| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Show definite integral equals value |
| Difficulty | Moderate -0.8 This is a highly scaffolded C2 integration question with four parts that guide students through each step: converting to index form, expanding, integrating using the power rule, and evaluating limits. The techniques are routine and the structure removes any problem-solving challenge, making it easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{x} = x^{\frac{1}{2}}\) | B1 | Accept \(k = 0.5\) |
| Total for 4(a): 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{x}(x-1) = x^{\frac{1}{2}}x - x^{\frac{1}{2}} = x^{\frac{3}{2}} - x^{\frac{1}{2}}\) | M1, A1 | Accept \(p = 1.5\), \(q = 0.5\) |
| Total for 4(b): 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\sqrt{x}(x-1)dx = \frac{x^{2.5}}{2.5} - \frac{x^{1.5}}{1.5} (+c)\) | M1, A1√, A1√ | Increases a power of \(x\) by 1; ft non-integer \(p\); ft non-integer \(q\) |
| Total for 4(c): 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^2 dx = \left(\frac{2^{2.5}}{2.5} - \frac{2^{1.5}}{1.5}\right) - \left(\frac{1}{2.5} - \frac{1}{1.5}\right)\); \(.... = \left(\frac{4\sqrt{2}}{2.5} - \frac{2\sqrt{2}}{1.5}\right) - \left(\frac{4}{15}\right)\) or \(= \text{pr. ans}\) | M1, m1, A1 | Limits; F(2) − F(1); Fractional powers to surds; CSO AG (be convinced) |
| Total for 4(d): 3 marks | ||
| Total for Question 4: 9 marks |
**4(a)**
| $\sqrt{x} = x^{\frac{1}{2}}$ | B1 | Accept $k = 0.5$ |
| Total for 4(a): 1 mark |
**4(b)**
| $\sqrt{x}(x-1) = x^{\frac{1}{2}}x - x^{\frac{1}{2}} = x^{\frac{3}{2}} - x^{\frac{1}{2}}$ | M1, A1 | Accept $p = 1.5$, $q = 0.5$ |
| Total for 4(b): 2 marks |
**4(c)**
| $\int\sqrt{x}(x-1)dx = \frac{x^{2.5}}{2.5} - \frac{x^{1.5}}{1.5} (+c)$ | M1, A1√, A1√ | Increases a power of $x$ by 1; ft non-integer $p$; ft non-integer $q$ |
| Total for 4(c): 3 marks |
**4(d)**
| $\int_1^2 dx = \left(\frac{2^{2.5}}{2.5} - \frac{2^{1.5}}{1.5}\right) - \left(\frac{1}{2.5} - \frac{1}{1.5}\right)$; $.... = \left(\frac{4\sqrt{2}}{2.5} - \frac{2\sqrt{2}}{1.5}\right) - \left(\frac{4}{15}\right)$ or $= \text{pr. ans}$ | M1, m1, A1 | Limits; F(2) − F(1); Fractional powers to surds; CSO AG (be convinced) |
| Total for 4(d): 3 marks |
| **Total for Question 4: 9 marks** |
---4
\begin{enumerate}[label=(\alph*)]
\item Write $\sqrt { x }$ in the form $x ^ { k }$, where $k$ is a fraction.
\item Hence express $\sqrt { x } ( x - 1 )$ in the form $x ^ { p } - x ^ { q }$.
\item Find $\int \sqrt { x } ( x - 1 ) \mathrm { d } x$.
\item Hence show that $\int _ { 1 } ^ { 2 } \sqrt { x } ( x - 1 ) \mathrm { d } x = \frac { 4 } { 15 } ( \sqrt { 2 } + 1 )$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q4 [9]}}