C4 June 2013 Q5
5.
| Scheme | Marks |
|---|---|
| \(\left\{x = u^2 \Rightarrow\right\} \dfrac{\mathrm{d}x}{\mathrm{d}u} = 2u\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{2}x^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{2\sqrt{x}}\) | B1 |
| \(\left\{\displaystyle\int \dfrac{1}{x(2\sqrt{x} - 1)}\,\mathrm{d}x\right\} = \displaystyle\int \dfrac{1}{u^2(2u - 1)}\,2u\,\mathrm{d}u\) | M1 |
| \(= \displaystyle\int \dfrac{2}{u(2u - 1)}\,\mathrm{d}u\) | A1 * cso |
| (3) |
Notes
B1: \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 2u\) or \(\mathrm{d}x = 2u\,\mathrm{d}u\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{2}x^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{2\sqrt{x}}\) or \(\mathrm{d}u = \dfrac{\mathrm{d}x}{2\sqrt{x}}\)
M1: A full substitution producing an integral in \(u\) only (including the \(\mathrm{d}u\)) (Integral sign not necessary).
The candidate needs to deal with the “\(x\)”, the “\((2\sqrt{x} - 1)\)” and the “\(\mathrm{d}x\)” and converts from an integral term in \(x\) to an integral in \(u\). (Remember the integral sign is not necessary for M1).
A1*: leading to the result printed on the question paper (including the \(\mathrm{d}u\)). (Integral sign is needed).
| Scheme | Marks |
|---|---|
| \(\dfrac{2}{u(2u - 1)} \equiv \dfrac{A}{u} + \dfrac{B}{(2u - 1)} \Rightarrow 2 \equiv A(2u - 1) + Bu\) \(u = 0 \Rightarrow 2 = -A \Rightarrow A = -2\) \(u = \tfrac{1}{2} \Rightarrow 2 = \tfrac{1}{2}B \Rightarrow B = 4\) See notes | M1 A1 |
| So \(\displaystyle\int \dfrac{2}{u(2u - 1)}\,\mathrm{d}u = \displaystyle\int \dfrac{-2}{u} + \dfrac{4}{(2u - 1)}\,\mathrm{d}u\) Integrates \(\dfrac{M}{u} + \dfrac{N}{(2u - 1)}\), \(M \neq 0\), \(N \neq 0\) to obtain any one of \(\pm\lambda\ln u\) or \(\pm\mu\ln(2u - 1)\) | M1 |
| \(= -2\ln u + 2\ln(2u - 1)\) At least one term correctly followed through \(-2\ln u + 2\ln(2u - 1)\). | A1 ft A1 cao |
| So, \(\left[-2\ln u + 2\ln(2u - 1)\right]_1^3\) \(= \left(-2\ln 3 + 2\ln(2(3) - 1)\right) - \left(-2\ln 1 + 2\ln(2(1) - 1)\right)\) Applies limits of 3 and 1 in \(u\) or 9 and 1 in \(x\) in their integrated function and subtracts the correct way round. | M1 |
| \(= -2\ln 3 + 2\ln 5 - (0)\) | |
| \(= 2\ln\left(\dfrac{5}{3}\right)\) \(2\ln\left(\dfrac{5}{3}\right)\) | A1 cso cao |
| (7) | |
| (10 marks) |
Notes
M1: Writing \(\dfrac{2}{u(2u - 1)} \equiv \dfrac{A}{u} + \dfrac{B}{(2u - 1)}\) or writing \(\dfrac{1}{u(2u - 1)} \equiv \dfrac{P}{u} + \dfrac{Q}{(2u - 1)}\) and a complete method for finding the value of at least one of their \(A\) or their \(B\) (or their \(P\) or their \(Q\)).
A1: Both their \(A = -2\) and their \(B = 4\). (Or their \(P = -1\) and their \(Q = 2\) with the multiplying factor of 2 in front of the integral sign).
M1: Integrates \(\dfrac{M}{u} + \dfrac{N}{(2u - 1)}\), \(M \neq 0\), \(N \neq 0\) (i.e. a two term partial fraction) to obtain any one of
\(\pm\lambda\ln u\) or \(\pm\mu\ln(2u - 1)\) or \(\pm\mu\ln\left(u - \tfrac{1}{2}\right)\)
A1ft: At least one term correctly followed through from their \(A\) or from their \(B\) (or their \(P\) and their \(Q\)).
A1: \(-2\ln u + 2\ln(2u - 1)\)
M1: Applies limits of 3 and 1 in \(u\) or 9 and 1 in \(x\) in their (i.e. any) changed function and subtracts the correct way round.
Note: If a candidate just writes \(\left(-2\ln 3 + 2\ln(2(3) - 1)\right)\) oe , this is ok for M1.
A1: \(2\ln\left(\dfrac{5}{3}\right)\) correct answer only. (Note: \(a = 5\), \(b = 3\)).
Important note: Award M0A0M1A1A0 for a candidate who writes
\(\displaystyle\int \dfrac{2}{u(2u - 1)}\,\mathrm{d}u = \displaystyle\int \dfrac{2}{u} + \dfrac{2}{(2u - 1)}\,\mathrm{d}u = 2\ln u + \ln(2u - 1)\)
AS EVIDENCE OF WRITING \(\boldsymbol{\dfrac{2}{u(2u - 1)}}\) AS PARTIAL FRACTIONS IS GIVEN.
Important note: Award M0A0M0A0A0 for a candidate who writes down either
\(\displaystyle\int \dfrac{2}{u(2u - 1)}\,\mathrm{d}u = 2\ln u + 2\ln(2u - 1)\) or \(\displaystyle\int \dfrac{2}{u(2u - 1)}\,\mathrm{d}u = 2\ln u + \ln(2u - 1)\)
WITHOUT ANY EVIDENCE OF WRITING \(\dfrac{2}{u(2u - 1)}\) as partial fractions.
Important note: Award M1A1M1A1A1 for a candidate who writes down
\(\displaystyle\int \dfrac{2}{u(2u - 1)}\,\mathrm{d}u = -2\ln u + 2\ln(2u - 1)\)
WITHOUT ANY EVIDENCE OF WRITING \(\dfrac{2}{u(2u - 1)}\) as partial fractions.
Note: In part (b) if they lose the “2” and find \(\displaystyle\int \dfrac{1}{u(2u - 1)}\,\mathrm{d}u\) we can allow a maximum of
M1A0 M1A1ftA0 M1A0.