| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.8 This is a multi-step AS-level integration problem requiring: finding the derivative, determining the normal equation, identifying intersection points, and computing area between curve and line using definite integration. The algebraic manipulation and setup are non-trivial but follow standard techniques, placing it moderately above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy: differentiate, use changed gradient for normal, find where normal cuts \(x\)-axis | M1 | 3.1a |
| \(y = \frac{32}{x^2} + 3x - 8 \Rightarrow \frac{dy}{dx} = -\frac{64}{x^3} + 3\) | M1, A1 | 1.1b, 1.1b |
| Substitute \(x=4\): \(\frac{dy}{dx} = -\frac{64}{64}+3 = 2\); use perpendicular gradient to find normal equation \(y - 6 = -\frac{1}{2}(x-4)\) | dM1 | 2.1 |
| Normal cuts the \(x\)-axis at \(x = 16\) | A1 | 1.1b |
| Complete strategy for finding two key areas: integrate between 2 and 4, find area of triangle | M1 | 3.1a |
| \(\int \frac{32}{x^2} + 3x - 8 \, dx = -\frac{32}{x} + \frac{3}{2}x^2 - 8x\) | M1, A1 | 1.1b, 1.1b |
| Area under curve \(= \left[-\frac{32}{x} + \frac{3}{2}x^2 - 8x\right]_2^4 = (-16)-(-26) = 10\) | dM1 | 1.1b |
| Total area \(= 10 + 36 = 46\) | A1* | 2.1 |
| Total | (10) |
## Question 15:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy: differentiate, use changed gradient for normal, find where normal cuts $x$-axis | M1 | 3.1a |
| $y = \frac{32}{x^2} + 3x - 8 \Rightarrow \frac{dy}{dx} = -\frac{64}{x^3} + 3$ | M1, A1 | 1.1b, 1.1b |
| Substitute $x=4$: $\frac{dy}{dx} = -\frac{64}{64}+3 = 2$; use perpendicular gradient to find normal equation $y - 6 = -\frac{1}{2}(x-4)$ | dM1 | 2.1 |
| Normal cuts the $x$-axis at $x = 16$ | A1 | 1.1b |
| Complete strategy for finding two key areas: integrate between 2 and 4, find area of triangle | M1 | 3.1a |
| $\int \frac{32}{x^2} + 3x - 8 \, dx = -\frac{32}{x} + \frac{3}{2}x^2 - 8x$ | M1, A1 | 1.1b, 1.1b |
| Area under curve $= \left[-\frac{32}{x} + \frac{3}{2}x^2 - 8x\right]_2^4 = (-16)-(-26) = 10$ | dM1 | 1.1b |
| Total area $= 10 + 36 = 46$ | A1* | 2.1 |
| **Total** | **(10)** | |
**Additional guidance for area section:**
- Alternative: integrate $\int_2^4 \left(-\frac{1}{2}x+8\right) - \left(\frac{32}{x^2}+3x-8\right)dx$
- A1: $\pm\left(-\frac{32}{x}+\frac{7}{4}x^2-16x\right)$ must be correct
- A1: Area $= 49 - 3 = 46$
- **Note:** Candidates who subtract area under curve from large triangle $= 56$ lose both strategy mark and answer mark15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f7935caa-6626-4ba8-87ef-e9bb59e1ac3e-44_595_977_242_536}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 32 } { x ^ { 2 } } + 3 x - 8 , \quad x > 0$$
The point $P ( 4,6 )$ lies on $C$.\\
The line $l$ is the normal to $C$ at the point $P$.\\
The region $R$, shown shaded in Figure 4, is bounded by the line $l$, the curve $C$, the line with equation $x = 2$ and the $x$-axis.
Show that the area of $R$ is 46\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q15 [10]}}