| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a standard A-level integration question requiring finding a tangent equation, determining intersection points, and calculating area between curve and line. While it involves multiple steps (differentiation, tangent equation, solving for intersection, integration with limits), each step uses routine techniques with no novel insight required. The 'show that' format provides a target to verify, making it slightly easier than average. |
| 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 |
| \(\frac{dy}{dx} = \frac{15}{2}x^{\frac{1}{2}} - 9\) | M1, A1 | Differentiates \(5x^{\frac{3}{2}} - 9x + 11\) to form \(Ax^{\frac{1}{2}} + B\); answer may not be simplified |
| Substitutes \(x = 4 \Rightarrow \frac{dy}{dx} = 6\) | M1 | Substitutes \(x=4\) in their \(\frac{dy}{dx}\) to find gradient of tangent |
| Uses \((4, 15)\) and gradient \(\Rightarrow y - 15 = 6(x-4)\) | M1 | Uses their gradient and the point \((4,15)\) |
| Equation of \(l\) is \(y = 6x - 9\) | A1 | |
| Area \(R = \int_0^4 \left(5x^{\frac{3}{2}} - 9x + 11\right) - (6x-9)\,dx\) | M1 | Uses Area \(R = \int_0^4 \left(5x^{\frac{3}{2}}-9x+11\right)-(6x-9)\,dx\) following through on their \(y=6x-9\) |
| \(= \left[2x^{\frac{5}{2}} - \frac{15}{2}x^2 + 20x\right]_0^4\) | A1 | Must be correct; look for form \(Ax^{\frac{5}{2}} + Bx^2 + Cx\) |
| Uses both limits of 4 and 0: \(2\times 4^{\frac{5}{2}} - \frac{15}{2}\times 4^2 + 20\times 4 - 0\) | M1 | Substitutes in both limits and subtracts |
| Area of \(R = 24\) | A1* | Correct area (given answer) |
| Correct notation with good explanations | A1 | Correct notation used consistently for differentiation and integration; correct explanation for equation of \(l\) and area of \(R\) |
## Question 15:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{15}{2}x^{\frac{1}{2}} - 9$ | M1, A1 | Differentiates $5x^{\frac{3}{2}} - 9x + 11$ to form $Ax^{\frac{1}{2}} + B$; answer may not be simplified |
| Substitutes $x = 4 \Rightarrow \frac{dy}{dx} = 6$ | M1 | Substitutes $x=4$ in their $\frac{dy}{dx}$ to find gradient of tangent |
| Uses $(4, 15)$ and gradient $\Rightarrow y - 15 = 6(x-4)$ | M1 | Uses their gradient and the point $(4,15)$ |
| Equation of $l$ is $y = 6x - 9$ | A1 | |
| Area $R = \int_0^4 \left(5x^{\frac{3}{2}} - 9x + 11\right) - (6x-9)\,dx$ | M1 | Uses Area $R = \int_0^4 \left(5x^{\frac{3}{2}}-9x+11\right)-(6x-9)\,dx$ following through on their $y=6x-9$ |
| $= \left[2x^{\frac{5}{2}} - \frac{15}{2}x^2 + 20x\right]_0^4$ | A1 | Must be correct; look for form $Ax^{\frac{5}{2}} + Bx^2 + Cx$ |
| Uses both limits of 4 and 0: $2\times 4^{\frac{5}{2}} - \frac{15}{2}\times 4^2 + 20\times 4 - 0$ | M1 | Substitutes in both limits and subtracts |
| Area of $R = 24$ | A1* | Correct area (given answer) |
| Correct notation with good explanations | A1 | Correct notation used consistently for differentiation and integration; correct explanation for equation of $l$ and area of $R$ |
---15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a75c9ef7-b648-47be-bad1-fc8b315be3df-22_796_974_244_548}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve $C$ with equation
$$y = 5 x ^ { \frac { 3 } { 2 } } - 9 x + 11 , x \geqslant 0$$
The point $P$ with coordinates $( 4,15 )$ lies on $C$.\\
The line $l$ is the tangent to $C$ at the point $P$.\\
The region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the line $l$ and the $y$-axis.
Show that the area of $R$ is 24 , making your method clear.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel Paper 2 Q15 [10]}}