| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule with stated number of strips |
| Difficulty | Standard +0.3 This is a standard two-part integration question combining numerical approximation (trapezium rule with 4 strips) and exact integration using a given substitution. The trapezium rule is routine calculation, and the substitution u=2x-3 is explicitly provided, making the integration straightforward. Both parts are textbook exercises requiring careful arithmetic but no novel problem-solving or insight, placing this slightly easier than average. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Strip width \(= 0.5\) | B1 | Correct value stated or used within the formula |
| \(\frac{11\sqrt{5}}{5}+\frac{13}{3}+2\left(\frac{23\sqrt{6}}{12}+\frac{12\sqrt{7}}{7}+\frac{25\sqrt{2}}{8}\right)\) or \((4.91...+4.33...+2(4.69...+4.53...+4.41...))\) | M1 | Correct structure for their \(y\) values; must have \(y\) values starting at \(x=4\) and ending at \(x=6\) |
| Area \(\approx \frac{1}{2}\times\frac{1}{2}\left(\frac{11\sqrt{5}}{5}+\frac{13}{3}+2\left(\frac{23\sqrt{6}}{12}+\frac{12\sqrt{7}}{7}+\frac{25\sqrt{2}}{8}\right)\right)\) | A1 | Correct numerical expression (allow decimal values to 2sf) |
| \(9.14\) | A1 | 9.14 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u=2x-3 \Rightarrow \frac{du}{dx}=2\) | B1 | Correct derivative; accept \(du=2\,dx\) |
| \(\int\frac{x+7}{\sqrt{2x-3}}\,dx = \int\frac{\frac{u+3}{2}+7}{\sqrt{u}}\cdot\frac{1}{2}\,du\) | M1A1 | M1: Fully substitutes, replacing \(dx\) with \(du\) with no evidence of where \(du\) came from; A1: Fully correct expression |
| \(\frac{1}{4}\left(\frac{2}{3}u^{\frac{3}{2}}+34u^{\frac{1}{2}}\right)(+c)\) | A1 | Fully correct integration in any form (\(+c\) not required) |
| \(x=4, u=5\quad x=6, u=9\) | B1 | Correct \(u\) limits seen anywhere |
| \(\frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}+34u^{\frac{1}{2}}\right]_5^9 = \frac{1}{4}\left\{\left(\frac{2}{3}(9)^{\frac{3}{2}}+34(9)^{\frac{1}{2}}\right)-\left(\frac{2}{3}(5)^{\frac{3}{2}}+34(5)^{\frac{1}{2}}\right)\right\}\) | M1 | Substitutes \(u\) limits into changed function and subtracts either way round |
| \(=30-\frac{28}{3}\sqrt{5}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u^2=2x-3 \Rightarrow 2u\frac{du}{dx}=2\) | B1 | Correct derivative; accept \(2u=2\frac{dx}{du}\), \(dx=u\,du\) |
| \(\int\frac{x+7}{\sqrt{2x-3}}\,dx=\int\frac{\frac{u^2+3}{2}+7}{u}\cdot u\,du\) | M1A1 | M1: Fully substitutes; A1: Fully correct expression |
| \(\frac{u^3}{6}+\frac{17}{2}u(+c)\) | A1 | Fully correct integration in any form |
| \(x=4, u=\sqrt{5}\quad x=6, u=3\) | B1 | Correct \(u\) limits seen anywhere |
| \(\left[\frac{1}{6}u^3+\frac{17}{2}u\right]_{\sqrt{5}}^3=\left\{\left(\frac{27}{6}+\frac{17}{2}(3)\right)-\left(\frac{1}{6}(\sqrt{5})^3+\frac{17}{2}(\sqrt{5})\right)\right\}\) | M1 | Substitutes \(u\) limits and subtracts either way round |
| \(=30-\frac{28}{3}\sqrt{5}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(x+7)(2x-3)^{-\frac{1}{2}}\,dx\) | B1 | Uses \(\frac{x+7}{\sqrt{2x-3}}\) as \((x+7)(2x-3)^{-\frac{1}{2}}\) and makes some progress attempting integration |
| \(\int(x+7)(2x-3)^{-\frac{1}{2}}\,dx=(x+7)(2x-3)^{\frac{1}{2}}-\int(2x-3)^{\frac{1}{2}}\,dx\) | M1A1 | M1: Integration by parts in correct direction; A1: Correct expression |
| \(\int(2x-3)^{\frac{1}{2}}\,dx=k(2x-3)^{\frac{3}{2}}\) | M1 | |
| \(\int(2x-3)^{\frac{1}{2}}\,dx=\frac{1}{3}(2x-3)^{\frac{3}{2}}\) | A1 | |
| \(\left[(x+7)(2x-3)^{\frac{1}{2}}-\frac{1}{3}(2x-3)^{\frac{3}{2}}\right]_4^6=\left\{\left(11(9)^{\frac{1}{2}}-\frac{1}{3}(9)^{\frac{3}{2}}\right)-\left(11(5)^{\frac{1}{2}}-\frac{1}{3}(5)^{\frac{3}{2}}\right)\right\}\) | M1 | Substitutes limits 4 and 6 and subtracts either way round |
| \(=30-\frac{28}{3}\sqrt{5}\) | A1 | cao |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Strip width $= 0.5$ | B1 | Correct value stated or used within the formula |
