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IAL 2024 May R Q2

A Level / Edexcel / P2

IAL 2024 May (R) Paper · Question 2

题目

Problem

The table shows corresponding values of xx and yy for a continuous curve with equation y=f(x)y=\text{f}(x) between x=−4x=-4 and x=5x=5, where aa is a constant.

x−4−2.5−10.523.55y4.162.91a1.731.371.432.28\begin{array}{c|ccccccc} x&-4&-2.5&-1&0.5&2&3.5&5\\ \hline y&4.16&2.91&a&1.73&1.37&1.43&2.28 \end{array}

The trapezium rule is used with all the yy values in the table to find an approximation for

∫−45f(x) dx\begin{align*} \int_{-4}^{5}\text{f}(x)\,\mathrm{d}x \end{align*}

Given that the value of this approximation is 19.319.3

(a) find the value of the constant aa to 3 significant figures.

(3)

(b) Use the given answer of 19.319.3 to find an approximate value for

∫−45(2f(x)−3)dx\begin{align*} \int_{-4}^{5}\left(2\text{f}(x)-3\right)\mathrm{d}x \end{align*}
(2)

解答

(a)

解法一

思路

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梯形法则的标准公式是 ∫aby dx≈h2[y0+2(y1+y2+⋯+yn−1)+yn]\int_a^b y\,\mathrm{d}x \approx \frac{h}{2}[y_0 + 2(y_1+y_2+\cdots+y_{n-1})+y_n]。这里 hh 是相邻 xx 值的间隔，代入已知数值后解出 aa。

答题过程

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The strip width is

h=−2.5−(−4)=1.5.\begin{align*} h=-2.5-(-4)=1.5. \end{align*}

The seven yy values give six strips. Applying the trapezium rule,

∫−45f(x) dx≈1.52[4.16+2(2.91+a+1.73+1.37+1.43)+2.28].\begin{align*} \int_{-4}^{5}\text{f}(x)\,\mathrm{d}x\approx\frac{1.5}{2}\Big[4.16+2\big(2.91+a+1.73+1.37+1.43\big)+2.28\Big]. \end{align*}

Simplifying the inner sum,

2.91+1.73+1.37+1.43=7.44,\begin{align*} 2.91+1.73+1.37+1.43=7.44, \end{align*}

so

1.52[4.16+2(7.44+a)+2.28]=19.3.\begin{align*} \frac{1.5}{2}\Big[4.16+2(7.44+a)+2.28\Big]=19.3. \end{align*} 0.75[4.16+14.88+2a+2.28]=19.3\begin{align*} 0.75\Big[4.16+14.88+2a+2.28\Big]=19.3 \end{align*} 0.75(21.32+2a)=19.3\begin{align*} 0.75\big(21.32+2a\big)=19.3 \end{align*} 15.99+1.5a=19.3\begin{align*} 15.99+1.5a=19.3 \end{align*} 1.5a=3.31\begin{align*} 1.5a=3.31 \end{align*} a=2.21(3 s.f.).\begin{align*} a=2.21\quad(3\text{ s.f.}). \end{align*}

(b)

解法一

思路

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利用积分的线性性质，把 ∫(2f(x)−3) dx\int(2\text{f}(x)-3)\,\mathrm{d}x 拆成 2∫f(x) dx−3∫dx2\int\text{f}(x)\,\mathrm{d}x-3\int\mathrm{d}x，再代入 (a) 的结果。

答题过程

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Using the linearity of integration,

∫−45(2f(x)−3)dx= 2∫−45f(x) dx−3∫−45dx.\begin{align*} \int_{-4}^{5}\big(2\text{f}(x)-3\big)\mathrm{d}x =&\, 2\int_{-4}^{5}\text{f}(x)\,\mathrm{d}x-3\int_{-4}^{5}\mathrm{d}x. \end{align*}

Substituting the given approximation 19.319.3 and evaluating the second integral,

= 2×19.3−3×(5−(−4))= 38.6−3×9= 38.6−27=11.6.\begin{align*} =&\, 2\times19.3-3\times\big(5-(-4)\big) \\[2mm] =&\, 38.6-3\times9 \\[2mm] =&\, 38.6-27=11.6. \end{align*}