| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with complex substitution requiring algebraic rearrangement |
| Difficulty | Standard +0.3 This is a standard P4 integration by substitution question with a given substitution. Students must find du/dx, change limits, simplify the integrand (which becomes a simple rational function), and integrate to get a logarithmic answer. While it requires careful algebraic manipulation and multiple steps, the substitution is provided and the techniques are routine for P4 level, making it slightly easier than average. |
| Spec | 1.08h Integration by substitution |
| VIIIV SIHI NI IIIIM ION OC | VIIN SIHI NI III M M O N OO | VIIV SIHI NI IIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 3+\sqrt{2x-1} \Rightarrow x = \frac{(u-3)^2+1}{2} \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} = u-3\) or \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{\sqrt{2x-1}} = \frac{1}{u-3}\) | M1 A1 | M1: differentiates to get \(\frac{\mathrm{d}u}{\mathrm{d}x}\) in terms of \(x\) then obtains \(\frac{\mathrm{d}x}{\mathrm{d}u}\) in terms of \(u\). Need \(\frac{\mathrm{d}u}{\mathrm{d}x} = k(2x-1)^{-\frac{1}{2}}\) or \(\frac{\mathrm{d}x}{\mathrm{d}u} = au+b\) |
| \(\int \frac{4}{3+\sqrt{2x-1}}\,\mathrm{d}x = \int \frac{4}{u} \times (u-3)\,\mathrm{d}u\) | M1 | Attempts to write integral completely in terms of \(u\) |
| \(\int \frac{4}{u}\times(u-3)\,\mathrm{d}u = \int\left(4 - \frac{12}{u}\right)\mathrm{d}u\) | dM1 | Divides to reach form \(\int\!\left(A + B \times \frac{1}{u}\right)\mathrm{d}u\). Depends on both previous M marks |
| \(\int\left(4-\frac{12}{u}\right)\mathrm{d}u = 4u - 12\ln u\) or \(k(4u-12\ln u)\) | ddM1 A1ft | Integrates to form \(Au + B\ln u\). Depends on previous M. A1ft follows through on \(\frac{\mathrm{d}x}{\mathrm{d}u} = k(u-3)\) only |
| \(\int_1^{13}\frac{4}{3+\sqrt{2x-1}}\,\mathrm{d}x = \left[4u-12\ln u\right]_4^8 = (4\times8-12\ln8)-(4\times4-12\ln4)\) | M1 | Substitutes limits 8 and 4 into \(4u-12\ln u\) and subtracts, or substitutes 13 and 1 with \(u=3+\sqrt{2x-1}\) |
| \(= 16 - 12\ln 2\) | A1 |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 3+\sqrt{2x-1} \Rightarrow x = \frac{(u-3)^2+1}{2} \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} = u-3$ **or** $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{\sqrt{2x-1}} = \frac{1}{u-3}$ | M1 A1 | M1: differentiates to get $\frac{\mathrm{d}u}{\mathrm{d}x}$ in terms of $x$ then obtains $\frac{\mathrm{d}x}{\mathrm{d}u}$ in terms of $u$. Need $\frac{\mathrm{d}u}{\mathrm{d}x} = k(2x-1)^{-\frac{1}{2}}$ or $\frac{\mathrm{d}x}{\mathrm{d}u} = au+b$ |
| $\int \frac{4}{3+\sqrt{2x-1}}\,\mathrm{d}x = \int \frac{4}{u} \times (u-3)\,\mathrm{d}u$ | M1 | Attempts to write integral completely in terms of $u$ |
| $\int \frac{4}{u}\times(u-3)\,\mathrm{d}u = \int\left(4 - \frac{12}{u}\right)\mathrm{d}u$ | dM1 | Divides to reach form $\int\!\left(A + B \times \frac{1}{u}\right)\mathrm{d}u$. Depends on both previous M marks |
| $\int\left(4-\frac{12}{u}\right)\mathrm{d}u = 4u - 12\ln u$ or $k(4u-12\ln u)$ | ddM1 A1ft | Integrates to form $Au + B\ln u$. Depends on previous M. A1ft follows through on $\frac{\mathrm{d}x}{\mathrm{d}u} = k(u-3)$ only |
| $\int_1^{13}\frac{4}{3+\sqrt{2x-1}}\,\mathrm{d}x = \left[4u-12\ln u\right]_4^8 = (4\times8-12\ln8)-(4\times4-12\ln4)$ | M1 | Substitutes limits 8 and 4 into $4u-12\ln u$ and subtracts, or substitutes 13 and 1 with $u=3+\sqrt{2x-1}$ |
| $= 16 - 12\ln 2$ | A1 | |
---5. In this question you should show all stages of your working.
Solutions relying on calculator technology are not acceptable.\\
Using the substitution $u = 3 + \sqrt { 2 x - 1 }$ find the exact value of
$$\int _ { 1 } ^ { 13 } \frac { 4 } { 3 + \sqrt { 2 x - 1 } } d x$$
giving your answer in the form $p + q \ln 2$, where $p$ and $q$ are integers to be found.\\
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VIIIV SIHI NI IIIIM ION OC & VIIN SIHI NI III M M O N OO & VIIV SIHI NI IIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel P4 2021 Q5 [8]}}