| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a standard two-part integration question requiring area between curves and volume of revolution. Part (i) uses straightforward integration with simple functions. Part (ii) requires rotating about the y-axis, which needs expressing x in terms of y (x = y-1 and x = y²-1), then applying the volume formula—methodical but routine for P1 level with no conceptual surprises. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{6}\) | M1, M1A1, DM1, A1 [5] | Dealing with line as a triangle or integral with correct limits; Attempt at integral of curve; Applying limits -1→0 or 0→1 to curve; \(\pi\) included loses last mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(V_1 = \frac{8}{15(\pi)}\) or \(0.533(\pi)\) (AWRT) | M1, A1, DM1, A1 | Attempt at \(\int x^2 \, dy\) for curve; Apply limits 0→1 |
| Answer | Marks | Guidance |
|---|---|---|
| Volume \(= \frac{8}{15}\pi - \frac{1}{3}\pi = \frac{1}{5}\pi\) (or 0.628) | M1, A1, A1 | Or \(\frac{1}{3}\pi(\times 1^2 \times 1)\); Vol of cone or attempt to \(\int x^2dy\) for line |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{5}\pi\) | M1, M1, A1,A1,A1, DM1, A1 [7] | Attempt to \(\int x^2 \, dy\); Attempt to \(\int(x_1^2 - x_2^2)\); Apply limits 0→1 dependent on first M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} - \frac{1}{3} - \frac{1}{6}\pi\) (0.524) | M1, M1, A1 | SC MR integrating about x axis; Use of -1,0 as limits |
**(i)** $\int(x+1)^{\frac{1}{2}} - (x+1)$ or $\int[(y^2-1) - (y-1)]$
$\frac{2}{3}(x+1)^{\frac{3}{2}} - \frac{1}{2}x^2 - x$ or $\frac{1}{3}y^3 - \frac{1}{2}y^2$
$\frac{2}{3}\left(0-\frac{1}{2}+1\right)$ or $\frac{1}{3}-\frac{1}{2}$
$\frac{1}{6}$ | M1, M1A1, DM1, A1 [5] | Dealing with line as a triangle or integral with correct limits; Attempt at integral of curve; Applying limits -1→0 or 0→1 to curve; $\pi$ included loses last mark
**(ii)** $V_1 = (\pi)\int(y^2-1)^2 = (\pi)\int(y^4 - 2y^2 + 1$
$(\pi)\left[\frac{y^5}{5} - \frac{2y^2}{3} + y\right]$
$(\pi)\left[\frac{1}{5} - \frac{2}{3} + 1\right]$
$V_1 = \frac{8}{15(\pi)}$ or $0.533(\pi)$ (AWRT) | M1, A1, DM1, A1 | Attempt at $\int x^2 \, dy$ for curve; Apply limits 0→1
**OR** $(\pi)\left[y^2 - y^2 + y\right]$
$V_2 = \frac{1}{3}\pi$
Volume $= \frac{8}{15}\pi - \frac{1}{3}\pi = \frac{1}{5}\pi$ (or 0.628) | M1, A1, A1 | Or $\frac{1}{3}\pi(\times 1^2 \times 1)$; Vol of cone or attempt to $\int x^2dy$ for line
**OR** $(y^4 - 2y^2 + 1) - (y^2 - 2y + 1)$
$(\pi)\int(y^4 - 3y^2 + 2y)$
$(\pi)[y^5/5 - y^3 \cdot 3 + y^2 \cdot 2]$
$(\pi)\left[\frac{1}{5} - 1 + 1\right]$
$\frac{1}{5}\pi$ | M1, M1, A1,A1,A1, DM1, A1 [7] | Attempt to $\int x^2 \, dy$; Attempt to $\int(x_1^2 - x_2^2)$; Apply limits 0→1 dependent on first M1
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## Additional Integration Problem
$\int_{-1}^{0} x + 1 - \int_{-1}^{0}(x+1)^2$
$\left[\frac{x^2}{2} + x\right] - \left[\frac{(x+1)^3}{3}\right]$
SC $= \left[(0) - \left(\frac{1}{2}-1\right)\right] - \left[\frac{1}{3} - 0\right]$
$\frac{1}{2} - \frac{1}{3} - \frac{1}{6}\pi$ (0.524) | M1, M1, A1 | SC MR integrating about x axis; Use of -1,0 as limits\includegraphics{figure_10}
The diagram shows the line $y = x + 1$ and the curve $y = \sqrt{(x + 1)}$, meeting at $(-1, 0)$ and $(0, 1)$.
\begin{enumerate}[label=(\roman*)]
\item Find the area of the shaded region. [5]
\item Find the volume obtained when the shaded region is rotated through $360°$ about the $y$-axis. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q10 [12]}}