| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Trapezium rule estimation |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing routine application of the trapezium rule (mechanical calculation with given ordinates) and standard integration of power functions with fractional indices. Both parts require only direct recall and application of formulas with no problem-solving or insight needed, making it easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(h = 1\) | B1 | PI |
| \(I \approx \frac{h}{2}\{f(1)+f(5)+2[f(2)+f(3)+f(4)]\}\) | M1 | OE summing of areas of four trapezoids |
| \(\frac{h}{2}\{\ldots\} = \frac{1}{2}+\frac{1}{26}+2\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{17}\right)\) | A1 | OE Accept 2dp (rounded or truncated) for non-terminating decimals |
| \(I \approx 0.628054\ldots = \frac{694}{1105} = 0.628\) (to 3sf) | A1 | CAO Must be 0.628. SC for 5 strips, max possible B0M1A1A0 |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\int\left(x^{-\frac{3}{2}}+6x^{\frac{1}{2}}\right)dx = \frac{x^{-\frac{1}{2}}}{-1/2}+\frac{6x^{\frac{3}{2}}}{3/2}\) \((+c)\) | M1, A1 | M1: one term correct; A1: both terms correct (even unsimplified) |
| \(= -2x^{-0.5}+4x^{1.5}\) \((+c)\) | A1 | Must be simplified |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\int_1^4\left(x^{-\frac{3}{2}}+6x^{\frac{1}{2}}\right)dx\) | ||
| \(= [-2(4^{-0.5})+4(4^{1.5})]-[-2(1^{-0.5})+4(1^{1.5})]\) | M1 | Attempt to calculate \(F(4)-F(1)\) where \(F(x)\) follows integration and is not just the integrand |
| \(= (-1+32)-(-2+4) = 29\) | A1 | Since 'Hence' NMS scores 0/2 |
| Total: 2 marks |
# Question 2:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $h = 1$ | B1 | PI |
| $I \approx \frac{h}{2}\{f(1)+f(5)+2[f(2)+f(3)+f(4)]\}$ | M1 | OE summing of areas of four trapezoids |
| $\frac{h}{2}\{\ldots\} = \frac{1}{2}+\frac{1}{26}+2\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{17}\right)$ | A1 | OE Accept 2dp (rounded or truncated) for non-terminating decimals |
| $I \approx 0.628054\ldots = \frac{694}{1105} = 0.628$ (to 3sf) | A1 | CAO Must be 0.628. SC for 5 strips, max possible B0M1A1A0 |
| **Total: 4 marks** | | |
## Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\int\left(x^{-\frac{3}{2}}+6x^{\frac{1}{2}}\right)dx = \frac{x^{-\frac{1}{2}}}{-1/2}+\frac{6x^{\frac{3}{2}}}{3/2}$ $(+c)$ | M1, A1 | M1: one term correct; A1: both terms correct (even unsimplified) |
| $= -2x^{-0.5}+4x^{1.5}$ $(+c)$ | A1 | Must be simplified |
| **Total: 3 marks** | | |
## Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\int_1^4\left(x^{-\frac{3}{2}}+6x^{\frac{1}{2}}\right)dx$ | | |
| $= [-2(4^{-0.5})+4(4^{1.5})]-[-2(1^{-0.5})+4(1^{1.5})]$ | M1 | Attempt to calculate $F(4)-F(1)$ where $F(x)$ follows integration and is not just the integrand |
| $= (-1+32)-(-2+4) = 29$ | A1 | Since 'Hence' NMS scores 0/2 |
| **Total: 2 marks** | | |
---2
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with five ordinates (four strips) to find an approximate value for
$$\int _ { 1 } ^ { 5 } \frac { 1 } { x ^ { 2 } + 1 } \mathrm {~d} x$$
giving your answer to three significant figures.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( x ^ { - \frac { 3 } { 2 } } + 6 x ^ { \frac { 1 } { 2 } } \right) \mathrm { d } x$, giving the coefficient of each term in its simplest form.
\item Hence find the value of $\int _ { 1 } ^ { 4 } \left( x ^ { - \frac { 3 } { 2 } } + 6 x ^ { \frac { 1 } { 2 } } \right) \mathrm { d } x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q2 [9]}}