C4 June 2011 Q4
4.
Figure 2 shows a sketch of the curve with equation \(y = x^3\ln(x^2 + 2),\ x \geqslant 0\).
The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line \(x = \sqrt{2}\).
The table below shows corresponding values of \(x\) and \(y\) for \(y = x^3\ln(x^2 + 2)\).
| \(x\) | 0 | \(\dfrac{\sqrt{2}}{4}\) | \(\dfrac{\sqrt{2}}{2}\) | \(\dfrac{3\sqrt{2}}{4}\) | \(\sqrt{2}\) |
|---|---|---|---|---|---|
| \(y\) | 0 | 0.3240 | 3.9210 |
(a) Complete the table above giving the missing values of \(y\) to 4 decimal places. (2)
(b) Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places. (3)
(c) Use the substitution \(u = x^2 + 2\) to show that the area of \(R\) is\[\frac{1}{2}\int_2^4 (u - 2)\ln u\ \mathrm{d}u\] (4)
(d) Hence, or otherwise, find the exact area of \(R\). (6)
| Scheme | Marks |
|---|---|
| 0.0333, 1.3596 awrt 0.0333, 1.3596 | B1 B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\text{Area}(R) \approx \dfrac{1}{2} \times \dfrac{\sqrt{2}}{4}[\ \ldots\ ]\) | B1 |
| \(\approx \ldots\ [0 + 2(0.0333 + 0.3240 + 1.3596) + 3.9210]\) | M1 |
| \(\approx 1.30\) Accept 1.3 | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(u = x^2 + 2 \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = 2x\) | B1 |
| \(\text{Area}(R) = \displaystyle\int_0^{\sqrt{2}} x^3\ln(x^2 + 2)\,\mathrm{d}x\) | B1 |
| \(\displaystyle\int x^3\ln(x^2 + 2)\,\mathrm{d}x = \int x^2\ln(x^2 + 2)\,x\,\mathrm{d}x = \int (u - 2)(\ln u)\tfrac{1}{2}\,\mathrm{d}u\) | M1 |
| Hence \(\qquad \text{Area}(R) = \dfrac{1}{2}\displaystyle\int_2^4 (u - 2)\ln u\,\mathrm{d}u\ \ *\) cso | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int (u - 2)\ln u\,\mathrm{d}u = \left(\frac{u^2}{2} - 2u\right)\ln u - \int \left(\frac{u^2}{2} - 2u\right)\frac{1}{u}\,\mathrm{d}u\) | M1 A1 |
| \(= \left(\dfrac{u^2}{2} - 2u\right)\ln u - \displaystyle\int \left(\frac{u}{2} - 2\right)\mathrm{d}u\) \(= \left(\dfrac{u^2}{2} - 2u\right)\ln u - \left(\dfrac{u^2}{4} - 2u\right) \quad (+C)\) | M1 A1 |
| \(\text{Area}(R) = \dfrac{1}{2}\left[\left(\dfrac{u^2}{2} - 2u\right)\ln u - \left(\dfrac{u^2}{4} - 2u\right)\right]_2^4\) \(= \tfrac{1}{2}\left[(8 - 8)\ln 4 - 4 + 8 - \left((2 - 4)\ln 2 - 1 + 4\right)\right]\) | M1 |
| \(= \tfrac{1}{2}(2\ln 2 + 1)\) \(\ln 2 + \tfrac{1}{2}\) | A1 |
| (6) | |
| (15 marks) |
Notes
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.