| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Standard +0.8 This is a standard P4/Further Pure parametric area question requiring the formula A = ∫y(dx/dt)dt, followed by a substitution integral. Part (a) involves straightforward differentiation and algebraic manipulation to find K=-2. Part (b) requires a standard substitution leading to ∫u^(1/2) and ∫u^(-1/2), which are routine A-level integration techniques. While it requires multiple steps and careful algebra, it follows a well-established template for this topic with no novel insights needed. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dx}{dt} = \frac{1}{2}(9-4t)^{-\frac{1}{2}} \times -4 = \left(\frac{-2}{\sqrt{9-4t}}\right)\) | M1 | Attempts \(\frac{dx}{dt}\) achieving the form \(k(9-4t)^{-\frac{1}{2}}\) |
| \(A = \int_{(x=0)}^{(x=3)} y\frac{dx}{dt}(\text{d}t) = \int_{\left(t=\frac{9}{4}\right)}^{(t=0)} \frac{t^3}{\sqrt{9+4t}} \times \frac{-2}{\sqrt{9-4t}}(\text{d}t)\) | M1 | Applies correct formula for area \(A = \int y\frac{dx}{dt}\text{d}t\), with \(y\) and their \(\frac{dx}{dt}\) (no need for limits or "d\(t\)") |
| \(= -2\int_{\left(\frac{9}{4}\right)}^{(0)} \frac{t^3}{\sqrt{81-16t^2}}(\text{d}t)\) or \(-2\int_{\left(\frac{9}{4}\right)}^{(0)} \frac{t^3}{\sqrt{(9-4t)(9+4t)}}(\text{d}t)\) | ddM1 | Combines terms to a single fraction with one square root. Depends on both previous M marks |
| \(= 2\int_0^{\frac{9}{4}} \frac{t^3}{\sqrt{81-16t^2}}\text{d}t\) | A1 | Applies difference of squares, deals with limits correctly. The "d\(t\)" must appear at least once before final line |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{du}{dt} = -32t\) or \(\text{d}u = -32t\,\text{d}t\) | B1 | Correct statement of \(\frac{du}{dt}\) or \(\frac{dt}{du}\) or any correct equation connecting d\(u\) with d\(t\) |
| \(A = 2\int_0^{\frac{9}{4}}\frac{t^2}{\sqrt{81-16t^2}}t\,\text{d}t = 2\int_{(81)}^{(0)}\frac{(81-u)}{16\sqrt{u}}\frac{\text{d}u}{-32}\) | M1 | Complete substitution made into integrand from part (a) with their \(K\) to obtain integral in terms of \(u\) only |
| \(= k\int_{(81)}^{(0)} 81u^{-\frac{1}{2}} - u^{\frac{1}{2}}\,\text{d}u\) | A1ft | For any multiple of \(\int\left(81u^{-\frac{1}{2}} - u^{\frac{1}{2}}\right)\text{d}u\) |
| \(= K\left[\frac{81u^{\frac{1}{2}}}{\frac{1}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{(81)}^{(0)}\) | M1 | Integrates expression of form \(ku^{-\frac{1}{2}} + lu^{\frac{1}{2}} \to Au^{\frac{1}{2}} + Bu^{\frac{3}{2}}\) |
| \(= -\frac{1}{256}\left(0 - \left(162\sqrt{81} - \frac{2}{3}\sqrt{81^3}\right)\right) = \ldots\) | ddM1 | Applies correct limits (0 and 81 in \(u\), or 0 and \(\frac{9}{4}\) in \(t\)). Depends on all previous M marks |
| \(= \frac{243}{64}\) | A1 | Correct answer. Note \(-\frac{243}{64}\) scores A0 but allow "hence area is \(\frac{243}{64}\)" |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \frac{1}{2}(9-4t)^{-\frac{1}{2}} \times -4 = \left(\frac{-2}{\sqrt{9-4t}}\right)$ | M1 | Attempts $\frac{dx}{dt}$ achieving the form $k(9-4t)^{-\frac{1}{2}}$ |
