| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under polynomial curve |
| Difficulty | Moderate -0.3 This is a straightforward C2 integration question requiring finding intersection points by solving a quadratic equation, then integrating a polynomial to find area. While it involves multiple steps, both parts use standard techniques (solving equations, basic integration) with no conceptual challenges beyond routine application of formulas. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 + 2x + 2 = 10 \Rightarrow x^2 + 2x - 8 = 0\) so \((x+4)(x-2) = 0 \Rightarrow x = \ldots\) | M1 | Set curve equal to 10, collect terms, solve quadratic |
| \(x = -4, \ 2\) | A1 | Both values correct; allow \(A = -4, B = 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(x^2 + 2x + 2)dx = \frac{x^3}{3} + \frac{2x^2}{2} + 2x \ (+C)\) | M1A1A1 | One correct integration; two correct integrations; all 3 terms correct |
| \(\left[\frac{x^3}{3} + \frac{2x^2}{2} + 2x\right]_{-4}^{2} = \left(\frac{8}{3} + \frac{8}{2} + 4\right) - \left(-\frac{64}{3} + \frac{32}{2} - 8\right) \ (= 24)\) | M1 | Substitute limits from (a) into integrated function and subtract |
| Rectangle: \(10 \times (2 - -4) = 60\) | B1 cao | Find area under line by integration or area of rectangle — should be 60 |
| \(R = \text{"60"} - \text{"24"}\) | M1 | Subtract one area from the other |
| \(= 36\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(8 - x^2 - 2x)dx = 8x - \frac{x^3}{3} - \frac{2x^2}{2} \ (+C)\) | M1 A1ft A1 | Penalise subtraction errors; all 3 terms correct |
| \(\left[8x - \frac{x^3}{3} - \frac{2x^2}{2}\right]_{-4}^{2} = \left(16 - \frac{8}{3} - 4\right) - \left(-32 + \frac{64}{3} - 16\right) = (9.3 - (-26.7))\) | M1 | |
| Implied by final answer of 36 after correct work | B1 | |
| \(10 - (x^2 + 2x + 2) = 8 - x^2 - 2x, = 36\) | M1, A1 | Must be 36 not \(-36\) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + 2x + 2 = 10 \Rightarrow x^2 + 2x - 8 = 0$ so $(x+4)(x-2) = 0 \Rightarrow x = \ldots$ | M1 | Set curve equal to 10, collect terms, solve quadratic |
| $x = -4, \ 2$ | A1 | Both values correct; allow $A = -4, B = 2$ |
## Part (b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(x^2 + 2x + 2)dx = \frac{x^3}{3} + \frac{2x^2}{2} + 2x \ (+C)$ | M1A1A1 | One correct integration; two correct integrations; all 3 terms correct |
| $\left[\frac{x^3}{3} + \frac{2x^2}{2} + 2x\right]_{-4}^{2} = \left(\frac{8}{3} + \frac{8}{2} + 4\right) - \left(-\frac{64}{3} + \frac{32}{2} - 8\right) \ (= 24)$ | M1 | Substitute limits from (a) into integrated function and subtract |
| Rectangle: $10 \times (2 - -4) = 60$ | B1 cao | Find area under line by integration or area of rectangle — should be 60 |
| $R = \text{"60"} - \text{"24"}$ | M1 | Subtract one area from the other |
| $= 36$ | A1 | |
## Part (b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(8 - x^2 - 2x)dx = 8x - \frac{x^3}{3} - \frac{2x^2}{2} \ (+C)$ | M1 A1ft A1 | Penalise subtraction errors; all 3 terms correct |
| $\left[8x - \frac{x^3}{3} - \frac{2x^2}{2}\right]_{-4}^{2} = \left(16 - \frac{8}{3} - 4\right) - \left(-32 + \frac{64}{3} - 16\right) = (9.3 - (-26.7))$ | M1 | |
| Implied by final answer of 36 after correct work | B1 | |
| $10 - (x^2 + 2x + 2) = 8 - x^2 - 2x, = 36$ | M1, A1 | Must be 36 not $-36$ |
---\begin{enumerate}[label=(\alph*)]
\item Find by calculation the $x$-coordinate of $A$ and the $x$-coordinate of $B$.
The shaded region $R$ is bounded by the line with equation $y = 10$ and the curve as shown in Figure 1.
\item Use calculus to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q7 [9]}}