| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Moderate -0.3 This is a straightforward volumes of revolution question requiring standard integration techniques. Part (i) involves integrating (4x+1)^(-1/2) using the chain rule in reverse, and part (ii) applies the standard formula V=π∫y²dx. Both are routine C3 exercises with no conceptual challenges beyond applying memorized formulas. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int_0^2 (4x + 1)^{-\frac{1}{2}} dx = \left[\frac{1}{3}(4x + 1)^{\frac{1}{2}}\right]_0^2 = \frac{1}{3}(3 - 1) = 1\) | M1, A1, M1, A1 | For integral of the form \(k(4x + 1)^{\frac{1}{2}}\); For correct indefinite integral; For correct use of limits; For given answer correctly shown |
| (ii) \(\pi\int_0^2 \frac{1}{4x + 1}dx = \pi\left[\frac{1}{4}\ln(4x + 1)\right]_0^2 = \frac{\pi}{4}\ln 9\) | M1, A1, M1, A1 | For integral of the form \(k\ln(4x + 1)\); For correct \(\frac{1}{4}\ln(4x + 1)\), with or without \(\pi\); Correct use of limits and \(\pi\); For correct (simplified) exact value |
**(i)** $\int_0^2 (4x + 1)^{-\frac{1}{2}} dx = \left[\frac{1}{3}(4x + 1)^{\frac{1}{2}}\right]_0^2 = \frac{1}{3}(3 - 1) = 1$ | M1, A1, M1, A1 | For integral of the form $k(4x + 1)^{\frac{1}{2}}$; For correct indefinite integral; For correct use of limits; For given answer correctly shown
**(ii)** $\pi\int_0^2 \frac{1}{4x + 1}dx = \pi\left[\frac{1}{4}\ln(4x + 1)\right]_0^2 = \frac{\pi}{4}\ln 9$ | M1, A1, M1, A1 | For integral of the form $k\ln(4x + 1)$; For correct $\frac{1}{4}\ln(4x + 1)$, with or without $\pi$; Correct use of limits and $\pi$; For correct (simplified) exact value
**Total for Question 4: 8 marks**
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-2_419_707_1576_660}
The diagram shows the curve
$$y = \frac { 1 } { \sqrt { } ( 4 x + 1 ) }$$
The region $R$ (shaded in the diagram) is enclosed by the curve, the axes and the line $x = 2$.\\
(i) Show that the exact area of $R$ is 1 .\\
(ii) The region $R$ is rotated completely about the $x$-axis. Find the exact volume of the solid formed.
\hfill \mbox{\textit{OCR C3 Q4 [8]}}