| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Standard +0.3 This is a straightforward multi-part C4 integration question combining standard techniques: completing a table (simple substitution), trapezium rule (routine application of formula), substitution (guided step-by-step), and integration by parts (standard te^t type). All parts are scaffolded with no novel insight required, making it slightly easier than average. |
| Spec | 1.08h Integration by substitution1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(y\) | \(\mathrm { e } ^ { 1 }\) | \(\mathrm { e } ^ { 2 }\) | \(\mathrm { e } ^ { 4 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\): 0, 1, 2, 3, 4, 5; \(y\): \(e^1\), \(e^2\), \(e^{\sqrt{7}}\), \(e^{\sqrt{10}}\), \(e^{\sqrt{13}}\), \(e^4\); or \(y\): 2.71828…, 7.38906…, 14.09403…, 23.62434…, 36.80197…, 54.59815… | B1, B1 | Either \(e^{\sqrt{7}}\), \(e^{\sqrt{10}}\) and \(e^{\sqrt{13}}\) or awrt 14.1, 23.6 and 36.8 or e to the power awrt 2.65, 3.16, 3.61 (or mixture of decimals and e's); At least two correct / All three correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I \approx \frac{1}{2} \times 1 \times \left\{e^1 + 2\left(e^2 + e^{\sqrt{7}} + e^{\sqrt{10}} + e^{\sqrt{13}}\right) + e^4\right\}\) | B1, M1\(\checkmark\) | Outside brackets \(\frac{1}{2} \times 1\); For structure of trapezium rule \(\{\ldots\}\) |
| \(= \frac{1}{2} \times 221.1352227\ldots = 110.5676113\ldots = \underline{110.6}\) (4sf) | A1 cao | 110.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = (3x+1)^{\frac{1}{2}} \Rightarrow \frac{dt}{dx} = \frac{1}{2}\cdot 3\cdot(3x+1)^{-\frac{1}{2}}\) or \(t^2 = 3x+1 \Rightarrow 2t\frac{dt}{dx} = 3\) | M1, A1 | \(A(3x+1)^{-\frac{1}{2}}\) or \(t\frac{dt}{dx} = A\); \(\frac{3}{2}(3x+1)^{-\frac{1}{2}}\) or \(2t\frac{dt}{dx} = 3\) |
| \(\frac{dt}{dx} = \frac{3}{2\cdot(3x+1)^{\frac{1}{2}}} = \frac{3}{2t} \Rightarrow \frac{dx}{dt} = \frac{2t}{3}\) | — | Candidate obtains either \(\frac{dt}{dx}\) or \(\frac{dx}{dt}\) in terms of \(t\) |
| \(I = \int e^{\sqrt{(3x+1)}}\,dx = \int e^t \frac{dx}{dt}\,dt = \int e^t \cdot \frac{2t}{3}\,dt\) | dM1 | Moves on to substitute into \(I\) to convert integral wrt \(x\) to integral wrt \(t\) |
| \(I = \int \frac{2}{3}te^t\,dt\) | A1 | \(\int \frac{2}{3}te^t\) |
| When \(x=0\), \(t=1\) and when \(x=5\), \(t=4\) | B1 | Changes limits \(x \to t\) so that \(0 \to 1\) and \(5 \to 4\) |
| Hence \(I = \int_1^4 \frac{2}{3}te^t\,dt\); where \(a=1\), \(b=4\), \(k=\frac{2}{3}\) | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = t \Rightarrow \frac{du}{dt} = 1\); \(\frac{dv}{dt} = e^t \Rightarrow v = e^t\) | — | Let \(k\) be any constant for first three marks |
| \(k\int te^t\,dt = k\left(te^t - \int e^t \cdot 1\,dt\right)\) | M1, A1 | Use of integration by parts formula in correct direction; Correct expression with constant factor \(k\) |
| \(= k\left(te^t - e^t\right) + c\) | A1 | Correct integration with/without constant factor \(k\) |
| \(\int_1^4 \frac{2}{3}te^t\,dt = \frac{2}{3}\left\{(4e^4 - e^4)-(e^1 - e^1)\right\}\) | dM1 oe | Substitutes changed limits into integrand and subtracts |
| \(= \frac{2}{3}(3e^4) = \underline{2e^4} = 109.1963\ldots\) | A1 | Either \(2e^4\) or awrt \(109.2\) |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x$: 0, 1, 2, 3, 4, 5; $y$: $e^1$, $e^2$, $e^{\sqrt{7}}$, $e^{\sqrt{10}}$, $e^{\sqrt{13}}$, $e^4$; or $y$: 2.71828…, 7.38906…, 14.09403…, 23.62434…, 36.80197…, 54.59815… | B1, B1 | Either $e^{\sqrt{7}}$, $e^{\sqrt{10}}$ and $e^{\sqrt{13}}$ or awrt 14.1, 23.6 and 36.8 or e to the power awrt 2.65, 3.16, 3.61 (or mixture of decimals and e's); At least two correct / All three correct |
