| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Area under curve using substitution |
| Difficulty | Standard +0.3 This is a straightforward integration by substitution question. Part (a) requires recognizing that sin(√x) = 0 when √x = π, giving x = π². Part (b) involves a standard substitution u = √x leading to ∫2u sin u du, which is routine integration by parts. The question is slightly easier than average as the substitution is given and the steps are mechanical. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \((a =) \pi^2\) | B1 | Allow 32400, \(180^2\). Accept \((x =)\pi^2\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(dx = 2u\, du\) or equivalent | B1 | e.g. \(\frac{du}{dx} = \frac{1}{2\sqrt{x}}\). Incorrect statements e.g. \(du = \frac{1}{2\sqrt{x}}\) is B0 |
| Substitute for \(x\) and \(dx\) throughout the integral | M1 | |
| Obtain \(\int 2u\sin u\, du\) | A1 | Allow with missing \(du\) |
| Commence integration of \(\int ku\sin u\, du\) by parts and reach \(\mp ku\cos u \pm \int k\cos u\, du\) | \*M1 | |
| Obtain integral \(-ku\cos u + k\sin u\) | A1 | |
| Substitute limits \(u = 0\) and \(u = \sqrt{\text{their } a}\), \(a \neq 0\), \(a\) in radians, or \(x = 0\) and their \(a\) in the complete integral | DM1 | \(-2\pi\cos\pi + 2\sin\pi(+0 - 2\sin 0)\). Need limits stated but condone if zeros not shown in substitution |
| Obtain answer \(2\pi\) | A1 | |
| 7 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $(a =) \pi^2$ | **B1** | Allow 32400, $180^2$. Accept $(x =)\pi^2$ |
| | **1** | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $dx = 2u\, du$ or equivalent | **B1** | e.g. $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$. Incorrect statements e.g. $du = \frac{1}{2\sqrt{x}}$ is B0 |
| Substitute for $x$ and $dx$ throughout the integral | **M1** | |
| Obtain $\int 2u\sin u\, du$ | **A1** | Allow with missing $du$ |
| Commence integration of $\int ku\sin u\, du$ by parts and reach $\mp ku\cos u \pm \int k\cos u\, du$ | **\*M1** | |
| Obtain integral $-ku\cos u + k\sin u$ | **A1** | |
| Substitute limits $u = 0$ and $u = \sqrt{\text{their } a}$, $a \neq 0$, $a$ in radians, or $x = 0$ and their $a$ in the complete integral | **DM1** | $-2\pi\cos\pi + 2\sin\pi(+0 - 2\sin 0)$. Need limits stated but condone if zeros not shown in substitution |
| Obtain answer $2\pi$ | **A1** | |
| | **7** | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{3c63c42a-2658-4984-93e8-b2a8d18eb37a-12_473_839_274_644}
The diagram shows part of the curve $y = \sin \sqrt { x }$. This part of the curve intersects the $x$-axis at the point where $x = a$.
\begin{enumerate}[label=(\alph*)]
\item State the exact value of $a$.
\item Using the substitution $u = \sqrt { x }$, find the exact area of the shaded region in the first quadrant bounded by this part of the curve and the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q8 [8]}}