| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Exponential and logarithmic integration |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 integration techniques (exponential functions), coordinate geometry (finding intercepts, normals), and area calculations. While it requires multiple steps and careful organization, each individual component uses routine methods: basic integration of e^(2x), substituting limits, finding gradients/normals, and computing areas using definite integrals. The 'show that' in part (a)(ii) provides the answer to check against, reducing difficulty. This is slightly easier than the average A-level question due to its highly scaffolded structure. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int(4 - e^{2x})\,dx\) | ||
| \(= 4x - \frac{1}{2}e^{2x}\ (+c)\) | B1 | \(4x\) |
| B1 | \(-\frac{1}{2}e^{2x}\) | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_0^{\ln 2} = \left[4x - \frac{1}{2}e^{2x}\right]_0^{\ln 2}\) | ||
| \(= \left[4\ln 2 - \frac{1}{2}e^{2\ln 2}\right] - \left[(0) - \frac{1}{2}(e^0)\right]\) | M1 | Substitute both \(\ln 2\) and \(0\) correctly into an integrated expression |
| \(= 4\ln 2 - 2 + \frac{1}{2}\) | Convincing | |
| \(= 4\ln 2 - \frac{3}{2}\) | A1 | AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 0,\ y = 4 - 1 = 3\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(B\), \(y = 0\): \(4 - e^{2x} = 0\) | M1 | Or reverse argument |
| \(e^{2x} = 4 \Rightarrow x = \ln 2\) | A1 | AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = -2e^{2x}\) | B1 | |
| \(x = \ln 2\), Gradient \(= -2e^{2\ln 2} = -8\) | M1 | \(x = \ln 2\) into \(ke^{2x}\) |
| Gradient of normal \(= \frac{1}{8} = \frac{1}{2e^{2\ln 2}}\) | A1 | OE |
| Equation \(y = \frac{1}{8}x - \frac{1}{8}\ln 2\) | A1 | OE |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(x = 0\): \(y = -\frac{1}{8}\ln 2\) | M1 | Attempt to integrate their line and substitute \(x = 0,\ \ln 2\) |
| Area \(\Delta = \frac{1}{16}(\ln 2)^2\) condone \(-\)ve sign \(= 0.03\) | A1\(\checkmark\) | \(\frac{1}{2}(\text{their } y) \times \ln 2\) |
| Total area \(= 4\ln 2 - \frac{3}{2} + \frac{1}{16}(\ln 2)^2 = 1.30\) | A1 | CSO |
| Total: 3 |
# Question 9:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int(4 - e^{2x})\,dx$ | | |
| $= 4x - \frac{1}{2}e^{2x}\ (+c)$ | B1 | $4x$ |
| | B1 | $-\frac{1}{2}e^{2x}$ |
| **Total: 2** | | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^{\ln 2} = \left[4x - \frac{1}{2}e^{2x}\right]_0^{\ln 2}$ | | |
| $= \left[4\ln 2 - \frac{1}{2}e^{2\ln 2}\right] - \left[(0) - \frac{1}{2}(e^0)\right]$ | M1 | Substitute both $\ln 2$ and $0$ correctly into an integrated expression |
| $= 4\ln 2 - 2 + \frac{1}{2}$ | | Convincing |
| $= 4\ln 2 - \frac{3}{2}$ | A1 | **AG** |
| **Total: 2** | | |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 0,\ y = 4 - 1 = 3$ | B1 | |
| **Total: 1** | | |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $B$, $y = 0$: $4 - e^{2x} = 0$ | M1 | Or reverse argument |
| $e^{2x} = 4 \Rightarrow x = \ln 2$ | A1 | **AG** |
| **Total: 2** | | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = -2e^{2x}$ | B1 | |
| $x = \ln 2$, Gradient $= -2e^{2\ln 2} = -8$ | M1 | $x = \ln 2$ into $ke^{2x}$ |
| Gradient of normal $= \frac{1}{8} = \frac{1}{2e^{2\ln 2}}$ | A1 | OE |
| Equation $y = \frac{1}{8}x - \frac{1}{8}\ln 2$ | A1 | OE |
| **Total: 4** | | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $x = 0$: $y = -\frac{1}{8}\ln 2$ | M1 | Attempt to integrate their line and substitute $x = 0,\ \ln 2$ |
| Area $\Delta = \frac{1}{16}(\ln 2)^2$ condone $-$ve sign $= 0.03$ | A1$\checkmark$ | $\frac{1}{2}(\text{their } y) \times \ln 2$ |
| Total area $= 4\ln 2 - \frac{3}{2} + \frac{1}{16}(\ln 2)^2 = 1.30$ | A1 | CSO |
| **Total: 3** | | |9 The sketch shows the graph of $y = 4 - \mathrm { e } ^ { 2 x }$. The curve crosses the $y$-axis at the point $A$ and the $x$-axis at the point $B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-5_711_921_466_557}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( 4 - \mathrm { e } ^ { 2 x } \right) \mathrm { d } x$.\\
(2 marks)
\item Hence show that $\int _ { 0 } ^ { \ln 2 } \left( 4 - \mathrm { e } ^ { 2 x } \right) \mathrm { d } x = 4 \ln 2 - \frac { 3 } { 2 }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down the $y$-coordinate of $A$.
\item Show that $x = \ln 2$ at $B$.
\end{enumerate}\item Find the equation of the normal to the curve $y = 4 - \mathrm { e } ^ { 2 x }$ at the point $B$.
\item Find the area of the region enclosed by the curve $y = 4 - \mathrm { e } ^ { 2 x }$, the normal to the curve at $B$ and the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2007 Q9 [14]}}