| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Trapezium Rule Approximation with Area |
| Difficulty | Moderate -0.3 This is a multi-part C2 question requiring standard techniques: describing simple transformations (translation and horizontal stretch), using linearity of integration with a given result, applying the trapezium rule with clear instructions, and finding area between curves. While it has multiple parts (7 marks total), each component is routine and well-practiced at this level, making it slightly easier than average but not trivial due to the multi-step nature and need to reason about over/underestimation. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Translation by \(\begin{pmatrix}0\\1\end{pmatrix}\) (translation 1 unit in positive \(y\)-direction) | B2 | B1 for translation, B1 for correct vector/description |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Stretch scale factor \(\frac{1}{9}\) in the \(x\)-direction (horizontal stretch, s.f. \(\frac{1}{9}\)) | B2 | B1 for stretch, B1 for correct direction and scale factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^9 (1+\sqrt{x})\,dx = 9 + 18 = 27\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = \frac{9-0}{4} = 2.25\), ordinates at \(x = 0, 2.25, 4.5, 6.75, 9\) | M1 | Correct strip width |
| \(y\) values: \(1,\ 4^{0.25},\ 4^{0.5},\ 4^{0.75},\ 4\) i.e. \(1,\ 1.414...,\ 2,\ 2.828...,\ 4\) | M1 | At least 3 correct ordinates |
| \(\approx \frac{2.25}{2}(1 + 4 + 2(1.414 + 2 + 2.828))\) | M1 | Correct trapezium rule structure |
| \(\approx 22.9\) | A1 | Accept 22.8–23.0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Approximate area \(\approx 27 - 22.9 = 4.1\) | B1 | Follow through from (b)(i) and (b)(ii) |
| The approximation is an overestimate | B1 | Correct conclusion |
| Because \(y = 4^{x/9}\) is convex (curves upwards), so trapezium rule overestimates the integral | B1 | Correct reason relating to shape of curve |
## Question 7:
### Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Translation by $\begin{pmatrix}0\\1\end{pmatrix}$ (translation 1 unit in positive $y$-direction) | B2 | B1 for translation, B1 for correct vector/description |
### Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Stretch scale factor $\frac{1}{9}$ in the $x$-direction (horizontal stretch, s.f. $\frac{1}{9}$) | B2 | B1 for stretch, B1 for correct direction and scale factor |
### Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^9 (1+\sqrt{x})\,dx = 9 + 18 = 27$ | B1 | cao |
### Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = \frac{9-0}{4} = 2.25$, ordinates at $x = 0, 2.25, 4.5, 6.75, 9$ | M1 | Correct strip width |
| $y$ values: $1,\ 4^{0.25},\ 4^{0.5},\ 4^{0.75},\ 4$ i.e. $1,\ 1.414...,\ 2,\ 2.828...,\ 4$ | M1 | At least 3 correct ordinates |
| $\approx \frac{2.25}{2}(1 + 4 + 2(1.414 + 2 + 2.828))$ | M1 | Correct trapezium rule structure |
| $\approx 22.9$ | A1 | Accept 22.8–23.0 |
### Part (b)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Approximate area $\approx 27 - 22.9 = 4.1$ | B1 | Follow through from (b)(i) and (b)(ii) |
| The approximation is an **overestimate** | B1 | Correct conclusion |
| Because $y = 4^{x/9}$ is convex (curves upwards), so trapezium rule overestimates the integral | B1 | Correct reason relating to shape of curve |7 The diagram shows a sketch of two curves.\\
\includegraphics[max width=\textwidth, alt={}, center]{24641e66-b73b-4323-98c8-349727151aba-14_448_527_370_762}
The equations of the two curves are $y = 1 + \sqrt { x }$ and $y = 4 ^ { \frac { x } { 9 } }$.\\
The curves meet at the points $P ( 0,1 )$ and $Q ( 9,4 )$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Describe the geometrical transformation that maps the graph of $y = \sqrt { x }$ onto the graph of $y = 1 + \sqrt { x }$.
\item Describe the geometrical transformation that maps the graph of $y = 4 ^ { x }$ onto the graph of $y = 4 ^ { \frac { x } { 9 } }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $\int _ { 0 } ^ { 9 } \sqrt { x } \mathrm {~d} x = 18$, find the value of $\int _ { 0 } ^ { 9 } ( 1 + \sqrt { x } ) \mathrm { d } x$.
\item Use the trapezium rule with five ordinates (four strips) to find an approximate value for $\int _ { 0 } ^ { 9 } 4 ^ { \frac { x } { 9 } } \mathrm {~d} x$. Give your answer to one decimal place.
\item Hence find an approximate value for the area of the shaded region bounded by the two curves and state, with an explanation, whether your approximation will be an overestimate or an underestimate of the true value for the area of the shaded region.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2015 Q7 [14]}}