M3 June 2015 Q2
2. The finite region bounded by the \(x\)-axis, the curve with equation \(y = 2\mathrm{e}^x\), the \(y\)-axis and the line \(x = 1\) is rotated through one complete revolution about the \(x\)-axis to form a uniform solid.
Use algebraic integration to
| Scheme | Marks |
|---|---|
| \(\text{Vol} = \pi\displaystyle\int_0^1 4\mathrm{e}^{2x}\,\mathrm{d}x\) | M1 |
| \(= \pi\left[2\mathrm{e}^{2x}\right]_0^1\) | DM1A1 |
| \(= 2\pi\left(\mathrm{e}^2 - 1\right)\) * | A1cso |
| (4) |
Notes
M1 Using \(\pi\displaystyle\int y^2\,\mathrm{d}x\) with the equation of the curve, no limits needed
DM1 Integrating their expression for the volume
A1 Correct integration inc limits now
A1 Substituting the limits to obtain the GIVEN answer
| Scheme | Marks |
|---|---|
| \(\pi\displaystyle\int_0^1 4x\mathrm{e}^{2x}\,\mathrm{d}x\) | M1 |
| \(= 4\pi\left\{\left[x \times \dfrac{1}{2}\mathrm{e}^{2x}\right]_0^1 - \displaystyle\int_0^1 \frac{1}{2}\mathrm{e}^{2x}\,\mathrm{d}x\right\}\) | DM1 |
| \(= 4\pi\left[\dfrac{1}{2}\mathrm{e}^2 - 0\right] - 4\pi\left[\dfrac{1}{4}\mathrm{e}^{2x}\right]_0^1\) | A1 |
| \(= \pi\left(\mathrm{e}^2 + 1\right)\) | A1 |
| \(x\) coord \(= \dfrac{\pi\left(\mathrm{e}^2 + 1\right)}{2\pi\left(\mathrm{e}^2 - 1\right)},\ \ = \dfrac{e^2 + 1}{2\left(e^2 - 1\right)}\) oe | M1A1 |
| (6) | |
| (10 marks) |
Notes
M1 Using \((\pi)\displaystyle\int xy^2\,\mathrm{d}x\) with the equation of the curve, no limits needed, \(\pi\) can be omitted
DM1 Attempting to use integration by parts; allow \(\pm\) between the two parts. No limits needed
A1 Correct integration, including limits; no substitution needed for this mark
A1 Correct after limits substituted
M1 Use of \(\dfrac{\pi\int xy^2\,\mathrm{d}x}{\pi\int y^2\,\mathrm{d}x}\) with their \(\pi\displaystyle\int xy^2\,\mathrm{d}x\). \(\pi\) must be seen in both numerator and denominator or in neither. This mark is not dependent on the previous M marks
A1cao Correct answer.