| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Integration with given constant |
| Difficulty | Moderate -0.5 This is a straightforward C2 integration question requiring basic polynomial integration and solving a linear equation for k. While it involves a definite integral with limits, the technique is routine: integrate the polynomial, apply limits, and solve for k. Slightly easier than average due to being purely procedural with no conceptual challenges. |
| Spec | 1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^3 (x^2 - 2x + k) \, dx = \left[\frac{1}{3}x^3 - x^2 + kx\right]_1^3\) | M1 A2 | |
| \(= (9 - 9 + 3k) - \left(\frac{1}{3} - 1 + k\right) = 2k + \frac{2}{3}\) | M1 | |
| \(\therefore 2k + \frac{2}{3} = 8\frac{2}{3}, \quad k = 4\) | M1 A1 | (6 marks) |
$\int_1^3 (x^2 - 2x + k) \, dx = \left[\frac{1}{3}x^3 - x^2 + kx\right]_1^3$ | M1 A2 |
$= (9 - 9 + 3k) - \left(\frac{1}{3} - 1 + k\right) = 2k + \frac{2}{3}$ | M1 |
$\therefore 2k + \frac{2}{3} = 8\frac{2}{3}, \quad k = 4$ | M1 A1 | **(6 marks)**\begin{enumerate}
\item Given that
\end{enumerate}
$$\int _ { 1 } ^ { 3 } \left( x ^ { 2 } - 2 x + k \right) d x = 8 \frac { 2 } { 3 }$$
find the value of the constant $k$.\\
\hfill \mbox{\textit{Edexcel C2 Q2 [6]}}