| $A = \int_{(x=0)}^{(x=3)} y\frac{dx}{dt}(\text{d}t) = \int_{\left(t=\frac{9}{4}\right)}^{(t=0)} \frac{t^3}{\sqrt{9+4t}} \times \frac{-2}{\sqrt{9-4t}}(\text{d}t)$ | M1 | Applies correct formula for area $A = \int y\frac{dx}{dt}\text{d}t$, with $y$ and their $\frac{dx}{dt}$ (no need for limits or "d$t$") |
| $= -2\int_{\left(\frac{9}{4}\right)}^{(0)} \frac{t^3}{\sqrt{81-16t^2}}(\text{d}t)$ or $-2\int_{\left(\frac{9}{4}\right)}^{(0)} \frac{t^3}{\sqrt{(9-4t)(9+4t)}}(\text{d}t)$ | ddM1 | Combines terms to a single fraction with one square root. Depends on both previous M marks |
| $= 2\int_0^{\frac{9}{4}} \frac{t^3}{\sqrt{81-16t^2}}\text{d}t$ | A1 | Applies difference of squares, deals with limits correctly. The "d$t$" must appear at least once before final line |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{du}{dt} = -32t$ or $\text{d}u = -32t\,\text{d}t$ | B1 | Correct statement of $\frac{du}{dt}$ or $\frac{dt}{du}$ or any correct equation connecting d$u$ with d$t$ |
| $A = 2\int_0^{\frac{9}{4}}\frac{t^2}{\sqrt{81-16t^2}}t\,\text{d}t = 2\int_{(81)}^{(0)}\frac{(81-u)}{16\sqrt{u}}\frac{\text{d}u}{-32}$ | M1 | Complete substitution made into integrand from part (a) with their $K$ to obtain integral in terms of $u$ only |
| $= k\int_{(81)}^{(0)} 81u^{-\frac{1}{2}} - u^{\frac{1}{2}}\,\text{d}u$ | A1ft | For any multiple of $\int\left(81u^{-\frac{1}{2}} - u^{\frac{1}{2}}\right)\text{d}u$ |
| $= K\left[\frac{81u^{\frac{1}{2}}}{\frac{1}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{(81)}^{(0)}$ | M1 | Integrates expression of form $ku^{-\frac{1}{2}} + lu^{\frac{1}{2}} \to Au^{\frac{1}{2}} + Bu^{\frac{3}{2}}$ |
| $= -\frac{1}{256}\left(0 - \left(162\sqrt{81} - \frac{2}{3}\sqrt{81^3}\right)\right) = \ldots$ | ddM1 | Applies correct limits (0 and 81 in $u$, or 0 and $\frac{9}{4}$ in $t$). Depends on all previous M marks |
| $= \frac{243}{64}$ | A1 | Correct answer. Note $-\frac{243}{64}$ scores A0 but allow "hence area is $\frac{243}{64}$" |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fe07afad-9cfc-48c0-84f1-5717f81977d4-14_688_691_251_630}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with parametric equations
$$x = \sqrt { 9 - 4 t } \quad y = \frac { t ^ { 3 } } { \sqrt { 9 + 4 t } } \quad 0 \leqslant t \leqslant \frac { 9 } { 4 }$$
The curve touches the $x$-axis when $t = 0$ and meets the $y$-axis when $t = \frac { 9 } { 4 }$\\
The region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the $y$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of $R$ is given by
$$K \int _ { 0 } ^ { \frac { 9 } { 4 } } \frac { t ^ { 3 } } { \sqrt { 81 - 16 t ^ { 2 } } } \mathrm {~d} t$$
where $K$ is a constant to be found.
\item Using the substitution $u = 81 - 16 t ^ { 2 }$, or otherwise, find the exact area of $R$.\\
(Solutions relying on calculator technology are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q5 [10]}}