| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area with parametric/substitution |
| Difficulty | Standard +0.3 This is a straightforward integration question with two standard parts: (a) applying the trapezium rule with given strips (routine calculation), and (b) using a suggested substitution u=√x to integrate e^√x. The substitution is explicitly provided and leads directly to a standard exponential integral. While it requires careful algebraic manipulation, both parts are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds \(y\) for \(x = 4, 5, 6, 7, 8, 9\): \(e^2, e^{\sqrt{5}}, e^{\sqrt{6}}, e^{\sqrt{7}}, e^{\sqrt{8}}, e^3\) i.e. \(7.389056..., 9.356469..., 11.582435..., 14.094030..., 16.918828..., 20.085536...\) | M1 | Needs six \(y\) values. Allow one slip. Must be accurate to 2 significant figures if decimals only |
| Outside brackets \(\frac{1}{2} \times 1\) or \(h = 1\) stated | B1 oe | Independent of method marks |
| \(\frac{1}{2} \times 1 \times \left\{e^2 + e^3 + 2\left(e^{\sqrt{5}} + e^{\sqrt{6}} + e^{\sqrt{7}} + e^{\sqrt{8}}\right)\right\}\) | M1 | Correct structure for \(\{\ldots\}\) using their \(y\) values; allow 5 or 6 \(y\) values |
| \(= 65.6890595... = 65.69\) (2 dp) | A1 | Wrong brackets e.g. \(\frac{1}{2}\times1\times(e^2+e^3)+2(\ldots)\) is M0 unless followed by correct answer 65.69 |
| Special case: \(h = \frac{5}{4}\), 5 ordinates giving 65.76 | — | Award M0B0M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = \sqrt{x} \Rightarrow \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\) or \(\frac{dx}{du} = 2u\) | B1 | States or uses either form |
| \(\int e^{\sqrt{x}}\, dx = \int e^u \cdot 2u\, du\) | M1 A1 | M1: obtains \(\pm\lambda\int ue^u\, du\) for constant \(\lambda\); A1: obtains \(2\int ue^u\, du\) |
| \(= \{2\}\left(ue^u - \int e^u\, du\right)\) | M1 | Attempt integration by parts in correct direction on \(\lambda u e^u\). Accept \(\int ue^u\, du = ue^u - \int e^u\, du\) |
| \(= \{2\}\left(ue^u - e^u\right)\) | A1 | \(\lambda ue^u \rightarrow \lambda ue^u - \lambda e^u\) |
| \(\left[2(ue^u - e^u)\right]_2^3 = 2(3e^3 - e^3) - 2(2e^2 - e^2)\) | ddM1 | Substitutes limits 3 and 2 in \(u\) (or 9 and 4 in \(x\)) into their integrand, correct way round. Depends on both previous M marks |
| \(4e^3 - 2e^2\) or \(2e^2(2e-1)\) etc. | A1 | Accept decimal equivalent |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds $y$ for $x = 4, 5, 6, 7, 8, 9$: $e^2, e^{\sqrt{5}}, e^{\sqrt{6}}, e^{\sqrt{7}}, e^{\sqrt{8}}, e^3$ i.e. $7.389056..., 9.356469..., 11.582435..., 14.094030..., 16.918828..., 20.085536...$ | M1 | Needs **six** $y$ values. Allow one slip. Must be accurate to 2 significant figures if decimals only |
| Outside brackets $\frac{1}{2} \times 1$ or $h = 1$ stated | B1 oe | Independent of method marks |
| $\frac{1}{2} \times 1 \times \left\{e^2 + e^3 + 2\left(e^{\sqrt{5}} + e^{\sqrt{6}} + e^{\sqrt{7}} + e^{\sqrt{8}}\right)\right\}$ | M1 | Correct structure for $\{\ldots\}$ using their $y$ values; allow 5 or 6 $y$ values |
| $= 65.6890595... = 65.69$ (2 dp) | A1 | Wrong brackets e.g. $\frac{1}{2}\times1\times(e^2+e^3)+2(\ldots)$ is M0 unless followed by correct answer 65.69 |
| Special case: $h = \frac{5}{4}$, 5 ordinates giving 65.76 | — | Award M0B0M1A1 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \sqrt{x} \Rightarrow \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ or $\frac{dx}{du} = 2u$ | B1 | States or uses either form |
| $\int e^{\sqrt{x}}\, dx = \int e^u \cdot 2u\, du$ | M1 A1 | M1: obtains $\pm\lambda\int ue^u\, du$ for constant $\lambda$; A1: obtains $2\int ue^u\, du$ |
| $= \{2\}\left(ue^u - \int e^u\, du\right)$ | M1 | Attempt integration by parts in correct direction on $\lambda u e^u$. Accept $\int ue^u\, du = ue^u - \int e^u\, du$ |
| $= \{2\}\left(ue^u - e^u\right)$ | A1 | $\lambda ue^u \rightarrow \lambda ue^u - \lambda e^u$ |
| $\left[2(ue^u - e^u)\right]_2^3 = 2(3e^3 - e^3) - 2(2e^2 - e^2)$ | ddM1 | Substitutes limits 3 and 2 in $u$ (or 9 and 4 in $x$) into **their** integrand, correct way round. Depends on both previous M marks |
| $4e^3 - 2e^2$ or $2e^2(2e-1)$ etc. | A1 | Accept decimal equivalent |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{423eb549-0873-4185-8faf-12dedafcd108-13_849_841_214_571}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \mathrm { e } ^ { \sqrt { x } } , x > 0$\\
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the lines $x = 4$ and $x = 9$
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule, with 5 strips of equal width, to obtain an estimate for the area of $R$, giving your answer to 2 decimal places.
\item Use the substitution $u = \sqrt { x }$ to find, by integrating, the exact value for the area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q9 [11]}}