| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Apply trapezium rule to given table |
| Difficulty | Standard +0.3 This is a standard multi-part integration question combining numerical methods (trapezium rule) with substitution and integration by parts. Part (a) is routine arithmetic, part (b) is a guided substitution with clear instructions, and part (c) requires integration by parts of u^(1/2)ln(2u) which is a standard technique. The question provides significant scaffolding and uses familiar C3/C4 techniques without requiring novel insight. |
| Spec | 1.08h Integration by substitution1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.25 | 1.5 | 1.75 | 2 |
| \(y\) | 8.32 | 21.4 | 40.6 | 66.6 | 99.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.25\) | B1 | Correct strip width |
| \(\text{Area} = \frac{0.25}{2}\{8.32+99.8+2\times(21.4+40.6+66.6)\}\) | M1 | Correct trapezium rule structure: \(\frac{h}{2}\{\text{correct }y\text{-value structure}\}\); or may see separate trapezia |
| Awrt 46 | A1 | Correct answer of awrt 46 with no incorrect working; calculator gives 45.1028... |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = x^2 \Rightarrow \frac{du}{dx} = 2x\) | B1 | Correct \(\frac{du}{dx}\) or equivalent such as \(x=u^{\frac{1}{2}} \Rightarrow \frac{dx}{du} = \frac{1}{2}u^{-\frac{1}{2}}\) |
| \(\int 12x^2\ln(2x^2)\,dx = \int 12u\ln(2u)\cdot\frac{1}{2}u^{-\frac{1}{2}}\,du\) | M1 | Uses substitution replacing at least the "\(dx\)" in terms of \(du\) and changes \(\ln(2x^2)\) to \(\ln(2u)\) |
| \(= \int_1^4 6u^{\frac{1}{2}}\ln(2u)\,du\) | A1* | Completes to printed answer with reference to limits; limits 1 and 2 in \(x\) becoming 1 and 4 in \(u\); \(12x^2\) and \(dx\) must appear in terms of same variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = x^2 \Rightarrow \frac{du}{dx} = 2x\), or equivalent such as \(x = u^{\frac{1}{2}} \Rightarrow \frac{dx}{du} = \frac{1}{2}u^{-\frac{1}{2}}\) | B1 | |
| \(\int 6u^{\frac{1}{2}}\ln(2u)\,du = \int 6x\ln(2x^2)2x\,dx\) or equivalent forms showing substitution | M1 | Uses the substitution and replaces at least the "\(du\)" in terms of \(dx\) and changes \(\ln(u)\) to \(\ln(2x^2)\) |
| \(= \int_1^2 12x^2\ln(2x^2)\,dx\) | A1* | Completes to printed answer with a conclusion. Must reference limits. At some point \(6u^{\frac{1}{2}}\) and \(du\) must appear in terms of same variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 6u^{\frac{1}{2}}\ln 2u\,du = 4u^{\frac{3}{2}}\ln 2u - \int 4u^{\frac{3}{2}} \times \frac{1}{u}\,du\) | M1A1 | M1: Integration by parts correct way achieving \(Pu^{\frac{3}{2}}\ln 2u - \int Qu^{\frac{3}{2}} \times \frac{1}{u}\,du\); A1: \(4u^{\frac{3}{2}}\ln 2u - \int 4u^{\frac{3}{2}} \times \frac{1}{u}\,du\) (or if "6" omitted: \(\frac{2}{3}u^{\frac{3}{2}}\ln 2u - \int \frac{2}{3}u^{\frac{3}{2}} \times \frac{1}{u}\,du\)) |
| \(= 4u^{\frac{3}{2}}\ln 2u - \frac{8}{3}u^{\frac{3}{2}}\) | A1 | (or if "6" omitted: \(\frac{2}{3}u^{\frac{3}{2}}\ln 2u - \frac{4}{9}u^{\frac{3}{2}}\)) |
| Area \(= \left[4u^{\frac{3}{2}}\ln 2u - \frac{8}{3}u^{\frac{3}{2}}\right]_1^4 = \left(32\ln 8 - \frac{64}{3}\right) - \left(4\ln 2 - \frac{8}{3}\right)\) | dM1 | Dependent on previous M. Putting in limits 4 and 1 and subtracting either way. Alternatively limits of 1 and 2 with substituted function in \(x\) |
