| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Show definite integral equals value |
| Difficulty | Moderate -0.8 This is a straightforward C2 integration question requiring only basic power rule integration and substitution of limits. The algebra is slightly more involved than the most routine questions due to the fractional coefficients and the need to express the answer in surd form, but it requires no problem-solving insight—just mechanical application of standard techniques. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int\left(\frac{x^3}{6} + \frac{1}{3x^2}\right)dx = \frac{x^4}{6(4)} + \frac{x^{-1}}{(3)(-1)}\) | M1 A1 A1 | M1: \(x^n \to x^{n+1}\). 1st A1: At least one of \(\frac{x^4}{6(4)}\) or \(\frac{x^{-1}}{(3)(-1)}\). 2nd A1: Both terms correct or equivalent |
| \(\left[\frac{(\sqrt{3})^4}{24} + \frac{(\sqrt{3})^{-1}}{-1(3)}\right] - \left[\frac{(1)^4}{24} + \frac{(1)^{-1}}{-1(3)}\right]\) | dM1 | Using limits \(\sqrt{3}\) and \(1\) on integrated expression, subtracting correct way. Dependent on 1st M1 |
| \(= \left(\frac{9}{24} - \frac{1}{3\sqrt{3}}\right) - \left(\frac{1}{24} - \frac{1}{3}\right) = \frac{2}{3} - \frac{1}{9}\sqrt{3}\) | A1 cso | \(a = \frac{2}{3}\) and \(b = -\frac{1}{9}\). Allow equivalent fractions but do not allow \(\frac{6-\sqrt{3}}{9}\). cao and cso — no previous errors |
# Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int\left(\frac{x^3}{6} + \frac{1}{3x^2}\right)dx = \frac{x^4}{6(4)} + \frac{x^{-1}}{(3)(-1)}$ | M1 A1 A1 | M1: $x^n \to x^{n+1}$. 1st A1: At least one of $\frac{x^4}{6(4)}$ or $\frac{x^{-1}}{(3)(-1)}$. 2nd A1: Both terms correct or equivalent |
| $\left[\frac{(\sqrt{3})^4}{24} + \frac{(\sqrt{3})^{-1}}{-1(3)}\right] - \left[\frac{(1)^4}{24} + \frac{(1)^{-1}}{-1(3)}\right]$ | dM1 | Using limits $\sqrt{3}$ and $1$ on integrated expression, subtracting correct way. Dependent on 1st M1 |
| $= \left(\frac{9}{24} - \frac{1}{3\sqrt{3}}\right) - \left(\frac{1}{24} - \frac{1}{3}\right) = \frac{2}{3} - \frac{1}{9}\sqrt{3}$ | A1 cso | $a = \frac{2}{3}$ and $b = -\frac{1}{9}$. Allow equivalent fractions but do **not** allow $\frac{6-\sqrt{3}}{9}$. cao and cso — no previous errors |\begin{enumerate}
\item Use integration to find
\end{enumerate}
$$\int _ { 1 } ^ { \sqrt { 3 } } \left( \frac { x ^ { 3 } } { 6 } + \frac { 1 } { 3 x ^ { 2 } } \right) \mathrm { d } x$$
giving your answer in the form $a + b \sqrt { 3 }$, where $a$ and $b$ are constants to be determined.\\
\hfill \mbox{\textit{Edexcel C2 2014 Q4 [5]}}