C4 June 2008 Q2
2.
| Scheme | Marks |
|---|---|
| \(\left\{\begin{aligned} u &= x &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = 1\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= \mathrm{e}^x &&\Rightarrow\ v = \mathrm{e}^x\end{aligned}\right\}\) | |
| \(\displaystyle\int x\mathrm{e}^x\,\mathrm{d}x = x\mathrm{e}^x - \int\mathrm{e}^x.1\,\mathrm{d}x\) | M1 A1 |
| \(= x\mathrm{e}^x - \displaystyle\int\mathrm{e}^x\,\mathrm{d}x\) | |
| \(= x\mathrm{e}^x - \mathrm{e}^x\ (+c)\) | A1 |
| (3) |
Notes
M1: Use of ‘integration by parts’ formula in the correct direction. (See note.) A1: Correct expression. (Ignore d\(x\))
A1: Correct integration with/without \(+\,c\)
Note integration by parts in the correct direction means that \(u\) and \(\tfrac{\mathrm{d}v}{\mathrm{d}x}\) must be assigned/used as \(u = x\) and \(\tfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^x\) in part (a) for example.
\(+\,c\) is not required in part (a).
| Scheme | Marks |
|---|---|
| \(\left\{\begin{aligned} u &= x^2 &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = 2x\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= \mathrm{e}^x &&\Rightarrow\ v = \mathrm{e}^x\end{aligned}\right\}\) | |
| \(\displaystyle\int x^2\mathrm{e}^x\,\mathrm{d}x = x^2\mathrm{e}^x - \int\mathrm{e}^x.2x\,\mathrm{d}x\) | M1 A1 |
| \(= x^2\mathrm{e}^x - 2\displaystyle\int x\mathrm{e}^x\,\mathrm{d}x\) | |
| \(= x^2\mathrm{e}^x - 2\left(x\mathrm{e}^x - \mathrm{e}^x\right) + c\) | A1 ISW |
| \(\left\{\begin{aligned}&= x^2\mathrm{e}^x - 2x\mathrm{e}^x + 2\mathrm{e}^x + c\\ &= \mathrm{e}^x\left(x^2 - 2x + 2\right) + c\end{aligned}\right\}\) Ignore subsequent working | |
| (3) | |
| (6 marks) |
Notes
M1: Use of ‘integration by parts’ formula in the correct direction. A1: Correct expression. (Ignore d\(x\))
A1 ISW: Correct expression including + c. (seen at any stage! in part (b)) You can ignore subsequent working.
\(+\,c\) is required in part (b). (This line of the printed note is partly cut off.)