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IAL 2024 Oct Q10

A Level / Edexcel / P4

IAL 2024 Oct Paper · Question 10

题目

Problem

Figure 5 shows a sketch of the curve with parametric equations

x=3t2y=sin⁡tsin⁡2t0≤t≤π2x=3t^2\qquad y=\sin t\sin 2t\qquad 0\le t\le \frac{\pi}{2}

The region RR, shown shaded in Figure 5, is bounded by the curve and the xx-axis.

(a) Show that the area of RR is

k∫0π2tsin⁡2tcos⁡t dtk\int_0^{\frac{\pi}{2}} t\sin^2 t\cos t\,dt

where kk is a constant to be found.

(3)

(b) Hence, using algebraic integration, find the exact area of RR, giving your answer in the form

pπ+qp\pi+q

where pp and qq are constants.

(5)
题目中文翻译

图 5 给出了曲线的草图，其参数方程为

x=3t2y=sin⁡tsin⁡2t0≤t≤π2x=3t^2\qquad y=\sin t\sin 2t\qquad 0\le t\le \frac{\pi}{2}

阴影区域 RR 由该曲线与 xx 轴围成。

(a) 证明区域 RR 的面积为

k∫0π2tsin⁡2tcos⁡t dtk\int_0^{\frac{\pi}{2}} t\sin^2 t\cos t\,dt

其中 kk 是待求常数。

(b) 进而使用代数积分，求区域 RR 的精确面积，并将答案写成

pπ+qp\pi+q

的形式，其中 p,qp,q 为常数。

解答

(a)

For a parametric curve, the area under the curve is

∫y dxdt dt\int y\,\frac{dx}{dt}\,\mathrm{d}t

Here

x=3t2x=3t^2

so

dxdt=6t\frac{dx}{dt}=6t

Also,

y=sin⁡tsin⁡2ty=\sin t\sin 2t

Using

sin⁡2t=2sin⁡tcos⁡t\sin2t=2\sin t\cos t

we get

y=2sin⁡2tcos⁡ty=2\sin^2t\cos t

Therefore the area of RR is

∫0π2y dxdt dt=∫0π2(2sin⁡2tcos⁡t)(6t) dt=12∫0π2tsin⁡2tcos⁡t dt\begin{aligned} \int_0^{\frac{\pi}{2}} y\,\frac{dx}{dt}\,\mathrm{d}t &=\int_0^{\frac{\pi}{2}} \left(2\sin^2t\cos t\right)(6t)\,\mathrm{d}t \\ &=12\int_0^{\frac{\pi}{2}}t\sin^2t\cos t\,\mathrm{d}t \end{aligned}

Hence

k=12\boxed{k=12}

(b)

From part (a), the area is

12∫0π2tsin⁡2tcos⁡t dt12\int_0^{\frac{\pi}{2}}t\sin^2t\cos t\,\mathrm{d}t

Use integration by parts with

u=t,dvdt=sin⁡2tcos⁡tu=t,\qquad \frac{dv}{dt}=\sin^2t\cos t

Then

dudt=1,v=13sin⁡3t\frac{du}{dt}=1,\qquad v=\frac13\sin^3t

So

∫tsin⁡2tcos⁡t dt=13tsin⁡3t−13∫sin⁡3t dt\begin{aligned} \int t\sin^2t\cos t\,\mathrm{d}t &=\frac13t\sin^3t-\frac13\int\sin^3t\,\mathrm{d}t \end{aligned}

Therefore the area is

Area=[4tsin⁡3t]0π2−4∫0π2sin⁡3t dt\text{Area} =\left[4t\sin^3t\right]_0^{\frac{\pi}{2}} -4\int_0^{\frac{\pi}{2}}\sin^3t\,\mathrm{d}t

Now

sin⁡3t=sin⁡t(1−cos⁡2t)\sin^3t=\sin t(1-\cos^2t)

So

∫sin⁡3t dt=∫sin⁡t(1−cos⁡2t) dt=∫sin⁡t dt−∫sin⁡tcos⁡2t dt=−cos⁡t+13cos⁡3t\begin{aligned} \int\sin^3t\,\mathrm{d}t &=\int\sin t(1-\cos^2t)\,\mathrm{d}t \\ &=\int \sin t\,\mathrm{d}t-\int\sin t\cos^2t\,\mathrm{d}t \\ &=-\cos t+\frac13\cos^3t \end{aligned}

Thus

Area=[4tsin⁡3t+4cos⁡t−43cos⁡3t]0π2\begin{aligned} \text{Area} &=\left[4t\sin^3t+4\cos t-\frac43\cos^3t\right]_0^{\frac{\pi}{2}} \end{aligned}

Apply the limits:

Area=(4⋅π2⋅1+4⋅0−43⋅0)−(0+4⋅1−43⋅1)=2π−(4−43)=2π−83\begin{aligned} \text{Area} &=\left(4\cdot\frac{\pi}{2}\cdot1+4\cdot0-\frac43\cdot0\right) -\left(0+4\cdot1-\frac43\cdot1\right) \\ &=2\pi-\left(4-\frac43\right) \\ &=2\pi-\frac83 \end{aligned}

Therefore the exact area is

2π−83\boxed{2\pi-\frac83}