C4 June 2013 (R) Q3
3. Using the substitution \(u = 2 + \sqrt{(2x + 1)}\), or other suitable substitutions, find the exact value of \[\int_0^4 \frac{1}{2 + \sqrt{(2x + 1)}}\,\mathrm{d}x\] giving your answer in the form \(A + 2\ln B\), where \(A\) is an integer and \(B\) is a positive constant. (8)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^4 \dfrac{1}{2 + \sqrt{(2x + 1)}}\,\mathrm{d}x,\quad u = 2 + \sqrt{(2x + 1)}\) | |
| \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = (2x + 1)^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = u - 2\) Either \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \pm K(2x + 1)^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \pm\lambda(u - 2)\) Either \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = (2x + 1)^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = (u - 2)\) | M1 A1 |
| \(\left\{\displaystyle\int \dfrac{1}{2 + \sqrt{(2x + 1)}}\,\mathrm{d}x\right\} = \displaystyle\int \dfrac{1}{u}(u - 2)\,\mathrm{d}u\) Correct substitution (Ignore integral sign and \(\mathrm{d}u\)). | A1 |
| \(= \displaystyle\int \left(1 - \dfrac{2}{u}\right)\mathrm{d}u\) An attempt to divide each term by \(u\). | dM1 |
| \(= u - 2\ln u\) \(\pm Au \pm B\ln u\) \(u - 2\ln u\) | ddM1 A1 ft |
| \(\left\{\text{So } \left[u - 2\ln u\right]_3^5\right\} = \left(5 - 2\ln 5\right) - \left(3 - 2\ln 3\right)\) Applies limits of 5 and 3 in \(u\) or 4 and 0 in \(x\) in their integrated function and subtracts the correct way round. | M1 |
| \(= 2 + 2\ln\left(\dfrac{3}{5}\right)\) \(2 + 2\ln\left(\dfrac{3}{5}\right)\) | A1 cao cso |
| (8) | |
| (8 marks) |
Notes
M1: Also allow \(\mathrm{d}u = \pm\lambda\dfrac{1}{(u - 2)}\mathrm{d}x\) or \((u - 2)\mathrm{d}u = \pm\lambda\,\mathrm{d}x\)
Note: The expressions must contain \(\mathrm{d}u\) and \(\mathrm{d}x\). They can be simplified or un-simplified.
A1: Also allow \(\mathrm{d}u = \dfrac{1}{(u - 2)}\mathrm{d}x\) or \((u - 2)\mathrm{d}u = \pm\lambda\,\mathrm{d}x\)
Note: The expressions must contain \(\mathrm{d}u\) and \(\mathrm{d}x\). They can be simplified or un-simplified.
A1: \(\displaystyle\int \dfrac{1}{u}(u - 2)\,\mathrm{d}u\). (Ignore integral sign and \(\mathrm{d}u\)).
dM1: An attempt to divide each term by \(u\).
Note that this mark is dependent on the previous M1 mark being awarded.
Note that this mark can be implied by later working.
ddM1: \(\pm Au \pm B\ln u,\ A \neq 0,\ B \neq 0\)
Note that this mark is dependent on the two previous M1 marks being awarded.
A1ft: \(u - 2\ln u\) or \(\pm Au \pm B\ln u\) being correctly followed through, \(A \neq 0,\ B \neq 0\)
M1: Applies limits of 5 and 3 in \(u\) or 4 and 0 in \(x\) in their integrated function and subtracts the correct way round.
A1: cso and cao. \(2 + 2\ln\left(\dfrac{3}{5}\right)\) or \(2 + 2\ln(0.6)\), \(\left(= A + 2\ln B, \text{ so } A = 2,\ B = \dfrac{3}{5}\right)\)
Note: \(2 - 2\ln\left(\dfrac{3}{5}\right)\) is A0.
Note: \(\displaystyle\int \dfrac{1}{u}(u - 2)\,\mathrm{d}u = u - 2\ln u\) with no working is 2nd M1, 3rd M1, 3rd A1.
but Note: \(\displaystyle\int \dfrac{1}{u}(u - 2)\,\mathrm{d}u = (u - 2)\ln u\) with no working is 2nd M0, 3rd M0, 3rd A0.