| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Standard +0.3 This is a standard parametric area question requiring routine application of the formula A = ∫y(dx/dt)dt, followed by straightforward integration using substitution (u = sin t), conversion to Cartesian form using double angle formula, and reading range from given domain. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.03g Parametric equations: of curves and conversion to cartesian1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(R = \int y\frac{dx}{dt}dt = \int 4\sin t \times -4\sin 2t \, dt\) | M1 A1 | Attempts to multiply \(y\) by \(\frac{dx}{dt}\) to achieve integrand of form \(\pm k\sin t\sin 2t\) |
| \(= -\int 32\sin^2 t\cos t \, dt\) | dM1 | Substitutes \(\sin 2t = 2\sin t\cos t\); dependent on previous M |
| \(x=0 \Rightarrow t=\frac{\pi}{4}\) and \(y=0 \Rightarrow t=0\); Area \(= -\int_{\pi/4}^{0} 32\sin^2 t\cos t \, dt = \int_0^{\pi/4} 32\sin^2 t\cos t \, dt\) | A1* | Must see evidence of \(dt\); correct limit application showing \(-\int_{\pi/4}^{0}\to\int_0^{\pi/4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= \left[\frac{32}{3}\sin^3 t\right]_0^{\pi/4} = \frac{32}{3}\times\frac{2\sqrt{2}}{8} = \frac{8\sqrt{2}}{3}\) | M1 A1 | M1: integrates to form \(\left[A\sin^3 t\right]\) with attempt to apply limits; A1: correct answer \(\frac{8\sqrt{2}}{3}\) following correct algebraic integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 2t = 1-2\sin^2 t \Rightarrow \frac{x}{2} = 1-2\left(\frac{y}{4}\right)^2\) | M1 A1 | M1: attempts double angle formula \(\cos 2t = \pm 1\pm 2\sin^2 t\); A1: any correct unsimplified equation |
| \(y = \sqrt{8-4x}\) | A1 | \(y=\sqrt{8-4x}\) or \(y=\sqrt{-4x+8}\) ONLY |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Range is \(0\leqslant f\leqslant 4\) | B1 | Allow \(0\leqslant y\leqslant 4\), \(0\leqslant f(x)\leqslant 4\), \([0,4]\); unacceptable: \(0\leqslant x\leqslant 4\), \([0,4)\) |
# Question 6:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $R = \int y\frac{dx}{dt}dt = \int 4\sin t \times -4\sin 2t \, dt$ | M1 A1 | Attempts to multiply $y$ by $\frac{dx}{dt}$ to achieve integrand of form $\pm k\sin t\sin 2t$ |
| $= -\int 32\sin^2 t\cos t \, dt$ | dM1 | Substitutes $\sin 2t = 2\sin t\cos t$; dependent on previous M |
| $x=0 \Rightarrow t=\frac{\pi}{4}$ and $y=0 \Rightarrow t=0$; Area $= -\int_{\pi/4}^{0} 32\sin^2 t\cos t \, dt = \int_0^{\pi/4} 32\sin^2 t\cos t \, dt$ | A1* | Must see evidence of $dt$; correct limit application showing $-\int_{\pi/4}^{0}\to\int_0^{\pi/4}$ |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= \left[\frac{32}{3}\sin^3 t\right]_0^{\pi/4} = \frac{32}{3}\times\frac{2\sqrt{2}}{8} = \frac{8\sqrt{2}}{3}$ | M1 A1 | M1: integrates to form $\left[A\sin^3 t\right]$ with attempt to apply limits; A1: correct answer $\frac{8\sqrt{2}}{3}$ following correct algebraic integration |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 2t = 1-2\sin^2 t \Rightarrow \frac{x}{2} = 1-2\left(\frac{y}{4}\right)^2$ | M1 A1 | M1: attempts double angle formula $\cos 2t = \pm 1\pm 2\sin^2 t$; A1: any correct unsimplified equation |
| $y = \sqrt{8-4x}$ | A1 | $y=\sqrt{8-4x}$ or $y=\sqrt{-4x+8}$ ONLY |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range is $0\leqslant f\leqslant 4$ | B1 | Allow $0\leqslant y\leqslant 4$, $0\leqslant f(x)\leqslant 4$, $[0,4]$; unacceptable: $0\leqslant x\leqslant 4$, $[0,4)$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{960fe82f-c180-422c-b409-a5cdc5fae924-18_563_844_255_552}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$ with parametric equations
$$x = 2 \cos 2 t \quad y = 4 \sin t \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
The region $R$, shown shaded in Figure 3, is bounded by the curve, the $x$-axis and the $y$-axis.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show, making your working clear, that the area of $R = \int _ { 0 } ^ { \frac { \pi } { 4 } } 32 \sin ^ { 2 } t \cos t d t$
\item Hence find, by algebraic integration, the exact value of the area of $R$.
\end{enumerate}\item Show that all points on $C$ satisfy $y = \sqrt { a x + b }$, where $a$ and $b$ are constants to be found.
The curve $C$ has equation $y = \mathrm { f } ( x )$ where f is the function
$$f ( x ) = \sqrt { a x + b } \quad - 2 \leqslant x \leqslant 2$$
and $a$ and $b$ are the constants found in part (b).
\item State the range of f.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q6 [10]}}