C4 June 2007 Q2
2. Use the substitution \(u = 2^x\) to find the exact value of\[\int_0^1 \frac{2^x}{(2^x + 1)^2}\,\mathrm{d}x.\](6)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^1 \frac{2^x}{(2^x+1)^2}\,\mathrm{d}x\), with substitution \(u = 2^x\) | |
| \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = 2^x.\ln 2 \ \Rightarrow\ \dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{2^x.\ln 2}\) | B1 |
| \(\displaystyle\int \frac{2^x}{(2^x+1)^2}\,\mathrm{d}x = \left(\frac{1}{\ln 2}\right)\int \frac{1}{(u+1)^2}\,\mathrm{d}u\) | M1* |
| \(= \left(\dfrac{1}{\ln 2}\right)\left(\dfrac{-1}{(u+1)}\right) + c\) | M1 A1 |
| change limits: when \(x = 0\) & \(x = 1\) then \(u = 1\) & \(u = 2\) | |
| \(\displaystyle\int_0^1 \frac{2^x}{(2^x+1)^2}\,\mathrm{d}x = \frac{1}{\ln 2}\left[\frac{-1}{(u+1)}\right]_1^2\) | |
| \(= \dfrac{1}{\ln 2}\left[\left(-\dfrac{1}{3}\right) - \left(-\dfrac{1}{2}\right)\right]\) | depM1* |
| \(= \dfrac{1}{6\ln 2}\) | A1 aef |
| (6) | |
| (6 marks) |
Notes
B1: \(\tfrac{\mathrm{d}u}{\mathrm{d}x} = 2^x.\ln 2\) or \(\tfrac{\mathrm{d}u}{\mathrm{d}x} = u.\ln 2\) or \(\left(\tfrac{1}{u}\right)\tfrac{\mathrm{d}u}{\mathrm{d}x} = \ln 2\)
M1*: \(k\displaystyle\int \frac{1}{(u+1)^2}\,\mathrm{d}u\) where \(k\) is constant
M1: \((u+1)^{-2} \to a(u+1)^{-1}\) A1: \((u+1)^{-2} \to -1.(u+1)^{-1}\)
If you see this integration applied anywhere in a candidate’s working then you can award M1, A1
depM1*: Correct use of limits \(u = 1\) and \(u = 2\)
A1 aef: \(\tfrac{1}{6\ln 2}\) or \(\tfrac{1}{\ln 4} - \tfrac{1}{\ln 8}\) or \(\tfrac{1}{2\ln 2} - \tfrac{1}{3\ln 2}\). Exact value only! (corrected from the printed mark scheme: the first answer box prints \(\tfrac{1}{3\mathrm{n}2}\) for \(\tfrac{1}{3\ln 2}\))
There are other acceptable answers for A1, eg: \(\tfrac{1}{2\ln 8}\) or \(\tfrac{1}{\ln 64}\)
NB: Use your calculator to check eg. 0.240449…
Alternative
Alternatively candidate can revert back to \(x\) …
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^1 \frac{2^x}{(2^x+1)^2}\,\mathrm{d}x = \frac{1}{\ln 2}\left[\frac{-1}{(2^x+1)}\right]_0^1\) | |
| \(= \dfrac{1}{\ln 2}\left[\left(-\dfrac{1}{3}\right) - \left(-\dfrac{1}{2}\right)\right]\) | depM1* |
| \(= \underline{\dfrac{1}{6\ln 2}}\) | A1 aef |
depM1*: Correct use of limits \(x = 0\) and \(x = 1\)
A1 aef: \(\tfrac{1}{6\ln 2}\) or \(\tfrac{1}{\ln 4} - \tfrac{1}{\ln 8}\) or \(\tfrac{1}{2\ln 2} - \tfrac{1}{3\ln 2}\). Exact value only!