| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area with logarithmic functions |
| Difficulty | Challenging +1.2 This requires integration by parts of (k-x)ln(x), finding roots of the function to determine limits, and careful algebraic manipulation. While technically demanding with multiple steps, it follows a standard integration by parts pattern that A-level students practice extensively. The conceptual challenge is moderate—recognizing the technique and handling the algebra carefully—but doesn't require novel insight beyond applying learned methods. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve crosses \(x\)-axis when \(y = 0\), so \((k-x)\ln x = 0\) | M1 | Attempt to solve \(y = 0\) |
| Either \(k - x = 0\) or \(\ln x = 0\), giving \(x = k\) or \(x = 1\) | A1 | Both roots required |
| Area \(= \int_1^k (k-x)\ln x \, dx\); let \(u = \ln x\), \(\frac{dv}{dx} = k-x\), \(\frac{du}{dx} = \frac{1}{x}\), \(v = kx - \frac{1}{2}x^2\) | M1 | Integration by parts with \(u = \ln x\), \(\frac{dv}{dx} = k-x\) clearly argued |
| Area \(= \left[(kx - \frac{1}{2}x^2)\ln x\right]_1^k - \int_1^k \frac{1}{x}(kx - \frac{1}{2}x^2)\,dx\) | A1 | Allow without limits |
| \(\left[(kx - \frac{1}{2}x^2)\ln x\right]_1^k - \int_1^k \left(k - \frac{1}{2}x\right)dx\) | M1 | Simplifying the integrand |
| \(\left[(kx - \frac{1}{2}x^2)\ln x - \left(kx - \frac{1}{4}x^2\right)\right]_1^k\) | A1 | Second part correct |
| \(\left((k^2 - \frac{1}{2}k^2)\ln k - (k^2 - \frac{1}{4}k^2)\right) - \left((k - \frac{1}{2})\ln 1 - (k - \frac{1}{4})\right)\) | M1dep | Using limits; dependent on M mark for integration by parts |
| \(= \frac{1}{2}k^2 \ln k - \frac{3}{4}k^2 + k - \frac{1}{4}\) | A1 [8] | cao |
# Question 10:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve crosses $x$-axis when $y = 0$, so $(k-x)\ln x = 0$ | M1 | Attempt to solve $y = 0$ |
| Either $k - x = 0$ or $\ln x = 0$, giving $x = k$ or $x = 1$ | A1 | Both roots required |
| Area $= \int_1^k (k-x)\ln x \, dx$; let $u = \ln x$, $\frac{dv}{dx} = k-x$, $\frac{du}{dx} = \frac{1}{x}$, $v = kx - \frac{1}{2}x^2$ | M1 | Integration by parts with $u = \ln x$, $\frac{dv}{dx} = k-x$ clearly argued |
| Area $= \left[(kx - \frac{1}{2}x^2)\ln x\right]_1^k - \int_1^k \frac{1}{x}(kx - \frac{1}{2}x^2)\,dx$ | A1 | Allow without limits |
| $\left[(kx - \frac{1}{2}x^2)\ln x\right]_1^k - \int_1^k \left(k - \frac{1}{2}x\right)dx$ | M1 | Simplifying the integrand |
| $\left[(kx - \frac{1}{2}x^2)\ln x - \left(kx - \frac{1}{4}x^2\right)\right]_1^k$ | A1 | Second part correct |
| $\left((k^2 - \frac{1}{2}k^2)\ln k - (k^2 - \frac{1}{4}k^2)\right) - \left((k - \frac{1}{2})\ln 1 - (k - \frac{1}{4})\right)$ | M1dep | Using limits; dependent on M mark for integration by parts |
| $= \frac{1}{2}k^2 \ln k - \frac{3}{4}k^2 + k - \frac{1}{4}$ | A1 [8] | cao |
---10 Fig. 10 shows the graph of $y = ( k - x ) \ln x$ where $k$ is a constant ( $k > 1$ ).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{904025c9-6d68-4344-bd41-8c0fccfcf92f-06_454_1266_1564_395}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
Find, in terms of $k$, the area of the finite region between the curve and the $x$-axis.
\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q10 [8]}}