| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Moderate -0.8 Part (i) is a straightforward integration of polynomials requiring only basic power rule. Part (ii) requires recognizing that the curve crosses the x-axis and computing two definite integrals, but the symmetry property makes this a standard C2 exercise with no novel insight needed. This is easier than average for A-level. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{x^4}{4} - x^2 + c\) | M1 A1 | 2 marks |
| (ii) \(\int_0^2 (x^3 - 2x)dx = \left[\frac{x^4}{4} - x^2\right]_0^2 = \frac{16}{4} - 4 = 0\) Therefore Area above = Area below | M1 A1 E1 | 3 marks |
**(i)** $\frac{x^4}{4} - x^2 + c$ | M1 A1 | 2 marks
**(ii)** $\int_0^2 (x^3 - 2x)dx = \left[\frac{x^4}{4} - x^2\right]_0^2 = \frac{16}{4} - 4 = 0$ Therefore Area above = Area below | M1 A1 E1 | 3 marks | Alternatively show: $\int_0^{\sqrt{2}} = \int_{\sqrt{2}}^2$1 (i) Find $\int \left( x ^ { 3 } - 2 x \right) \mathrm { d } x$.
The graph below shows part of the curve $y = x ^ { 3 } - 2 x$ for $0 \leq x \leq 2$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c55a5f04-3573-4f36-a12c-3755bdd4a45b-2_528_1019_520_321}\\
(ii) Show that the area of the shaded region $P$ is the same as the area of the shaded region $Q$.
\hfill \mbox{\textit{OCR MEI C2 Q1 [5]}}