FP3 June 2017 Q4
4. Use the substitution \(x + 2 = u^2\), where \(u > 0\), to show that \[\int_{-1}^{7}\frac{(x + 2)^{\frac{1}{2}}}{x + 5}\,\mathrm{d}x = a + b\pi\sqrt{3}\] where \(a\) and \(b\) are rational numbers to be found. (9)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 2u\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{2}(x + 2)^{-\frac{1}{2}}\) | B1 |
| \(\displaystyle\int\frac{(x + 2)^{\frac{1}{2}}}{x + 5}\,\mathrm{d}x = \int\frac{\left(u^2\right)^{\frac{1}{2}}}{u^2 - 2 + 5}2u\,(\mathrm{d}u)\) or \(\displaystyle\int\frac{(x + 2)^{\frac{1}{2}}}{x + 5}\,\mathrm{d}x = \int\frac{(x + 2)^{\frac{1}{2}}}{u^2 - 2 + 5} \times \frac{2}{(x + 2)^{\frac{1}{2}}}\,(\mathrm{d}u)\) | M1 |
| \(\displaystyle = 2\int\frac{u^2}{u^2 + 3}\,(\mathrm{d}u)\) or \(\displaystyle\int\frac{2u^2}{u^2 + 3}\,(\mathrm{d}u)\) | A1 |
| \(\displaystyle(2)\int\frac{u^2}{u^2 + 3}\,\mathrm{d}u = (2)\int\left(1 - \frac{3}{u^2 + 3}\right)\mathrm{d}u\) | M1 |
| \(= (2)\left[u - \dfrac{3}{\sqrt{3}}\arctan\dfrac{u}{\sqrt{3}}\right]\) | A1 A1 |
| \(x = -1 \Rightarrow u = 1,\quad x = 7 \Rightarrow u = 3\) | B1 |
| \(= 2\left[\left(3 - \dfrac{3}{\sqrt{3}}\dfrac{\pi}{3}\right) - \left(1 - \dfrac{3}{\sqrt{3}}\dfrac{\pi}{6}\right)\right]\) | M1 |
| \(= 4 - \dfrac{\sqrt{3}}{3}\pi\) | A1 |
| (9) | |
| (9 marks) |
Notes
B1: Or equivalent correct derivative in any form. May be implied by their substitution.
M1: Complete substitution including their “\(\mathrm{d}x\)”. Allow the omission of “\(\mathrm{d}u\)” if it is implied by later work.
A1: Correct integral
M1: Splits the fraction into \(A + \dfrac{B}{u^2 + 3}\)
A1: \(u\)
A1: \(-\dfrac{3}{\sqrt{3}}\arctan\dfrac{u}{\sqrt{3}}\)
B1: Correct limits.
M1: Substitutes \(u\) limits correctly into an expression of the form \(\pm\alpha u \pm \beta\arctan(ku),\ \alpha, \beta \neq 0\) and subtracts the right way round.
A1: Cao (oe)
Alternative using substitution again for last 6 marks
| Scheme | Marks |
|---|---|
| \(\displaystyle u = \sqrt{3}\tan\theta \Rightarrow (2)\int\frac{u^2}{u^2 + 3}\,\mathrm{d}u = (2)\int\frac{3\tan^2\theta}{3\tan^2\theta + 3}\sqrt{3}\sec^2\theta\,\mathrm{d}\theta\) Use of \(u = \sqrt{3}\tan\theta\) and a complete substitution. | M1 |
| \(\displaystyle = \left(2\sqrt{3}\right)\int\tan^2\theta\,\mathrm{d}\theta = \left(2\sqrt{3}\right)\int\left(\sec^2\theta - 1\right)\mathrm{d}\theta\) \(= \left(2\sqrt{3}\right)\left[\tan\theta - \theta\right]\) | A1A1 |
| \(u = 1 \Rightarrow \theta = \dfrac{\pi}{6},\quad u = 3 \Rightarrow \theta = \dfrac{\pi}{3}\) | B1 |
| \(= 2\sqrt{3}\left[\left(\sqrt{3} - \dfrac{\pi}{3}\right) - \left(\dfrac{1}{\sqrt{3}} - \dfrac{\pi}{6}\right)\right]\) | M1 |
| \(= 4 - \dfrac{\sqrt{3}}{3}\pi\) | A1 |
A1: \(\theta\)
A1: \(\tan\theta\)
B1: Correct limits
M1: Substitutes \(\theta\) limits correctly into an expression of the form \(\pm\alpha\tan\theta \pm \beta\theta,\ \alpha, \beta \neq 0\) and subtracts the right way round.
A1: cao