| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Challenging +1.2 This is a standard parametric area question requiring finding the parameter value where x=4, then applying the formula A = ∫y(dx/dt)dt with trigonometric manipulation. Part (a) involves routine differentiation and double-angle identities to reach the given form. Part (b) requires integrating standard trigonometric functions using cos²t = (1+cos2t)/2. While multi-step, all techniques are standard A-level Further Maths content with no novel insight required, making it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y \cdot \frac{dx}{dt} = (2\sin 2t + 3\sin t) \times 16\sin t \cos t\) and uses \(\sin 2t = 2\sin t \cos t\) | M1 | Attempt at key step; \(\frac{dx}{dt}\) must be of form \(k\sin t \cos t\) |
| Correct expanded integrand, usually one of: \(\int 48\sin^2 t \cos t + 16\sin^2 2t \, dt\), or \(\int 48\sin^2 t \cos t + 64\sin^2 t \cos^2 t \, dt\), or \(\int 24\sin 2t \sin t + 16\sin^2 2t \, dt\) | A1 | Correct expanded integrand in \(t\); ignore absence of \(\int\) or \(dt\) or limits |
| Attempts to use \(\cos 4t = 1 - 2\sin^2 2t = (1 - 8\sin^2 t \cos^2 t)\) | M1 | |
| \(R = \int_0^a 8 - 8\cos 4t + 48\sin^2 t \cos t \, dt\) | A1* | Proceeds to given answer with correct working; \(\int\) sign and \(dt\) must be seen |
| Deduces \(a = \dfrac{\pi}{4}\) | B1 | May be awarded from upper limit or from part (b) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int 8 - 8\cos 4t + 48\sin^2 t \cos t \, dt = 8t - 2\sin 4t + 16\sin^3 t\) | M1 | Correct approach to integrating trig terms; \(\int \ldots = \ldots \pm P\sin 4t \pm Q\sin^3 t\) |
| \(= 8t - 2\sin 4t + 16\sin^3 t\) | A1 | Correct integration \((+c)\) |
| \(\left[8t - 2\sin 4t + 16\sin^3 t\right]_0^{\frac{\pi}{4}} = 2\pi + 4\sqrt{2}\) | M1 | Uses limits \(a\) and \(0\) where \(a = \frac{\pi}{6}, \frac{\pi}{4}\) or \(\frac{\pi}{3}\) in expression of form \(kt \pm P\sin 4t \pm Q\sin^3 t\) |
| \(= 2\pi + 4\sqrt{2}\) | A1 | CSO; accept exact simplified equivalents such as \(2\pi + \frac{8}{\sqrt{2}}\) or \(2\pi + \sqrt{32}\) |
| (4) |
# Question 16:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y \cdot \frac{dx}{dt} = (2\sin 2t + 3\sin t) \times 16\sin t \cos t$ and uses $\sin 2t = 2\sin t \cos t$ | M1 | Attempt at key step; $\frac{dx}{dt}$ must be of form $k\sin t \cos t$ |
| Correct expanded integrand, **usually** one of: $\int 48\sin^2 t \cos t + 16\sin^2 2t \, dt$, or $\int 48\sin^2 t \cos t + 64\sin^2 t \cos^2 t \, dt$, or $\int 24\sin 2t \sin t + 16\sin^2 2t \, dt$ | A1 | Correct expanded integrand in $t$; ignore absence of $\int$ or $dt$ or limits |
| Attempts to use $\cos 4t = 1 - 2\sin^2 2t = (1 - 8\sin^2 t \cos^2 t)$ | M1 | |
| $R = \int_0^a 8 - 8\cos 4t + 48\sin^2 t \cos t \, dt$ | A1* | Proceeds to given answer with correct working; $\int$ sign and $dt$ must be seen |
| Deduces $a = \dfrac{\pi}{4}$ | B1 | May be awarded from upper limit or from part (b) |
| | **(5)** | |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int 8 - 8\cos 4t + 48\sin^2 t \cos t \, dt = 8t - 2\sin 4t + 16\sin^3 t$ | M1 | Correct approach to integrating trig terms; $\int \ldots = \ldots \pm P\sin 4t \pm Q\sin^3 t$ |
| $= 8t - 2\sin 4t + 16\sin^3 t$ | A1 | Correct integration $(+c)$ |
| $\left[8t - 2\sin 4t + 16\sin^3 t\right]_0^{\frac{\pi}{4}} = 2\pi + 4\sqrt{2}$ | M1 | Uses limits $a$ and $0$ where $a = \frac{\pi}{6}, \frac{\pi}{4}$ or $\frac{\pi}{3}$ in expression of form $kt \pm P\sin 4t \pm Q\sin^3 t$ |
| $= 2\pi + 4\sqrt{2}$ | A1 | CSO; accept exact simplified equivalents such as $2\pi + \frac{8}{\sqrt{2}}$ or $2\pi + \sqrt{32}$ |
| | **(4)** | |16.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{129adfbb-98fa-4e88-b636-7b4d111f3349-46_770_999_242_534}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
Figure 6 shows a sketch of the curve $C$ with parametric equations
$$x = 8 \sin ^ { 2 } t \quad y = 2 \sin 2 t + 3 \sin t \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
The region $R$, shown shaded in Figure 6, is bounded by $C$, the $x$-axis and the line with equation $x = 4$
\begin{enumerate}[label=(\alph*)]
\item Show that the area of $R$ is given by
$$\int _ { 0 } ^ { a } \left( 8 - 8 \cos 4 t + 48 \sin ^ { 2 } t \cos t \right) \mathrm { d } t$$
where $a$ is a constant to be found.
\item Hence, using algebraic integration, find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q16 [9]}}