| \(= 96\ln 2 - 4\ln 2 - \frac{64}{3} + \frac{8}{3} = \alpha\ln 2 + \ldots\) | M1 | For correct log work on \(\ln 8\) term and combining correctly with \(\ln 2\) term to obtain single \(\ln 2\) term having substituted into an integrated function |
| \(= -\frac{56}{3} + 92\ln 2\) (or \(92\ln 2 - \frac{56}{3}\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 12x^2\ln(2x^2)\,dx = \frac{12x^3}{3}\ln(2x^2) - \int\frac{12x^3}{3} \times \frac{4x}{2x^2}\,dx\) | M1A1 | M1: \(Px^3\ln(2x^2) - \int Qx^3 \times \frac{1}{x}\,dx\); A1: exact equivalent shown (or if "12" omitted: \(\frac{x^3}{3}\ln(2x^2) - \int\frac{x^3}{3} \times \frac{4x}{2x^2}\,dx\)) |
| \(= 4x^3\ln(2x^2) - \frac{8}{3}x^3\) | A1 | (or if "12" omitted: \(\frac{1}{3}x^3\ln(2x^2) - \frac{2}{9}x^3\)) |
| Area \(= \left[4x^3\ln(2x^2) - \frac{8}{3}x^3\right]_1^2 = \left(32\ln 8 - \frac{64}{3}\right) - \left(4\ln 2 - \frac{8}{3}\right)\) | dM1 | Dependent on previous M. Limits of 2 and 1, subtracting either way |
| \(= 96\ln 2 - 4\ln 2 - \frac{64}{3} + \frac{8}{3}\) | M1 | Correct log work on \(\ln 8\) and combining with \(\ln 2\) term having substituted into integrated function |
| \(= 92\ln 2 - \frac{56}{3}\) | A1 | CSO |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.25$ | B1 | Correct strip width |
| $\text{Area} = \frac{0.25}{2}\{8.32+99.8+2\times(21.4+40.6+66.6)\}$ | M1 | Correct trapezium rule structure: $\frac{h}{2}\{\text{correct }y\text{-value structure}\}$; or may see separate trapezia |
| Awrt 46 | A1 | Correct answer of awrt 46 with no incorrect working; calculator gives 45.1028... |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x^2 \Rightarrow \frac{du}{dx} = 2x$ | B1 | Correct $\frac{du}{dx}$ or equivalent such as $x=u^{\frac{1}{2}} \Rightarrow \frac{dx}{du} = \frac{1}{2}u^{-\frac{1}{2}}$ |
| $\int 12x^2\ln(2x^2)\,dx = \int 12u\ln(2u)\cdot\frac{1}{2}u^{-\frac{1}{2}}\,du$ | M1 | Uses substitution replacing at least the "$dx$" in terms of $du$ **and** changes $\ln(2x^2)$ to $\ln(2u)$ |
| $= \int_1^4 6u^{\frac{1}{2}}\ln(2u)\,du$ | A1* | Completes to printed answer with reference to limits; limits 1 and 2 in $x$ becoming 1 and 4 in $u$; $12x^2$ and $dx$ must appear in terms of same variable |
## Question 13(b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x^2 \Rightarrow \frac{du}{dx} = 2x$, or equivalent such as $x = u^{\frac{1}{2}} \Rightarrow \frac{dx}{du} = \frac{1}{2}u^{-\frac{1}{2}}$ | B1 | |
| $\int 6u^{\frac{1}{2}}\ln(2u)\,du = \int 6x\ln(2x^2)2x\,dx$ or equivalent forms showing substitution | M1 | Uses the substitution and replaces at least the "$du$" in terms of $dx$ **and** changes $\ln(u)$ to $\ln(2x^2)$ |
| $= \int_1^2 12x^2\ln(2x^2)\,dx$ | A1* | Completes to printed answer **with a conclusion**. Must reference limits. At some point $6u^{\frac{1}{2}}$ and $du$ must appear in terms of same variable |
---
