C4 June 2009 Q6
6.
(a) Find \(\displaystyle\int \sqrt{(5 - x)}\,\mathrm{d}x\). (2)
Figure 3 shows a sketch of the curve with equation \[y = (x - 1)\sqrt{(5 - x)}, \qquad 1 \leqslant x \leqslant 5\]
(b)
(i) Using integration by parts, or otherwise, find \[\int (x - 1)\sqrt{(5 - x)}\,\mathrm{d}x\] (4)
(ii) Hence find \(\displaystyle\int_1^5 (x - 1)\sqrt{(5 - x)}\,\mathrm{d}x\). (2)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \sqrt{(5 - x)}\,\mathrm{d}x = \int (5 - x)^{\frac{1}{2}}\,\mathrm{d}x = \frac{(5 - x)^{\frac{3}{2}}}{-\frac{3}{2}}\quad (+C)\) \(\left(= -\dfrac{2}{3}(5 - x)^{\frac{3}{2}} + C\right)\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| (i) \(\displaystyle\int (x - 1)\sqrt{(5 - x)}\,\mathrm{d}x = -\frac{2}{3}(x - 1)(5 - x)^{\frac{3}{2}} + \frac{2}{3}\int (5 - x)^{\frac{3}{2}}\,\mathrm{d}x\) | M1 A1ft |
| \(= \quad \ldots \quad + \dfrac{2}{3} \times \dfrac{(5 - x)^{\frac{5}{2}}}{-\frac{5}{2}}\quad (+C)\) | M1 |
| \(= -\dfrac{2}{3}(x - 1)(5 - x)^{\frac{3}{2}} - \dfrac{4}{15}(5 - x)^{\frac{5}{2}}\quad (+C)\) | A1 |
| (4) | |
| (ii) \(\left[-\dfrac{2}{3}(x - 1)(5 - x)^{\frac{3}{2}} - \dfrac{4}{15}(5 - x)^{\frac{5}{2}}\right]_1^5 = (0 - 0) - \left(0 - \dfrac{4}{15} \times 4^{\frac{5}{2}}\right)\) \(= \dfrac{128}{15}\left(= 8\dfrac{8}{15} \approx 8.53\right)\) awrt 8.53 | M1 A1 |
| (2) | |
| (8 marks) |
Alternatives for (b) and (c)
(b)
| \(u^2 = 5 - x \Rightarrow 2u\dfrac{\mathrm{d}u}{\mathrm{d}x} = -1\quad \left(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}u} = -2u\right)\) | |
| \(\displaystyle\int (x - 1)\sqrt{(5 - x)}\,\mathrm{d}x = \int (4 - u^2)u\frac{\mathrm{d}x}{\mathrm{d}u}\,\mathrm{d}u = \int (4 - u^2)u(-2u)\,\mathrm{d}u\) | M1 A1 |
| \(\displaystyle = \int \left(2u^4 - 8u^2\right)\mathrm{d}u = \frac{2}{5}u^5 - \frac{8}{3}u^3\quad (+C)\) | M1 |
| \(= \dfrac{2}{5}(5 - x)^{\frac{5}{2}} - \dfrac{8}{3}(5 - x)^{\frac{3}{2}}\quad (+C)\) | A1 |
(c)
| \(x = 1 \Rightarrow u = 2,\quad x = 5 \Rightarrow u = 0\) | |
| \(\left[\dfrac{2}{5}u^5 - \dfrac{8}{3}u^3\right]_2^0 = (0 - 0) - \left(\dfrac{64}{5} - \dfrac{64}{3}\right)\) | M1 |
| \(= \dfrac{128}{15}\left(= 8\dfrac{8}{15} \approx 8.53\right)\) awrt 8.53 | A1 |
| (2) |