| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.3 This is a standard C2 question testing understanding that integration gives signed area when the curve goes below the x-axis. Students must identify where the curve crosses the axis (x=3), split the integral, and use absolute values. The algebra is straightforward (integrating x²-3x), requiring only basic integration and arithmetic. Slightly above average difficulty due to the conceptual element, but this is a textbook exercise on a core C2 topic. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Some of the area is below the \(x\)-axis | B1 [1] | Refer to area/curve below \(x\)-axis or 'negative area' |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt integration | M1 | Attempt integration with any one term correct |
| \(\frac{1}{3}x^3 - \frac{3}{2}x^2\) | A1 | Obtain \(\frac{1}{3}x^3 - \frac{3}{2}x^2\) |
| \(\left[\frac{1}{3}x^3 - \frac{3}{2}x^2\right]_0^3 = (9 - \frac{27}{2}) - (0-0)\) | M1 | Use limits 3 and 0 — correct order/subtraction |
| \(= -4\frac{1}{2}\) | A1 | Obtain \((-) 4\frac{1}{2}\) |
| \(\left[\frac{1}{3}x^3 - \frac{3}{2}x^2\right]_3^5 = (\frac{125}{3} - \frac{75}{2}) - (9 - \frac{27}{2})\) | M1 | Use limits 5 and 3 — correct order/subtraction |
| \(= 8\frac{2}{3}\) | A1 | Obtain \(8^2/_3\) (allow 8.7 or better) |
| Hence total area is \(13\frac{1}{6}\) | A1 [7] | Obtain total area as \(13^1/_6\), or exact equiv. SR: if no longer \(\int f(x)dx\), then B1 for using \([0,3]\) and \([3,5]\) |
# Question 7:
## Part (i)
| Some of the area is below the $x$-axis | B1 **[1]** | Refer to area/curve below $x$-axis or 'negative area' |
## Part (ii)
| Attempt integration | M1 | Attempt integration with any one term correct |
| $\frac{1}{3}x^3 - \frac{3}{2}x^2$ | A1 | Obtain $\frac{1}{3}x^3 - \frac{3}{2}x^2$ |
| $\left[\frac{1}{3}x^3 - \frac{3}{2}x^2\right]_0^3 = (9 - \frac{27}{2}) - (0-0)$ | M1 | Use limits 3 and 0 — correct order/subtraction |
| $= -4\frac{1}{2}$ | A1 | Obtain $(-) 4\frac{1}{2}$ |
| $\left[\frac{1}{3}x^3 - \frac{3}{2}x^2\right]_3^5 = (\frac{125}{3} - \frac{75}{2}) - (9 - \frac{27}{2})$ | M1 | Use limits 5 and 3 — correct order/subtraction |
| $= 8\frac{2}{3}$ | A1 | Obtain $8^2/_3$ (allow 8.7 or better) |
| Hence total area is $13\frac{1}{6}$ | A1 **[7]** | Obtain total area as $13^1/_6$, or exact equiv. SR: if no longer $\int f(x)dx$, then B1 for using $[0,3]$ and $[3,5]$ |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{2ae05b46-6c9f-4aaa-9cba-1116c0ec27d4-3_579_557_858_794}
The diagram shows part of the curve $y = x ^ { 2 } - 3 x$ and the line $x = 5$.\\
(i) Explain why $\int _ { 0 } ^ { 5 } \left( x ^ { 2 } - 3 x \right) \mathrm { d } x$ does not give the total area of the regions shaded in the diagram.\\
(ii) Use integration to find the exact total area of the shaded regions.
\hfill \mbox{\textit{OCR C2 2008 Q7 [8]}}