| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with numerical methods |
| Difficulty | Standard +0.8 This question combines numerical methods (Simpson's rule with 6 strips) and volumes of revolution requiring algebraic manipulation to reach an exact form involving logarithms. Part (i) is routine C3 application, but part (ii) requires integrating x/(x+1) by algebraic division or substitution, then simplifying to the given exact answer—this is more demanding than typical C3 volume questions which usually have simpler integrands. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\): 0, 0.5, 1, 1.5, 2, 2.5, 3; \(y\): 0, 0.5774, 0.7071, 0.7746, 0.8165, 0.8452, 0.8660 | M1 A1 | |
| area \(\approx \frac{1}{3} \times 0.5 \times [0 + 0.8660 + 4(0.5774 + 0.7746 + 0.8452) + 2(0.7071 + 0.8165)]\) | M1 | |
| \(= 2.12\) (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \pi\int_0^3 \frac{x}{x+1}\, dx\) | M1 | |
| \(= \pi\int_0^3 \frac{x+1-1}{x+1}\, dx = \pi\int_0^3 \left(1 - \frac{1}{x+1}\right)\, dx\) | M1 | |
| \(= \pi[x - \ln | x+1 | ]_0^3\) |
| \(= \pi\{(3 - \ln 4) - (0)\} = \pi(3 - \ln 4)\) | M1 A1 | (10) |
# Question 6:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x$: 0, 0.5, 1, 1.5, 2, 2.5, 3; $y$: 0, 0.5774, 0.7071, 0.7746, 0.8165, 0.8452, 0.8660 | M1 A1 | |
| area $\approx \frac{1}{3} \times 0.5 \times [0 + 0.8660 + 4(0.5774 + 0.7746 + 0.8452) + 2(0.7071 + 0.8165)]$ | M1 | |
| $= 2.12$ (3sf) | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \pi\int_0^3 \frac{x}{x+1}\, dx$ | M1 | |
| $= \pi\int_0^3 \frac{x+1-1}{x+1}\, dx = \pi\int_0^3 \left(1 - \frac{1}{x+1}\right)\, dx$ | M1 | |
| $= \pi[x - \ln|x+1|]_0^3$ | M1 A1 | |
| $= \pi\{(3 - \ln 4) - (0)\} = \pi(3 - \ln 4)$ | M1 A1 | **(10)** |
---6.\\
\includegraphics[max width=\textwidth, alt={}, center]{687756c0-2038-4077-8c5c-fe0ca0f6ce65-2_444_825_1571_516}
The diagram shows the curve with equation $y = \sqrt { \frac { x } { x + 1 } }$.\\
The shaded region is bounded by the curve, the $x$-axis and the line $x = 3$.\\
(i) Use Simpson's rule with six strips to estimate the area of the shaded region.
The shaded region is rotated through four right angles about the $x$-axis.\\
(ii) Show that the volume of the solid formed is $\pi ( 3 - \ln 4 )$.\\
\hfill \mbox{\textit{OCR C3 Q6 [10]}}