| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Integration with given constant |
| Difficulty | Standard +0.3 This is a straightforward integration problem requiring evaluation of a definite integral with a parameter, then solving a quadratic equation. The integration is routine (power rule), and solving 2A² - A - 14 = 0 is standard. Slightly above average only because it involves algebraic manipulation with a parameter rather than pure numerical calculation. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(3x^{0.5}+A)\,dx = 2x^{1.5}+Ax(+c)\) | M1, A1 | Integrates achieving form \(kx^{1.5}+Ax(+c)\) where \(k\) is non-zero; correct answer (may be unsimplified) |
| Uses limits and sets \(= 2A^2 \Rightarrow (2\times8+4A)-(2\times1+A)=2A^2\) | M1 | Substitutes limits and subtracts; only if \(\int A\,dx = Ax\) (not \(\frac{A^2}{2}\)) |
| Sets up quadratic and attempts to solve | M1 | Sets up quadratic in \(A\) and either attempts to solve or attempts \(b^2-4ac\) |
| \(A = -2, \frac{7}{2}\) and states there are two roots | A1 | Either states \(A=-2,\frac{7}{2}\) and two roots exist, or states \(b^2-4ac=121>0\) hence two roots |
# Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(3x^{0.5}+A)\,dx = 2x^{1.5}+Ax(+c)$ | M1, A1 | Integrates achieving form $kx^{1.5}+Ax(+c)$ where $k$ is non-zero; correct answer (may be unsimplified) |
| Uses limits and sets $= 2A^2 \Rightarrow (2\times8+4A)-(2\times1+A)=2A^2$ | M1 | Substitutes limits and subtracts; only if $\int A\,dx = Ax$ (not $\frac{A^2}{2}$) |
| Sets up quadratic and attempts to solve | M1 | Sets up quadratic in $A$ and either attempts to solve or attempts $b^2-4ac$ |
| $A = -2, \frac{7}{2}$ **and** states there are two roots | A1 | Either states $A=-2,\frac{7}{2}$ and two roots exist, **or** states $b^2-4ac=121>0$ hence two roots |
---\begin{enumerate}
\item Given that $A$ is constant and
\end{enumerate}
$$\int _ { 1 } ^ { 4 } ( 3 \sqrt { x } + A ) \mathrm { d } x = 2 A ^ { 2 }$$
show that there are exactly two possible values for $A$.\\
\hfill \mbox{\textit{Edexcel Paper 2 Q9 [5]}}