| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward integration question requiring finding a tangent equation (standard differentiation), identifying intersection points (basic algebra), calculating a triangle area (formula application), and finding an area between curve and line (standard integration setup). All steps are routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 3x^2\) | M1 | 3.1a |
| Gradient at \((1, 1)\) is \(3\) | M1 | 1.1 |
| Tangent is \((y-1) = 3(x-1)\) | A1 | 1.1 |
| \(OE = 2\) and \(OD = \frac{2}{3}\) | M1 | 3.1a |
| Both \(OE\) and \(OD\) correct | A1 | 1.1 |
| Area triangle \(ODE = \frac{1}{2} \times \frac{2}{3} \times 2 = \frac{2}{3}\) | A1 | 1.1 |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= \int_0^1 \left(x^3 - (3x-2)\right) dx\) | M1, M1 | 3.1a, 1.1 |
| \(\left[\frac{x^4}{4} - \frac{3x^2}{2} + 2x\right]_0^1\) | M1 | 1.1 |
| \(\frac{3}{4}\) | A1 | 2.2a |
| Alternative method: | ||
| Area between curve and \(x\)-axis: \(\int_0^1 x^3\, dx = \left[\frac{x^4}{4}\right]_0^1 = \left[\frac{1}{4}\right]\) | M1 | |
| Area triangle \(ABD = \frac{1}{2} \times \frac{1}{3} \times 1 = \frac{1}{6}\) | M1 | |
| Area \(=\) area between curve and \(x\)-axis \(+\) triangle \(ODE\) \(-\) triangle \(ABD\) | M1 | Using *their* values |
| \(\frac{3}{4}\) | A1 | |
| [4] |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 3x^2$ | M1 | 3.1a |
| Gradient at $(1, 1)$ is $3$ | M1 | 1.1 |
| Tangent is $(y-1) = 3(x-1)$ | A1 | 1.1 | $y = 3x-2$; simplified form not essential; or $\frac{AB}{AD} = 3$ |
| $OE = 2$ and $OD = \frac{2}{3}$ | M1 | 3.1a | Finding either length; soi by $E(0,-2)$ or $D\left(\frac{2}{3}, 0\right)$; or $AD = \frac{1}{3}$; Or Triangle ODE is similar to ABD [with scale factor 2] |
| Both $OE$ and $OD$ correct | A1 | 1.1 | |
| Area triangle $ODE = \frac{1}{2} \times \frac{2}{3} \times 2 = \frac{2}{3}$ | A1 | 1.1 | |
| **[6]** | | |
---
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= \int_0^1 \left(x^3 - (3x-2)\right) dx$ | M1, M1 | 3.1a, 1.1 | Integrand (FT their tangent); Limits (may be seen later) |
| $\left[\frac{x^4}{4} - \frac{3x^2}{2} + 2x\right]_0^1$ | M1 | 1.1 | Allow one error |
| $\frac{3}{4}$ | A1 | 2.2a | |
| **Alternative method:** | | |
| Area between curve and $x$-axis: $\int_0^1 x^3\, dx = \left[\frac{x^4}{4}\right]_0^1 = \left[\frac{1}{4}\right]$ | M1 | |
| Area triangle $ABD = \frac{1}{2} \times \frac{1}{3} \times 1 = \frac{1}{6}$ | M1 | |
| Area $=$ area between curve and $x$-axis $+$ triangle $ODE$ $-$ triangle $ABD$ | M1 | Using *their* values |
| $\frac{3}{4}$ | A1 | |
| **[4]** | | |
---8 In this question you must show detailed reasoning.
A is the point $( 1,0 ) , B$ is the point $( 1,1 )$ and $D$ is the point where the tangent to the curve $y = x ^ { 3 }$ at B crosses the $x$-axis, as shown in Fig. 8. The tangent meets the $y$-axis at E.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{99485c27-9ff8-4bdb-a7e6-49dfcaedc579-6_1154_832_450_242}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find the area of triangle ODE.
\item Find the area of the region bounded by the curve $y = x ^ { 3 }$, the tangent at B and the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q8 [10]}}