| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| 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 routine integration of a polynomial, evaluation at limits, and understanding that area between curve and line equals integral of curve minus area of triangle. The 'hence' in part (b) guides students to the method. Slightly easier than average due to straightforward polynomial integration and clear geometric setup, but requires multiple steps and careful handling of the triangle area calculation. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{-1}^{1}(x^3 - 2x^2 + 3)\,dx\) | ||
| \(= \left[\frac{x^4}{4} - \frac{2x^3}{3} + 3x\right]_{-1}^{1}\) | M1 | one term correct |
| A1 | another term correct | |
| A1 | all correct (condone \(+c\)) | |
| \(= \left(\frac{1}{4} - \frac{2}{3} + 3\right) - \left(-\frac{1}{4} + \frac{2}{3} - 3\right)\) | B1 \(\checkmark\) | 'their' \(F(1) - F(-1)\) with \((-1)^n\) etc evaluated correctly but must have earned M1 |
| \(= 4\frac{2}{3}\) | A1 also | \(\frac{14}{3}, \frac{56}{12}\) etc but combined as single fraction |
| Answer | Marks | Guidance |
|---|---|---|
| Area of \(\triangle = \left(\frac{1}{2} \times 2 \times 2\right) = 2\) | B1 | PI |
| Shaded region has area \(4\frac{2}{3} - 2 = 2\frac{2}{3}\) | M1 | \(\pm\) their (a) \(\pm\) their \(\triangle\) area |
| A1 also | \(\frac{8}{3}, \frac{32}{12}\) etc but combined as single fraction |
**6(a)**
$\int_{-1}^{1}(x^3 - 2x^2 + 3)\,dx$ | |
$= \left[\frac{x^4}{4} - \frac{2x^3}{3} + 3x\right]_{-1}^{1}$ | M1 | one term correct
| A1 | another term correct
| A1 | all correct (condone $+c$)
$= \left(\frac{1}{4} - \frac{2}{3} + 3\right) - \left(-\frac{1}{4} + \frac{2}{3} - 3\right)$ | B1 $\checkmark$ | 'their' $F(1) - F(-1)$ with $(-1)^n$ etc evaluated correctly but must have earned M1
$= 4\frac{2}{3}$ | A1 also | $\frac{14}{3}, \frac{56}{12}$ etc but combined as single fraction
**6(b)**
Area of $\triangle = \left(\frac{1}{2} \times 2 \times 2\right) = 2$ | B1 | PI
Shaded region has area $4\frac{2}{3} - 2 = 2\frac{2}{3}$ | M1 | $\pm$ their (a) $\pm$ their $\triangle$ area
| A1 also | $\frac{8}{3}, \frac{32}{12}$ etc but combined as single fraction
---6 The curve with equation $y = x ^ { 3 } - 2 x ^ { 2 } + 3$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{c44c229e-44b2-4799-9c9c-bfccdd09d450-4_590_787_365_625}
The curve cuts the $x$-axis at the point $A ( - 1,0 )$ and passes through the point $B ( 1,2 )$.
\begin{enumerate}[label=(\alph*)]
\item Find $\int _ { - 1 } ^ { 1 } \left( x ^ { 3 } - 2 x ^ { 2 } + 3 \right) \mathrm { d } x$.\\
(5 marks)
\item Hence find the area of the shaded region bounded by the curve $y = x ^ { 3 } - 2 x ^ { 2 } + 3$ and the line $A B$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q6 [8]}}