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IAL 2023 Jan Q1

A Level / Edexcel / P2

IAL 2023 Jan Paper · Question 1

题目

Problem

Figure 1 shows a sketch of part of the curve with equation y=f(x)y = \mathrm{f}(x).

The table below shows some corresponding values of xx and yy for this curve. The values of yy are given to 3 decimal places.

xx−1-1−0.5-0.5000.50.511
yy2.2872.2874.4704.4706.7196.7197.2917.2912.8342.834

Using the trapezium rule with all the values of yy in the given table,

(a) obtain an estimate for

∫−11f(x) dx\int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x

giving your answer to 2 decimal places.

(3)

(b) Use your answer to part (a) to estimate

(i) ∫−11(f(x)−2) dx\int_{-1}^1 (\mathrm{f}(x) - 2) \, \mathrm{d}x

(ii) ∫24f(x−3) dx\int_{2}^{4} \mathrm{f}(x-3) \, \mathrm{d}x

(3)

解答

(a)

解法一

思路

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梯形法则（Trapezium Rule）的公式为：

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

其中，步长 hh 是相邻 xx 值的差：

h=−0.5−(−1)=0.5h = -0.5 - (-1) = 0.5

将表中的 yy 值依次代入公式计算即可。首项 y0=2.287y_0 = 2.287，末项 yn=2.834y_n = 2.834，中间项为 4.470,6.719,7.2914.470, 6.719, 7.291。

答题过程

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The step size hh is given by:

h=  0.5.\begin{align*} h =&\,\, 0.5. \end{align*}

Using the trapezium rule:

∫−11f(x) dx≈  h2[y0+y4+2(y1+y2+y3)]=  0.52[2.287+2.834+2(4.470+6.719+7.291)]=  0.25[5.121+2(18.480)]=  0.25[5.121+36.960]=  0.25[42.081]=  10.52025.\begin{align*} \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x \approx&\,\, \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right] \\[3mm] =&\,\, \frac{0.5}{2} \left[ 2.287 + 2.834 + 2(4.470 + 6.719 + 7.291) \right] \\[3mm] =&\,\, 0.25 \left[ 5.121 + 2(18.480) \right] \\[3mm] =&\,\, 0.25 \left[ 5.121 + 36.960 \right] \\[3mm] =&\,\, 0.25 \left[ 42.081 \right] \\[3mm] =&\,\, 10.52025. \end{align*}

Therefore,

∫−11f(x) dx≈  10.52(to 2 d.p.).\begin{align*} \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x \approx&\,\, 10.52 \quad \text{(to 2 d.p.)}. \end{align*}

(b)

解法一

思路

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(i) 利用定积分的線性性質拆分：

∫−11(f(x)−2) dx=∫−11f(x) dx−∫−112 dx\int_{-1}^1 (\mathrm{f}(x) - 2) \, \mathrm{d}x = \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x - \int_{-1}^1 2 \, \mathrm{d}x

其中，常数定积分部分為 2×(1−(−1))=42 \times (1 - (-1)) = 4。因此只需將 (a) 小題的結果減去 44。

(ii) 觀察被積函數 f(x−3)\mathrm{f}(x-3)，這是一個水平向右平移 33 單位後的函數。 積分區間也從 [−1,1][-1, 1] 平移到了 [2,4][2, 4]。由於平移不改變曲線下的面積，因此這個積分值與 (a) 小題完全相同。 也可以用換元法：設 u=x−3u = x-3，則 du=dx\mathrm{d}u = \mathrm{d}x，且當 x=2x=2 時 u=−1u=-1，當 x=4x=4 時 u=1u=1。故原式化為 ∫−11f(u) du\int_{-1}^1 \mathrm{f}(u) \, \mathrm{d}u。

答题过程

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(i)

Using the properties of integration:

∫−11(f(x)−2) dx=  ∫−11f(x) dx−∫−112 dx=  ∫−11f(x) dx−[2x]−11=  ∫−11f(x) dx−(2(1)−2(−1))=  ∫−11f(x) dx−4.\begin{align*} \int_{-1}^1 (\mathrm{f}(x) - 2) \, \mathrm{d}x =&\,\, \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x - \int_{-1}^1 2 \, \mathrm{d}x \\[3mm] =&\,\, \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x - [2x]_{-1}^1 \\[3mm] =&\,\, \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x - (2(1) - 2(-1)) \\[3mm] =&\,\, \int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x - 4. \end{align*}

Using the answer from part (a):

∫−11(f(x)−2) dx≈  10.52−4=  6.52(to 2 d.p.).\begin{align*} \int_{-1}^1 (\mathrm{f}(x) - 2) \, \mathrm{d}x \approx&\,\, 10.52 - 4 \\[3mm] =&\,\, 6.52 \quad \text{(to 2 d.p.)}. \end{align*}

(ii)

Let u=x−3u = x - 3, then du=dx\mathrm{d}u = \mathrm{d}x.

When x=2x = 2, u=2−3=−1u = 2 - 3 = -1.

When x=4x = 4, u=4−3=1u = 4 - 3 = 1.

So,

∫24f(x−3) dx=  ∫−11f(u) du.\begin{align*} \int_{2}^{4} \mathrm{f}(x-3) \, \mathrm{d}x =&\,\, \int_{-1}^1 \mathrm{f}(u) \, \mathrm{d}u. \end{align*}

Since ∫−11f(u) du\int_{-1}^1 \mathrm{f}(u) \, \mathrm{d}u is equivalent to ∫−11f(x) dx\int_{-1}^1 \mathrm{f}(x) \, \mathrm{d}x, we have:

∫24f(x−3) dx≈  10.52(to 2 d.p.).\begin{align*} \int_{2}^{4} \mathrm{f}(x-3) \, \mathrm{d}x \approx&\,\, 10.52 \quad \text{(to 2 d.p.)}. \end{align*}