| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.3 This is a straightforward multi-part integration question requiring standard techniques: verifying a root by substitution, finding a tangent equation using differentiation (including fractional powers), and calculating area between curves using definite integration. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8(4) - 4^{\frac{5}{2}} = 32 - 32 = 0\) | B1 | Substitutes \(x=4\) and verifies \(y=0\). Accept "\(8(4)-4^{\frac{5}{2}}=0\)". Alternatively sets \(8x - x^{\frac{5}{2}}=0\) and solves to get \(x=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8 - \frac{5}{2}x^{\frac{3}{2}}\) | B1 | Correct differentiation. \(\frac{dy}{dx}=\) need not be present |
| \(x=4 \Rightarrow \frac{dy}{dx} = 8 - \frac{5}{2}\times 8 = -12 \Rightarrow y\{-0\} = \text{``}-12\text{''}(x-4)\) | M1 | Correct method for equation of tangent at \(A(4,0)\). Requires substitution of \(x=4\) into their \(\frac{dy}{dx}\) and attempt at line equation. If \(y=mx+c\) used must reach \(c=\ldots\) |
| \(12x + y = 48\) * | A1* | Correct work leading to given equation having scored B1M1. Condone \(y+12x=48\). Do not condone \(12x+y-48=0\) unless correct equation seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts intersection: \(y=8x\), \(12x+y=48 \Rightarrow y=19.2\) (or \(x=2.4\)) | M1 | Attempts to find either \(x\) or \(y\) coordinate of intersection of \(l_1\) and \(l_2\) |
| Triangle area \(= \frac{1}{2}\times 4 \times \text{``}19.2\text{''} \left(= 38.4 \text{ or } \frac{192}{5}\right)\) or \(\int_0^{\text{``}2.4\text{''}} 8x\,dx + \int_{\text{``}2.4\text{''}}^{4} \text{``}(48-12x)\text{''}\,dx\) | dM1 | Correct method for area of triangle. Condone slips in rearrangement of \(12x+y=48\) to \(y=48-12x\) |
| \(\int\left(8x - x^{\frac{5}{2}}\right)dx = 4x^2 - \frac{2}{7}x^{\frac{7}{2}}\) | B1 | Correct integration of curve ignoring limits. Condone e.g. \(\frac{8x^2}{2} - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}\) |
| \(A = 38.4 - \left[``4x^2 - \frac{2}{7}x^{\frac{7}{2}}\text{''}\right]_0^4 = 38.4 - 64 + \frac{256}{7}\) | ddM1 | Fully correct strategy including substitution leading to exact area. Dependent on both previous M marks. Implied by \(38.4 - \frac{192}{7}\) or correct final answer \(\frac{384}{35}\) |
| \(= \frac{384}{35}\) | A1 | Correct exact value. Either \(\frac{384}{35}\) or \(10\frac{34}{35}\) |
## Question 10:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8(4) - 4^{\frac{5}{2}} = 32 - 32 = 0$ | B1 | Substitutes $x=4$ and verifies $y=0$. Accept "$8(4)-4^{\frac{5}{2}}=0$". Alternatively sets $8x - x^{\frac{5}{2}}=0$ and solves to get $x=4$ |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8 - \frac{5}{2}x^{\frac{3}{2}}$ | B1 | Correct differentiation. $\frac{dy}{dx}=$ need not be present |
| $x=4 \Rightarrow \frac{dy}{dx} = 8 - \frac{5}{2}\times 8 = -12 \Rightarrow y\{-0\} = \text{``}-12\text{''}(x-4)$ | M1 | Correct method for equation of tangent at $A(4,0)$. Requires substitution of $x=4$ into their $\frac{dy}{dx}$ and attempt at line equation. If $y=mx+c$ used must reach $c=\ldots$ |
| $12x + y = 48$ * | A1* | Correct work leading to given equation having scored B1M1. Condone $y+12x=48$. Do not condone $12x+y-48=0$ unless correct equation seen |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts intersection: $y=8x$, $12x+y=48 \Rightarrow y=19.2$ (or $x=2.4$) | M1 | Attempts to find either $x$ or $y$ coordinate of intersection of $l_1$ and $l_2$ |
| Triangle area $= \frac{1}{2}\times 4 \times \text{``}19.2\text{''} \left(= 38.4 \text{ or } \frac{192}{5}\right)$ or $\int_0^{\text{``}2.4\text{''}} 8x\,dx + \int_{\text{``}2.4\text{''}}^{4} \text{``}(48-12x)\text{''}\,dx$ | dM1 | Correct method for area of triangle. Condone slips in rearrangement of $12x+y=48$ to $y=48-12x$ |
| $\int\left(8x - x^{\frac{5}{2}}\right)dx = 4x^2 - \frac{2}{7}x^{\frac{7}{2}}$ | B1 | Correct integration of curve ignoring limits. Condone e.g. $\frac{8x^2}{2} - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}$ |
| $A = 38.4 - \left[``4x^2 - \frac{2}{7}x^{\frac{7}{2}}\text{''}\right]_0^4 = 38.4 - 64 + \frac{256}{7}$ | ddM1 | Fully correct strategy including substitution leading to **exact** area. Dependent on both previous M marks. Implied by $38.4 - \frac{192}{7}$ or correct final answer $\frac{384}{35}$ |
| $= \frac{384}{35}$ | A1 | Correct exact value. Either $\frac{384}{35}$ or $10\frac{34}{35}$ |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e116a86f-63e0-4e80-b49c-d9f3c819ce15-24_872_1285_246_392}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
In this question you must show all stages of your working.\\
Solutions relying entirely on calculator technology are not acceptable.\\
Figure 3 shows a sketch of part of the curve with equation
$$y = 8 x - x ^ { \frac { 5 } { 2 } } \quad x \geqslant 0$$
The curve crosses the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Verify that the $x$ coordinate of $A$ is 4
The line $l _ { 1 }$ is the tangent to the curve at $A$.
\item Use calculus to show that an equation of line $l _ { 1 }$ is
$$12 x + y = 48$$
The line $l _ { 2 }$ has equation $y = 8 x$\\
The region $R$, shown shaded in Figure 3, is bounded by the curve, the line $l _ { 1 }$ and the line $l _ { 2 }$
\item Use algebraic integration to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q10 [9]}}