| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Exact area with surds |
| Difficulty | Standard +0.3 This is a standard C3 integration question testing area under a curve and volume of revolution. Part (i) requires integrating a power function (rewriting as (3x+1)^{-1/2}) with chain rule reversal—routine for C3. Part (ii) applies the standard volume formula with another straightforward integration leading to a logarithm. Both parts are 'show that' questions with given answers, reducing problem-solving demand. Slightly easier than average due to predictable techniques and scaffolding. |
| Spec | 1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Integrate to obtain form \(k(3x+1)^{\frac{1}{2}}\) | *M1 | any non-zero constant \(k\) |
| Obtain \(4(3x+1)^{\frac{1}{2}}\) | A1 | or (unsimplified) equiv; or \(4u^{\frac{1}{2}}\) following substitution |
| Apply the limits and subtract the right way round | M1 | dep *M |
| Obtain \(4\sqrt{28} - 4\sqrt{7}\) and show at least one intermediate step in confirming \(4\sqrt{7}\) | A1 | AG; necessary detail required; decimal verification is A0; |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply volume is \(\int \pi \left(\frac{6}{\sqrt{3x+1}}\right)^2 dx\) or equiv | B1 | merely stating \(\int \pi y^2 dx\) not enough; condone absence of \(dx\); no need for limits yet; \(\pi\) may be implied by its later appearance |
| Integrate to obtain \(k \ln(3x+1)\) | M1 | any non-zero constant with or without \(\pi\) |
| Obtain \(12\pi \ln(3x+1)\) or \(12\ln(3x+1)\) | A1 | or unsimplified equiv |
| Substitute limits correct way round and show each logarithm property correctly applied | M1 | allowing correct applications to incorrect result of integration providing natural logarithm involved; evidence of \(\ln 28 - \ln 7 = \frac{\text{ln2}}{0.021}\) error means M0 |
| Obtain \(24\pi \ln 2\) | A1 | no need for explicit statement of value of \(k\) |
## (i)
Integrate to obtain form $k(3x+1)^{\frac{1}{2}}$ | *M1 | any non-zero constant $k$
Obtain $4(3x+1)^{\frac{1}{2}}$ | A1 | or (unsimplified) equiv; or $4u^{\frac{1}{2}}$ following substitution
Apply the limits and subtract the right way round | M1 | dep *M
Obtain $4\sqrt{28} - 4\sqrt{7}$ and show at least one intermediate step in confirming $4\sqrt{7}$ | A1 | AG; necessary detail required; decimal verification is A0;
$$\left[\cdots\right]_1^7 = 4\sqrt{28} - 4\sqrt{7} = 4\sqrt{7} \text{ is A0}; \quad \left[\cdots\right]_1^7 = 8\sqrt{7} - 4\sqrt{7} = 4\sqrt{7} \text{ is A0}$$
**Total: [4]**
## (ii)
State or imply volume is $\int \pi \left(\frac{6}{\sqrt{3x+1}}\right)^2 dx$ or equiv | B1 | merely stating $\int \pi y^2 dx$ not enough; condone absence of $dx$; no need for limits yet; $\pi$ may be implied by its later appearance
Integrate to obtain $k \ln(3x+1)$ | M1 | any non-zero constant with or without $\pi$
Obtain $12\pi \ln(3x+1)$ or $12\ln(3x+1)$ | A1 | or unsimplified equiv
Substitute limits correct way round and show each logarithm property correctly applied | M1 | allowing correct applications to incorrect result of integration providing natural logarithm involved; evidence of $\ln 28 - \ln 7 = \frac{\text{ln2}}{0.021}$ error means M0
Obtain $24\pi \ln 2$ | A1 | no need for explicit statement of value of $k$
**Total: [5]**
---\includegraphics{figure_5}
The diagram shows the curve $y = \frac{6}{\sqrt{3x + 1}}$. The shaded region is bounded by the curve and the lines $x = 2$, $x = 9$ and $y = 0$.
\begin{enumerate}[label=(\roman*)]
\item Show that the area of the shaded region is $4\sqrt{7}$ square units. [4]
\item The shaded region is rotated completely about the $x$-axis. Show that the volume of the solid produced can be written in the form $k\ln 2$, where the exact value of the constant $k$ is to be determined. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2013 Q5 [9]}}