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IAL 2025 May Q3

A Level / Edexcel / P2

IAL 2025 May Paper · Question 3

题目

Problem

Figure 1

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 1 shows a sketch of the curve CC with equation

y=3x+16x2−8,x>0y=3x+\frac{16}{x^2}-8,\qquad x>0

The points P(1,11)P(1,11) and Q(4,5)Q(4,5) lie on CC and are shown in Figure 1. The region RR, shown shaded in Figure 1, is bounded by CC and line segment PQPQ.

Use algebraic integration to find the area of RR.

(5)

解答

解法一

思路

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线段 PQPQ 和 xx-axis 形成一个 trapezium，曲线在下面。区域面积可以用梯形面积减去曲线下面积。

答题过程

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The area of the trapezium with parallel sides 1111 and 55 and width 33 is

12(11+5)(4−1)=24.\begin{align*} \frac12(11+5)(4-1)=24. \end{align*}

Now integrate the curve:

∫(3x+16x2−8) dx= ∫(3x+16x−2−8) dx= 32x2−16x−1−8x= 32x2−16x−8x.\begin{align*} \int\left(3x+\frac{16}{x^2}-8\right)\,dx =&\,\int(3x+16x^{-2}-8)\,dx\\ =&\,\frac32x^2-16x^{-1}-8x\\ =&\,\frac32x^2-\frac{16}{x}-8x. \end{align*}

The area under the curve from x=1x=1 to x=4x=4 is

[32x2−16x−8x]14= (24−4−32) −(32−16−8)= −12−(−452)= 212.\begin{align*} \left[\frac32x^2-\frac{16}{x}-8x\right]_1^4 =&\,\left(24-4-32\right)\\ &\,\hspace{2pt}-\left(\frac32-16-8\right)\\ =&\,-12-\left(-\frac{45}{2}\right)\\ =&\,\frac{21}{2}. \end{align*}

Therefore,

area of R= 24−212= 272.\begin{align*} \text{area of }R =&\,24-\frac{21}{2}\\ =&\,\frac{27}{2}. \end{align*}

解法二

思路

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也可以先求直线 PQPQ 的方程，然后直接积分“直线减曲线”。这样区域含义最直接。

答题过程

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The gradient of PQPQ is

5−114−1=−2.\begin{align*} \frac{5-11}{4-1}=-2. \end{align*}

Using P(1,11)P(1,11), the equation of PQPQ is

y−11= −2(x−1)y= 13−2x.\begin{align*} y-11=&\,-2(x-1)\\ y=&\,13-2x. \end{align*}

The area of RR is

∫14[(13−2x)−(3x+16x2−8)] dx.\begin{align*} \int_1^4\left[(13-2x)-\left(3x+\frac{16}{x^2}-8\right)\right]\,dx. \end{align*}

Simplify the integrand:

(13−2x)−(3x+16x2−8)=21−5x−16x2.(13-2x)-\left(3x+\frac{16}{x^2}-8\right) =21-5x-\frac{16}{x^2}.

So

area of R= ∫14(21−5x−16x−2) dx= [21x−52x2+16x−1]14= (84−40+4)−(21−52+16)= 48−692= 272.\begin{align*} \text{area of }R =&\,\int_1^4\left(21-5x-16x^{-2}\right)\,dx\\ =&\,\left[21x-\frac52x^2+16x^{-1}\right]_1^4\\ =&\,(84-40+4)-\left(21-\frac52+16\right)\\ =&\,48-\frac{69}{2}\\ =&\,\frac{27}{2}. \end{align*}