| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Moderate -0.3 This is a standard C1 integration question requiring finding a line equation (2 marks), evaluating a polynomial integral (straightforward application of power rule), and subtracting areas using a trapezium. All techniques are routine for C1 with no conceptual challenges, making it slightly easier than average but not trivial due to the multi-step nature. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Grad \(AC = \frac{15}{3} = 5\) | B1 | |
| Equation of \(AC\): \(y = m(x + 2)\) or \((y - 15) = m(x - 1)\) | M1 | |
| \(y = 5x + 10\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[16x - \frac{x^3}{5}\right]\) | M1, A1, A1 | Raise one power by 1, One term correct, All correct |
| \(\left(16 - \frac{1}{5}\right) - \left(-32 + \frac{32}{5}\right)\) | m1 | \(F(1) - F(-2)\) attempted |
| \(= 41\frac{4}{5}\) (or \(41.4\), \(\frac{207}{5}\) etc) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(\Delta = \frac{1}{2} \times 3 \times 15\) or \(22\frac{1}{2}\) or \(22.5\) | B1 | |
| Shaded area = "their (b)(i) answer" – correct triangle | M1 | |
| \(\Rightarrow\) shaded area \(= 18\frac{9}{10}\) | A1 | 3 |
**5(a)**
| Grad $AC = \frac{15}{3} = 5$ | B1 | | OE |
| Equation of $AC$: $y = m(x + 2)$ or $(y - 15) = m(x - 1)$ | M1 | | Or use of $y = mx + c$ with $(-2, 0)$ or $(1, 15)$ correctly substituted for $x$ and $y$ |
| $y = 5x + 10$ | A1 | 3 | OE eg $y - 15 = 5(x - 1)$, $y = 5(x + 2)$ |
**5(b)(i)**
| $\left[16x - \frac{x^3}{5}\right]$ | M1, A1, A1 | Raise one power by 1, One term correct, All correct |
| $\left(16 - \frac{1}{5}\right) - \left(-32 + \frac{32}{5}\right)$ | m1 | $F(1) - F(-2)$ attempted |
| $= 41\frac{4}{5}$ (or $41.4$, $\frac{207}{5}$ etc) | A1 | 5 | CSO; withhold if $+c$ added |
**5(b)(ii)**
| Area $\Delta = \frac{1}{2} \times 3 \times 15$ or $22\frac{1}{2}$ or $22.5$ | B1 | | Or $\int_{-2}^{1} (5x + 10) \, dx = 22.5$ |
| Shaded area = "their (b)(i) answer" – correct triangle | M1 | | Condone "difference" if $\Delta > \int$ |
| $\Rightarrow$ shaded area $= 18\frac{9}{10}$ | A1 | 3 | CSO; OE $(18.9$ etc$)$ |
---5 The curve with equation $y = 16 - x ^ { 4 }$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{fddf5016-a5bd-42db-b5c4-f4980b8d9d67-3_435_663_824_685}
The points $A ( - 2,0 ) , B ( 2,0 )$ and $C ( 1,15 )$ lie on the curve.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the straight line $A C$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int _ { - 2 } ^ { 1 } \left( 16 - x ^ { 4 } \right) \mathrm { d } x$.
\item Hence calculate the area of the shaded region bounded by the curve and the line $A C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2008 Q5 [11]}}