| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Exact area with surds |
| Difficulty | Moderate -0.8 This is a straightforward AS-level integration question requiring finding intersection points by solving a simple quadratic equation, then integrating a basic polynomial. The 'show that' format and exact answer requirement add minimal difficulty since the algebra is routine and the integration of 4x² + 3 is elementary. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Upper limit is \(\sqrt{5}\) | B1 | Score when seen as solution \(x = \sqrt{5}\) |
| \(\int 4x^2 + 3 \, dx = \frac{4}{3}x^3 + 3x\) | M1, A1 | Attempts to integrate \(4x^2 + 3\) or \(\pm(23-(4x^2+3))\); look for \(x^n \to x^{n+1}\); correct integration, ignore \(+c\) |
| \(23\sqrt{5} - \left[\frac{4}{3}x^3 + 3x\right]_0^{\sqrt{5}} = \ldots\) or \(\left[20x - \frac{4}{3}x^3\right]_0^{\sqrt{5}} = \ldots\) | M1 | Full method for area of \(R\); upper limit from solving \(4x^2+3=23\); lower limit should be 0 |
| Area \(R = \frac{40}{3}\sqrt{5}\) | A1 | Following correct algebraic integration; if curve\(-\)line used, must make \(-\frac{40}{3}\sqrt{5}\) positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Limits 3 and 23 | B1 | Must be clear attempt to integrate w.r.t. \(y\) |
| Rearrange to \(x=\) and integrate \(\sqrt{\frac{y-3}{4}}\) | M1 | Condone slips on rearrangement; look for \(\ldots(y\pm3)^{\frac{1}{2}} \to \ldots(y\pm3)^{\frac{3}{2}}\) |
| \(\int \frac{(y-3)^{\frac{1}{2}}}{2}\{dy\} = \frac{1}{3}(y-3)^{\frac{3}{2}}\) | A1 | Correct integration; ignore \(+c\) |
| Substitute limits 23 and 3 | M1 | Full method including substitution of limits |
| \(\frac{40}{3}\sqrt{5}\) | A1 | Following correct algebraic integration |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Upper limit is $\sqrt{5}$ | B1 | Score when seen as solution $x = \sqrt{5}$ |
| $\int 4x^2 + 3 \, dx = \frac{4}{3}x^3 + 3x$ | M1, A1 | Attempts to integrate $4x^2 + 3$ **or** $\pm(23-(4x^2+3))$; look for $x^n \to x^{n+1}$; correct integration, ignore $+c$ |
| $23\sqrt{5} - \left[\frac{4}{3}x^3 + 3x\right]_0^{\sqrt{5}} = \ldots$ **or** $\left[20x - \frac{4}{3}x^3\right]_0^{\sqrt{5}} = \ldots$ | M1 | Full method for area of $R$; upper limit from solving $4x^2+3=23$; lower limit should be 0 |
| Area $R = \frac{40}{3}\sqrt{5}$ | A1 | Following correct algebraic integration; if curve$-$line used, must make $-\frac{40}{3}\sqrt{5}$ positive |
**Alternative using $\int x \, dy$:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Limits 3 and 23 | B1 | Must be clear attempt to integrate w.r.t. $y$ |
| Rearrange to $x=$ and integrate $\sqrt{\frac{y-3}{4}}$ | M1 | Condone slips on rearrangement; look for $\ldots(y\pm3)^{\frac{1}{2}} \to \ldots(y\pm3)^{\frac{3}{2}}$ |
| $\int \frac{(y-3)^{\frac{1}{2}}}{2}\{dy\} = \frac{1}{3}(y-3)^{\frac{3}{2}}$ | A1 | Correct integration; ignore $+c$ |
| Substitute limits 23 and 3 | M1 | Full method including substitution of limits |
| $\frac{40}{3}\sqrt{5}$ | A1 | Following correct algebraic integration |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ce4f8375-0d88-4e48-85de-35f7e90b014d-10_488_519_365_772}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The finite region $R$, shown shaded in Figure 2, is bounded by the curve with equation $y = 4 x ^ { 2 } + 3$, the $y$-axis and the line with equation $y = 23$
Show that the exact area of $R$ is $k \sqrt { 5 }$ where $k$ is a rational constant to be found.
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q5 [5]}}