| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Limit of sum as integral |
| Difficulty | Challenging +1.2 This question tests understanding of the limit definition of integration and applying it to derive a volume formula. Part (a) is a direct translation of Riemann sum notation to integral form (straightforward). Part (b) requires recognizing this relates to a cone volume and completing the derivation using the integral result, which involves substitution and geometric insight beyond routine integration practice. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\displaystyle\int_0^h (h^2 - y^2)\ \mathrm{d}y\) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \dfrac{\pi r^2}{h^2}\displaystyle\int_0^h (h^2 - y^2)\ \mathrm{d}y\) | M1 | 2.2a |
| \(\dfrac{\pi r^2}{h^2}\left[h^2 y - \dfrac{y^3}{3}\right]_0^h\) | M1 | 1.1a |
| \(\dfrac{\pi r^2}{h^2}\left(h^3 - \dfrac{h^3}{3}\right) = \dfrac{2}{3}\pi r^2 h\) | A1 | 2.1 |
## Question 14:
### Part (a):
$\displaystyle\int_0^h (h^2 - y^2)\ \mathrm{d}y$ | B1 | 1.1 | Condone $\mathrm{d}y$ missing or as $\delta y$ but nothing else; condone integration started provided seen or used in 14(b) |
**[1 mark]**
### Part (b):
$V = \dfrac{\pi r^2}{h^2}\displaystyle\int_0^h (h^2 - y^2)\ \mathrm{d}y$ | M1 | 2.2a | Multiplying their integral from 14(a) by $\dfrac{\pi r^2}{h^2}$; may be at a later stage |
$\dfrac{\pi r^2}{h^2}\left[h^2 y - \dfrac{y^3}{3}\right]_0^h$ | M1 | 1.1a | Integration (one term correct; condone omission of $\dfrac{\pi r^2}{h^2}$) |
$\dfrac{\pi r^2}{h^2}\left(h^3 - \dfrac{h^3}{3}\right) = \dfrac{2}{3}\pi r^2 h$ | A1 | 2.1 | Convincing completion (answer given) |
**[3 marks]**
---14
\begin{enumerate}[label=(\alph*)]
\item Express $\lim _ { \delta y \rightarrow 0 } \sum _ { 0 } ^ { h } \left( h ^ { 2 } - y ^ { 2 } \right) \delta y$ as an integral.
\item Hence show that $V = \frac { 2 } { 3 } \pi r ^ { 2 } h$, as given in line 41 .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q14 [4]}}