| $\frac{11\sqrt{5}}{5}+\frac{13}{3}+2\left(\frac{23\sqrt{6}}{12}+\frac{12\sqrt{7}}{7}+\frac{25\sqrt{2}}{8}\right)$ or $(4.91...+4.33...+2(4.69...+4.53...+4.41...))$ | M1 | Correct structure for their $y$ values; must have $y$ values starting at $x=4$ and ending at $x=6$ |
| Area $\approx \frac{1}{2}\times\frac{1}{2}\left(\frac{11\sqrt{5}}{5}+\frac{13}{3}+2\left(\frac{23\sqrt{6}}{12}+\frac{12\sqrt{7}}{7}+\frac{25\sqrt{2}}{8}\right)\right)$ | A1 | Correct numerical expression (allow decimal values to 2sf) |
| $9.14$ | A1 | 9.14 **only** |
## Part (b) — Substitution Method 1 ($u=2x-3$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=2x-3 \Rightarrow \frac{du}{dx}=2$ | B1 | Correct derivative; accept $du=2\,dx$ |
| $\int\frac{x+7}{\sqrt{2x-3}}\,dx = \int\frac{\frac{u+3}{2}+7}{\sqrt{u}}\cdot\frac{1}{2}\,du$ | M1A1 | M1: Fully substitutes, replacing $dx$ with $du$ with no evidence of where $du$ came from; A1: Fully correct expression |
| $\frac{1}{4}\left(\frac{2}{3}u^{\frac{3}{2}}+34u^{\frac{1}{2}}\right)(+c)$ | A1 | Fully correct integration in any form ($+c$ not required) |
| $x=4, u=5\quad x=6, u=9$ | B1 | Correct $u$ limits seen anywhere |
| $\frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}+34u^{\frac{1}{2}}\right]_5^9 = \frac{1}{4}\left\{\left(\frac{2}{3}(9)^{\frac{3}{2}}+34(9)^{\frac{1}{2}}\right)-\left(\frac{2}{3}(5)^{\frac{3}{2}}+34(5)^{\frac{1}{2}}\right)\right\}$ | M1 | Substitutes $u$ limits into changed function and subtracts either way round |
| $=30-\frac{28}{3}\sqrt{5}$ | A1 | cao |
## Part (b) — Substitution Method 2 ($u^2=2x-3$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u^2=2x-3 \Rightarrow 2u\frac{du}{dx}=2$ | B1 | Correct derivative; accept $2u=2\frac{dx}{du}$, $dx=u\,du$ |
| $\int\frac{x+7}{\sqrt{2x-3}}\,dx=\int\frac{\frac{u^2+3}{2}+7}{u}\cdot u\,du$ | M1A1 | M1: Fully substitutes; A1: Fully correct expression |
| $\frac{u^3}{6}+\frac{17}{2}u(+c)$ | A1 | Fully correct integration in any form |
| $x=4, u=\sqrt{5}\quad x=6, u=3$ | B1 | Correct $u$ limits seen anywhere |
| $\left[\frac{1}{6}u^3+\frac{17}{2}u\right]_{\sqrt{5}}^3=\left\{\left(\frac{27}{6}+\frac{17}{2}(3)\right)-\left(\frac{1}{6}(\sqrt{5})^3+\frac{17}{2}(\sqrt{5})\right)\right\}$ | M1 | Substitutes $u$ limits and subtracts either way round |
| $=30-\frac{28}{3}\sqrt{5}$ | A1 | cao |
## Part (b) — Integration by Parts:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(x+7)(2x-3)^{-\frac{1}{2}}\,dx$ | B1 | Uses $\frac{x+7}{\sqrt{2x-3}}$ as $(x+7)(2x-3)^{-\frac{1}{2}}$ and makes some progress attempting integration |
| $\int(x+7)(2x-3)^{-\frac{1}{2}}\,dx=(x+7)(2x-3)^{\frac{1}{2}}-\int(2x-3)^{\frac{1}{2}}\,dx$ | M1A1 | M1: Integration by parts in correct direction; A1: Correct expression |
| $\int(2x-3)^{\frac{1}{2}}\,dx=k(2x-3)^{\frac{3}{2}}$ | M1 | |
| $\int(2x-3)^{\frac{1}{2}}\,dx=\frac{1}{3}(2x-3)^{\frac{3}{2}}$ | A1 | |
| $\left[(x+7)(2x-3)^{\frac{1}{2}}-\frac{1}{3}(2x-3)^{\frac{3}{2}}\right]_4^6=\left\{\left(11(9)^{\frac{1}{2}}-\frac{1}{3}(9)^{\frac{3}{2}}\right)-\left(11(5)^{\frac{1}{2}}-\frac{1}{3}(5)^{\frac{3}{2}}\right)\right\}$ | M1 | Substitutes limits 4 and 6 and subtracts either way round |
| $=30-\frac{28}{3}\sqrt{5}$ | A1 | cao |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ae871952-f525-44e6-8bac-09308aa1964f-26_615_867_292_534}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation
$$y = \frac { x + 7 } { \sqrt { 2 x - 3 } } \quad x > \frac { 3 } { 2 }$$
The region $R$, shown shaded in Figure 1, is bounded by the curve, the line with equation $x = 4$, the $x$-axis and the line with equation $x = 6$
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with 4 strips of equal width to find an estimate for the area of $R$, giving your answer to 2 decimal places.
\item Using the substitution $u = 2 x - 3$, or otherwise, use calculus to find the exact area of $R$, giving your answer in the form $a + b \sqrt { 5 }$, where $a$ and $b$ are constants to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q7 [11]}}