Question 13 - A-Level Maths

Edexcel PMT Mocks — Question 13 7 marks

Exam Board	Edexcel
Module	PMT Mocks (PMT Mocks)
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration by Substitution
Type	Area under curve using substitution
Difficulty	Standard +0.3 This is a straightforward integration by substitution question with a given substitution. Students must change variables (u = 3x + 1), adjust limits, rewrite the integrand, and integrate using standard techniques. While it requires multiple steps and careful algebraic manipulation, the substitution is provided and the method is a standard A-level technique with no novel insight required. Slightly easier than average due to the explicit substitution given.
Spec	1.08h Integration by substitution

13. \includegraphics[max width=\textwidth, alt={}, center]{63d85737-99d4-4916-a479-fe44f77b1505-25_679_1043_413_607} Figure 5 shows a sketch of part of the curve with equation \(y = \frac { 6 x } { \sqrt { 3 x + 1 } } , \quad x \geq 0\) The finite region \(\mathbf { R }\), shown shaded in figure 5 is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 5\). Use the substitution \(u = 3 x + 1\) to find the exact area of \(\mathbf { R }\).
(Total for Question 13 is 7 marks)

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Question 13:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(u = 3x+1 \Rightarrow \frac{du}{dx} = 3 \Rightarrow dx = \frac{du}{3}\)	B1	\(\frac{du}{dx} = 3\)
\(\frac{6x}{\sqrt{3x+1}}\) becoming \(\frac{2(u-1)}{u^{\frac{1}{2}}}\)	M1	Correct substitution of function
\(\frac{6x}{\sqrt{3x+1}}\,dx\) becoming \(\frac{2(u-1)}{3u^{\frac{1}{2}}}\,du\)	A1	Including \(du\)
\(\frac{2}{3}\left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)\)	M1	Attempt to divide the two terms by \(u\)
\(\int \frac{2}{3}\left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)du = \frac{2}{3}\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right)\)	A1	Correct integration
Limits of 16 and 7 in \(u\)	M1	Some evidence of changed limits
\(R = \frac{16}{9}(13 - \sqrt{7})\)	A1	Exact answer

## Question 13:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 3x+1 \Rightarrow \frac{du}{dx} = 3 \Rightarrow dx = \frac{du}{3}$ | B1 | $\frac{du}{dx} = 3$ |
| $\frac{6x}{\sqrt{3x+1}}$ becoming $\frac{2(u-1)}{u^{\frac{1}{2}}}$ | M1 | Correct substitution of function |
| $\frac{6x}{\sqrt{3x+1}}\,dx$ becoming $\frac{2(u-1)}{3u^{\frac{1}{2}}}\,du$ | A1 | Including $du$ |
| $\frac{2}{3}\left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)$ | M1 | Attempt to divide the two terms by $u$ |
| $\int \frac{2}{3}\left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)du = \frac{2}{3}\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right)$ | A1 | Correct integration |
| Limits of 16 and 7 in $u$ | M1 | Some evidence of changed limits |
| $R = \frac{16}{9}(13 - \sqrt{7})$ | A1 | Exact answer |

Show LaTeX source

13.\\
\includegraphics[max width=\textwidth, alt={}, center]{63d85737-99d4-4916-a479-fe44f77b1505-25_679_1043_413_607}

Figure 5 shows a sketch of part of the curve with equation $y = \frac { 6 x } { \sqrt { 3 x + 1 } } , \quad x \geq 0$\\
The finite region $\mathbf { R }$, shown shaded in figure 5 is bounded by the curve, the $x$-axis and the lines $x = 2$ and $x = 5$.

Use the substitution $u = 3 x + 1$ to find the exact area of $\mathbf { R }$.\\
(Total for Question 13 is 7 marks)\\

\hfill \mbox{\textit{Edexcel PMT Mocks  Q13 [7]}}

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