int_{-2}^{7}\left(2^{-x}+2x\right)\,dx
(3)

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*}

Using the trapezium rule,

estimate= 1.52{12+0.023 +2(4.243+1.5+0.530+0.188+0.066)}= 18.80775.\begin{align*} \text{estimate} =&\,\frac{1.5}{2}\{12+0.023\\ &\,\hspace{2pt}+2(4.243+1.5+0.530+0.188+0.066)\}\\ =&\,18.80775. \end{align*}

Therefore, to one decimal place,

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

(b)(i)

解法一

思路

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(12)x+2=(12)x(12)2\left(\frac12\right)^{x+2}=\left(\frac12\right)^x\left(\frac12\right)^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,\left(\frac12\right)^{x+2} =\frac14\left(\frac12\right)^x,

we have

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

Therefore,

∫−273(12)x+2 dx= 14∫−273(12)x dx≈14(18.8)= 4.7.\begin{align*} \int_{-2}^{7}3\left(\frac12\right)^{x+2}\,dx =&\,\frac14\int_{-2}^{7}3\left(\frac12\right)^x\,dx\\ &\approx\frac14(18.8)\\ =&\,4.7. \end{align*}

(b)(ii)

解法一

思路

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因为 3(12)x=3⋅2−x3\left(\frac12\right)^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\,dx 可以精确计算。

答题过程

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Since

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

we get

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

So

∫−27(2−x+2x) dx= 13∫−273(12)x dx +∫−272x dx≈13(18.8)+[x2]−27= 18.83+(49−4)= 51.266….\begin{align*} \int_{-2}^{7}\left(2^{-x}+2x\right)\,dx =&\,\frac13\int_{-2}^{7}3\left(\frac12\right)^x\,dx\\ &\,\hspace{2pt}+\int_{-2}^{7}2x\,dx\\ &\approx\frac13(18.8)+\left[x^2\right]_{-2}^{7}\\ =&\,\frac{18.8}{3}+(49-4)\\ =&\,51.266\ldots. \end{align*}

Therefore, the estimate is

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

to one decimal place.