| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Moderate -0.3 This is a standard C2 integration question requiring finding intersection points by solving a quadratic equation, then integrating the difference of two functions. The algebra is straightforward (4-x²=x+2 gives x²+x-2=0), and the integration involves only basic polynomial terms. Slightly easier than average due to the simple functions and routine procedure, but not trivial as it requires correct setup and execution of multiple steps. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Intersect where \(x^2 + x - 2 = 0 \Rightarrow x = -2, 1\) | M1, A1 | For finding \(x\) at both intersections. For both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Hence shaded area is \(9 - 4\frac{1}{2} = 4\frac{1}{2}\) | M1, M1, A1, M1, A1, A1 | For integration attempt with any one term correct. For use of limits – subtraction and correct order. For correct area of 9. Attempt area of triangle (\(\frac{1}{2}bh\) or integration). Obtain area of triangle as \(4\frac{1}{2}\). Obtain correct final area of \(4\frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 4\frac{1}{2}\) | M1, M1, A1, M1, A1, A1 | Attempt subtraction – either order. For integration attempt with any one term correct. Obtain \(\pm [-\frac{1}{3}x^3 - \frac{1}{2}x^2 + 2x]\). For use of limits – subtraction and correct order. Obtain \(\pm 4\frac{1}{2}\) - consistent with their order of subtraction. Obtain \(4\frac{1}{2}\) only, following correct method only |
**(i)** Intersect where $x^2 + x - 2 = 0 \Rightarrow x = -2, 1$ | M1, A1 | For finding $x$ at both intersections. For both values correct | 2 marks
**(ii)** Area under curve is $\left[-\frac{1}{3}x^3\right]_{-2}^{1}$
i.e. $(4 - \frac{1}{3}) - (-8 + \frac{1}{3}) = 9$
Area of triangle is $4\frac{1}{2}$
Hence shaded area is $9 - 4\frac{1}{2} = 4\frac{1}{2}$ | M1, M1, A1, M1, A1, A1 | For integration attempt with any one term correct. For use of limits – subtraction and correct order. For correct area of 9. Attempt area of triangle ($\frac{1}{2}bh$ or integration). Obtain area of triangle as $4\frac{1}{2}$. Obtain correct final area of $4\frac{1}{2}$ | 6 marks
**OR**
Area under curve is $\int_{-2}^{1} (2 - x - x^2) dx$
$= [-\frac{1}{3}x^3 - \frac{1}{2}x^2 + 2x]_{-2}^{1}$
$= (-\frac{1}{3} - \frac{1}{2} + 2) - (\frac{8}{3} - 2 - 4)$
$= 4\frac{1}{2}$ | M1, M1, A1, M1, A1, A1 | Attempt subtraction – either order. For integration attempt with any one term correct. Obtain $\pm [-\frac{1}{3}x^3 - \frac{1}{2}x^2 + 2x]$. For use of limits – subtraction and correct order. Obtain $\pm 4\frac{1}{2}$ - consistent with their order of subtraction. Obtain $4\frac{1}{2}$ only, following correct method only | 8 marks
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{367db494-294e-4b53-b9e8-fd2a69fb6069-2_634_670_1123_740}
The diagram shows the curve $y = 4 - x ^ { 2 }$ and the line $y = x + 2$.\\
(i) Find the $x$-coordinates of the points of intersection of the curve and the line.\\
(ii) Use integration to find the area of the shaded region bounded by the line and the curve.
\hfill \mbox{\textit{OCR C2 2006 Q4 [8]}}