| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Integration inequality bounds |
| Difficulty | Standard +0.3 This is a multi-part question involving standard integration, numerical methods (rectangles and trapezium rule), and graphical work. While it has many parts (6 sub-questions), each individual component is routine C4 material: evaluating a polynomial integral, calculating rectangle areas, applying trapezium rule with different strip numbers, and making comparisons. The question is slightly easier than average because it's highly scaffolded with no conceptual surprises—students follow clear instructions through standard procedures without requiring novel insight or problem-solving. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^2 (2 + 2x - x^2)\,dx = \left[2x + x^2 - \frac{x^3}{3}\right]_0^2\) | M1 | |
| \(= \left(4 + 4 - \frac{8}{3}\right) = 5\frac{1}{3}\) | A1 | |
| Area is \(5.33\text{ m}^2\), to 3 s.f. | ||
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area estimate \(= 2 \times 0.25 \times 11.49571 = 5.747855\) | M1 A1 | A1 for 11.4... or equivalent |
| \(= 5.75\) to 3 d.p. | A1 | |
| This is an overestimate | B1 | |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch of both curves correct | B1 B1 | B1 for each curve |
| Labels correct | B1 | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| It is an underestimate, since the curve \(2+\sqrt{2x-x^2}\) is everywhere below the correct curve (except at \(x=1\)) | B1 B1 | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Trapezium rule values computed | M1 | For the values for the trapezium method |
| 2 strips: \((4 + 2\times3)\times\frac{1}{2} = 5\) | A1 | |
| 4 strips: \((4 + 2\times(3 + 5.732051))\times\frac{0.5}{2} = 5.3660255\) | A1 | |
| 8 strips: \((4 + 2\times(3 + 5.732051 + 11.25937))\times\frac{0.5}{2} = 5.49785525\) | A1 | |
| \(A\) is probably about \(5.5\) from the rate of convergence | B1 | |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The trapezium rule uses an approximation which follows the curve much more closely so it will be the better | B1 B1 | |
| Total: 2 marks |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^2 (2 + 2x - x^2)\,dx = \left[2x + x^2 - \frac{x^3}{3}\right]_0^2$ | M1 | |
| $= \left(4 + 4 - \frac{8}{3}\right) = 5\frac{1}{3}$ | A1 | |
| Area is $5.33\text{ m}^2$, to 3 s.f. | | |
| **Total: 2 marks** | | |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area estimate $= 2 \times 0.25 \times 11.49571 = 5.747855$ | M1 A1 | A1 for 11.4... or equivalent |
| $= 5.75$ to 3 d.p. | A1 | |
| This is an overestimate | B1 | |
| **Total: 4 marks** | | |
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch of both curves correct | B1 B1 | B1 for each curve |
| Labels correct | B1 | |
| **Total: 3 marks** | | |
## Question 8(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| It is an underestimate, since the curve $2+\sqrt{2x-x^2}$ is everywhere below the correct curve (except at $x=1$) | B1 B1 | |
| **Total: 2 marks** | | |
## Question 8(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Trapezium rule values computed | M1 | For the values for the trapezium method |
| 2 strips: $(4 + 2\times3)\times\frac{1}{2} = 5$ | A1 | |
| 4 strips: $(4 + 2\times(3 + 5.732051))\times\frac{0.5}{2} = 5.3660255$ | A1 | |
| 8 strips: $(4 + 2\times(3 + 5.732051 + 11.25937))\times\frac{0.5}{2} = 5.49785525$ | A1 | |
| $A$ is probably about $5.5$ from the rate of convergence | B1 | |
| **Total: 5 marks** | | |
## Question 8(vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The trapezium rule uses an approximation which follows the curve much more closely so it will be the better | B1 B1 | |
| **Total: 2 marks** | | |
---
8
\begin{enumerate}[label=(\roman*)]
\item Evaluate $A _ { 0 } = \int _ { 0 } ^ { 2 } \left( 2 + 2 x - x ^ { 2 } \right) \mathrm { d } x$.
Fig 8.1 illustrates the cross-section of a proposed tunnel. Lengths are in metres. The equation of the curved section is $y = 2 + \sqrt { 2 x - x ^ { 2 } }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23771896-942c-4a1d-ab95-6b6d3cc5643c-3_419_515_1155_836}
\captionsetup{labelformat=empty}
\caption{Fig. 8.1}
\end{center}
\end{figure}
The designers need to know the area of the cross-section, $A \mathrm {~m} ^ { 2 }$, so that they can work out the volume of the soil that will need to be removed when the tunnel is built.
\item An initial estimate, $A _ { 1 }$, is given by the area of the 8 rectangles shown in Fig 8.2. Calculate $A _ { 1 }$, and state whether it is an overestimate or underestimate for $A$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23771896-942c-4a1d-ab95-6b6d3cc5643c-3_520_645_2053_644}
\captionsetup{labelformat=empty}
\caption{Fig. 8.2}
\end{center}
\end{figure}
\item On graph paper, draw the graphs of
$$y = 2 + 2 x - x ^ { 2 } \text { and } y = 2 + \sqrt { 2 x - x ^ { 2 } } \text { for } 0 \leq x \leq 2 .$$
Make it clear which equation applies to which curve.
\item State whether $A _ { 0 }$, your answer to part (i), is an underestimate for $A$ or an overestimate. Give a reason for your answer.
\item The designers use the trapezium rule to estimate $A$. What values does this give when they take\\
(A) 2 strips,\\
(B) 4 strips,\\
(C) 8 strips?
What can you conclude about the value of $A$ ?
\item The best estimate from the trapezium rule is denoted by $A _ { 2 }$.
State, with a reason, whether the true value of $A$ is nearer $A _ { 1 }$ or $A _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 Q8 [18]}}