**[2]**
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## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I \approx \frac{1}{2} \times 1 \times \left\{e^1 + 2\left(e^2 + e^{\sqrt{7}} + e^{\sqrt{10}} + e^{\sqrt{13}}\right) + e^4\right\}$ | B1, M1$\checkmark$ | Outside brackets $\frac{1}{2} \times 1$; For structure of trapezium rule $\{\ldots\}$ |
| $= \frac{1}{2} \times 221.1352227\ldots = 110.5676113\ldots = \underline{110.6}$ (4sf) | A1 cao | 110.6 |
**[3]**
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## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = (3x+1)^{\frac{1}{2}} \Rightarrow \frac{dt}{dx} = \frac{1}{2}\cdot 3\cdot(3x+1)^{-\frac{1}{2}}$ or $t^2 = 3x+1 \Rightarrow 2t\frac{dt}{dx} = 3$ | M1, A1 | $A(3x+1)^{-\frac{1}{2}}$ or $t\frac{dt}{dx} = A$; $\frac{3}{2}(3x+1)^{-\frac{1}{2}}$ or $2t\frac{dt}{dx} = 3$ |
| $\frac{dt}{dx} = \frac{3}{2\cdot(3x+1)^{\frac{1}{2}}} = \frac{3}{2t} \Rightarrow \frac{dx}{dt} = \frac{2t}{3}$ | — | Candidate obtains either $\frac{dt}{dx}$ or $\frac{dx}{dt}$ in terms of $t$ |
| $I = \int e^{\sqrt{(3x+1)}}\,dx = \int e^t \frac{dx}{dt}\,dt = \int e^t \cdot \frac{2t}{3}\,dt$ | dM1 | Moves on to substitute into $I$ to convert integral wrt $x$ to integral wrt $t$ |
| $I = \int \frac{2}{3}te^t\,dt$ | A1 | $\int \frac{2}{3}te^t$ |
| When $x=0$, $t=1$ and when $x=5$, $t=4$ | B1 | Changes limits $x \to t$ so that $0 \to 1$ and $5 \to 4$ |
| Hence $I = \int_1^4 \frac{2}{3}te^t\,dt$; where $a=1$, $b=4$, $k=\frac{2}{3}$ | — | — |
**[5]**
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## Question 8(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = t \Rightarrow \frac{du}{dt} = 1$; $\frac{dv}{dt} = e^t \Rightarrow v = e^t$ | — | Let $k$ be any constant for first three marks |
| $k\int te^t\,dt = k\left(te^t - \int e^t \cdot 1\,dt\right)$ | M1, A1 | Use of integration by parts formula in correct direction; Correct expression with constant factor $k$ |
| $= k\left(te^t - e^t\right) + c$ | A1 | Correct integration with/without constant factor $k$ |
| $\int_1^4 \frac{2}{3}te^t\,dt = \frac{2}{3}\left\{(4e^4 - e^4)-(e^1 - e^1)\right\}$ | dM1 oe | Substitutes changed limits into integrand and subtracts |
| $= \frac{2}{3}(3e^4) = \underline{2e^4} = 109.1963\ldots$ | A1 | Either $2e^4$ or awrt $109.2$ |
**[5] [15 marks total]**8.
$$I = \int _ { 0 } ^ { 5 } \mathrm { e } ^ { \sqrt { } ( 3 x + 1 ) } \mathrm { d } x$$
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \mathrm { e } ^ { \sqrt { } ( 3 x + 1 ) }$, complete the table with the values of $y$ corresponding to $x = 2$, 3 and 4.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$y$ & $\mathrm { e } ^ { 1 }$ & $\mathrm { e } ^ { 2 }$ & & & & $\mathrm { e } ^ { 4 }$ \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate for the original integral $I$, giving your answer to 4 significant figures.
\item Use the substitution $t = \sqrt { } ( 3 x + 1 )$ to show that $I$ may be expressed as $\int _ { a } ^ { b } k t e ^ { t } \mathrm {~d} t$, giving the values of $a , b$ and $k$.
\item Use integration by parts to evaluate this integral, and hence find the value of $I$ correct to 4 significant figures, showing all the steps in your working.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q8 [15]}}