C4 June 2013 Q1
1.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int x^2\mathrm{e}^x\,\mathrm{d}x\), 1st Application: \(\left\{\begin{aligned} u &= x^2 &\Rightarrow\ \dfrac{\mathrm{d}u}{\mathrm{d}x} &= 2x\\ \dfrac{\mathrm{d}v}{\mathrm{d}x} &= \mathrm{e}^x &\Rightarrow\ v &= \mathrm{e}^x\end{aligned}\right\}\), 2nd Application: \(\left\{\begin{aligned} u &= x &\Rightarrow\ \dfrac{\mathrm{d}u}{\mathrm{d}x} &= 1\\ \dfrac{\mathrm{d}v}{\mathrm{d}x} &= \mathrm{e}^x &\Rightarrow\ v &= \mathrm{e}^x\end{aligned}\right\}\) | |
| \(= x^2\mathrm{e}^x - \displaystyle\int 2x\mathrm{e}^x\,\mathrm{d}x\) \(x^2\mathrm{e}^x - \displaystyle\int \lambda x\mathrm{e}^x\{\mathrm{d}x\},\ \lambda > 0\) \(x^2\mathrm{e}^x - \displaystyle\int 2x\mathrm{e}^x\{\mathrm{d}x\}\) | M1 A1 oe |
| \(= x^2\mathrm{e}^x - 2\left(x\mathrm{e}^x - \displaystyle\int \mathrm{e}^x\,\mathrm{d}x\right)\) Either \(\pm Ax^2\mathrm{e}^x \pm Bx\mathrm{e}^x \pm C\displaystyle\int \mathrm{e}^x\{\mathrm{d}x\}\) or for \(\pm K\displaystyle\int x\mathrm{e}^x\{\mathrm{d}x\} \to \pm K\left(x\mathrm{e}^x - \displaystyle\int \mathrm{e}^x\{\mathrm{d}x\}\right)\) | M1 |
| \(= x^2\mathrm{e}^x - 2(x\mathrm{e}^x - \mathrm{e}^x)\ \{+ c\}\) \(\pm Ax^2\mathrm{e}^x \pm Bx\mathrm{e}^x \pm C\mathrm{e}^x\) Correct answer, with/without \(+ c\) | M1 A1 |
| (5) |
Notes
M1: Integration by parts is applied in the form \(x^2\mathrm{e}^x - \displaystyle\int \lambda x\mathrm{e}^x\{\mathrm{d}x\}\), where \(\lambda > 0\). (must be in this form).
A1: \(x^2\mathrm{e}^x - \displaystyle\int 2x\mathrm{e}^x\{\mathrm{d}x\}\) or equivalent.
M1: Either achieving a result in the form \(\pm Ax^2\mathrm{e}^x \pm Bx\mathrm{e}^x \pm C\displaystyle\int \mathrm{e}^x\{\mathrm{d}x\}\) (can be implied)
(where \(A \neq 0\), \(B \neq 0\) and \(C \neq 0\)) or for \(\pm K\displaystyle\int x\mathrm{e}^x\{\mathrm{d}x\} \to \pm K\left(x\mathrm{e}^x - \displaystyle\int \mathrm{e}^x\{\mathrm{d}x\}\right)\)
M1: \(\pm Ax^2\mathrm{e}^x \pm Bx\mathrm{e}^x \pm C\mathrm{e}^x\) (where \(A \neq 0\), \(B \neq 0\) and \(C \neq 0\))
A1: \(x^2\mathrm{e}^x - 2(x\mathrm{e}^x - \mathrm{e}^x)\) or \(x^2\mathrm{e}^x - 2x\mathrm{e}^x + 2\mathrm{e}^x\) or \((x^2 - 2x + 2)\mathrm{e}^x\) or equivalent with/without \(+ c\).
| Scheme | Marks |
|---|---|
| \(\left\{\left[x^2\mathrm{e}^x - 2(x\mathrm{e}^x - \mathrm{e}^x)\right]_0^1\right\}\) | |
| \(= \left(1^2\mathrm{e}^1 - 2(1\mathrm{e}^1 - \mathrm{e}^1)\right) - \left(0^2\mathrm{e}^0 - 2(0\mathrm{e}^0 - \mathrm{e}^0)\right)\) Applies limits of 1 and 0 to an expression of the form \(\pm Ax^2\mathrm{e}^x \pm Bx\mathrm{e}^x \pm C\mathrm{e}^x\), \(A \neq 0\), \(B \neq 0\) and \(C \neq 0\) and subtracts the correct way round. | M1 |
| \(= \mathrm{e} - 2\) \(\mathrm{e} - 2\) cso | A1 oe |
| (2) | |
| (7 marks) |
Notes
M1: Complete method of applying limits of 1 and 0 to their part (a) answer in the form \(\pm Ax^2\mathrm{e}^x \pm Bx\mathrm{e}^x \pm C\mathrm{e}^x\),
(where \(A \neq 0\), \(B \neq 0\) and \(C \neq 0\)) and subtracting the correct way round.
Evidence of a proper consideration of the limit of 0 (as detailed above) is needed for M1.
So, just subtracting zero is M0.
A1: \(\mathrm{e} - 2\) or \(\mathrm{e}^1 - 2\) or \(-2 + \mathrm{e}\). Do not allow \(\mathrm{e} - 2\mathrm{e}^0\) unless simplified to give \(\mathrm{e} - 2\).
Note: that 0.718... without seeing \(\mathrm{e} - 2\) or equivalent is A0.
WARNING: Please note that this A1 mark is for correct solution only.
So incorrect \(\left[\ldots\ldots\right]_0^1\) leading to \(\mathrm{e} - 2\) is A0.
Note: If their part (a) is correct candidates can get M1A1 in part (b) for \(\mathrm{e} - 2\) from no working.
Note: 0.718... from no working is M0A0