## Question 13(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 6u^{\frac{1}{2}}\ln 2u\,du = 4u^{\frac{3}{2}}\ln 2u - \int 4u^{\frac{3}{2}} \times \frac{1}{u}\,du$ | M1A1 | M1: Integration by parts correct way achieving $Pu^{\frac{3}{2}}\ln 2u - \int Qu^{\frac{3}{2}} \times \frac{1}{u}\,du$; A1: $4u^{\frac{3}{2}}\ln 2u - \int 4u^{\frac{3}{2}} \times \frac{1}{u}\,du$ (or if "6" omitted: $\frac{2}{3}u^{\frac{3}{2}}\ln 2u - \int \frac{2}{3}u^{\frac{3}{2}} \times \frac{1}{u}\,du$) |
| $= 4u^{\frac{3}{2}}\ln 2u - \frac{8}{3}u^{\frac{3}{2}}$ | A1 | (or if "6" omitted: $\frac{2}{3}u^{\frac{3}{2}}\ln 2u - \frac{4}{9}u^{\frac{3}{2}}$) |
| Area $= \left[4u^{\frac{3}{2}}\ln 2u - \frac{8}{3}u^{\frac{3}{2}}\right]_1^4 = \left(32\ln 8 - \frac{64}{3}\right) - \left(4\ln 2 - \frac{8}{3}\right)$ | dM1 | Dependent on previous M. Putting in limits 4 and 1 and subtracting either way. Alternatively limits of 1 and 2 with substituted function in $x$ |
| $= 96\ln 2 - 4\ln 2 - \frac{64}{3} + \frac{8}{3} = \alpha\ln 2 + \ldots$ | M1 | For correct log work on $\ln 8$ term and combining correctly with $\ln 2$ term to obtain single $\ln 2$ term **having substituted into an integrated function** |
| $= -\frac{56}{3} + 92\ln 2$ (or $92\ln 2 - \frac{56}{3}$) | A1 | |
**Alternative in terms of $x$:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 12x^2\ln(2x^2)\,dx = \frac{12x^3}{3}\ln(2x^2) - \int\frac{12x^3}{3} \times \frac{4x}{2x^2}\,dx$ | M1A1 | M1: $Px^3\ln(2x^2) - \int Qx^3 \times \frac{1}{x}\,dx$; A1: exact equivalent shown (or if "12" omitted: $\frac{x^3}{3}\ln(2x^2) - \int\frac{x^3}{3} \times \frac{4x}{2x^2}\,dx$) |
| $= 4x^3\ln(2x^2) - \frac{8}{3}x^3$ | A1 | (or if "12" omitted: $\frac{1}{3}x^3\ln(2x^2) - \frac{2}{9}x^3$) |
| Area $= \left[4x^3\ln(2x^2) - \frac{8}{3}x^3\right]_1^2 = \left(32\ln 8 - \frac{64}{3}\right) - \left(4\ln 2 - \frac{8}{3}\right)$ | dM1 | Dependent on previous M. Limits of 2 and 1, subtracting either way |
| $= 96\ln 2 - 4\ln 2 - \frac{64}{3} + \frac{8}{3}$ | M1 | Correct log work on $\ln 8$ and combining with $\ln 2$ term **having substituted into integrated function** |
| $= 92\ln 2 - \frac{56}{3}$ | A1 | CSO |
---13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-42_649_709_242_614}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve with equation $y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right) , x > 0$
The finite region $R$, shown shaded in Figure 4, is bounded by the curve, the line with equation $x = 1$, the $x$-axis and the line with equation $x = 2$
The table below shows corresponding values of $x$ and $y$ for $y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right)$, with the values of $y$ given to 3 significant figures.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.25 & 1.5 & 1.75 & 2 \\
\hline
$y$ & 8.32 & 21.4 & 40.6 & 66.6 & 99.8 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule, with all the values of $y$, to obtain an estimate for the area of $R$, giving your answer to 2 significant figures.
\item Use the substitution $u = x ^ { 2 }$ to show that the area of $R$ is given by
$$\int _ { 1 } ^ { 4 } 6 u ^ { \frac { 1 } { 2 } } \ln ( 2 u ) \mathrm { d } u$$
\item Hence, using calculus, find the exact area of $R$, writing your answer in the form $a + b \ln 2$, where $a$ and $b$ are constants to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q13 [12]}}