FP2 June 2018 Q8
8.
Given that \(y = 0\) when \(x = 1\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int 2x^5\mathrm{e}^{-x^2}\,\mathrm{d}x\) | |
| \(t = x^2 \Rightarrow \mathrm{d}t = 2x\,\mathrm{d}x\) or \(\mathrm{d}x = \dfrac{1}{2}t^{-\frac{1}{2}}\mathrm{d}t\) oe May be implied by subsequent work | M1 |
| \(\displaystyle\int 2x^5\mathrm{e}^{-x^2}\,\mathrm{d}x = \int t^2\mathrm{e}^{-t}\,\mathrm{d}t\) Integral in terms of \(t\) only required. d\(t\) may be implied Must have attempted to change d\(x\) to d\(t\) (ie not just used d\(x\) = d\(t\)) | M1 |
| \(= -t^2\mathrm{e}^{-t} + 2\displaystyle\int t\mathrm{e}^{-t}\,\mathrm{d}t\) Use of integration by parts Reduce the power of \(t\). Sign errors are allowed. \(\displaystyle\int kt^p\mathrm{e}^{-t} \to \pm kt^p\mathrm{e}^{-t} \pm A\int t^{p-1}\mathrm{e}^{-t}\,\mathrm{d}t\) | M1 |
| \(= -t^2\mathrm{e}^{-t} - 2t\mathrm{e}^{-t} + 2\displaystyle\int \mathrm{e}^{-t}\,\mathrm{d}t\) Use of integration by parts again in the same direction | dM1 |
| \(= -t^2\mathrm{e}^{-t} - 2t\mathrm{e}^{-t} - 2\mathrm{e}^{-t}\ (+C)\) oe Correct integration, constant not needed | A1 |
| \(= -x^4\mathrm{e}^{-x^2} - 2x^2\mathrm{e}^{-x^2} - 2\mathrm{e}^{-x^2}\ (+C)\) oe Reverse substitution, constant not needed. This mark cannot be recovered in (b) | A1 |
| ALTs Attempts without substitution which may merit part marks – send to review. | |
| (6) |
Notes
Some common alternative forms for the answers:
NB: This list is not exhaustive.
1) \(-x^4\mathrm{e}^{-x^2} - 2x^2\mathrm{e}^{-x^2} - 2\mathrm{e}^{-x^2}\ (+C)\)
2) \(\mathrm{e}^{-x^2}\left(-x^4 - 2x^2 - 2\right)\ (+C)\)
3) \(-\mathrm{e}^{-x^2}\left(x^4 + 2x^2 + 2\right)\ (+C)\)
4) \(\dfrac{-\left(x^4 + 2x^2 + 2\right)}{\mathrm{e}^{x^2}}\ (+C)\)
| Scheme | Marks |
|---|---|
| \(x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 4y = 2x^2\mathrm{e}^{-x^2}\) | |
| Integrating Factor \(\mathrm{e}^{\int\frac{4}{x}\mathrm{d}x} = x^4\) Use of \(x^4\) seen | B1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^4y\right) = 2x^5\mathrm{e}^{-x^2}\) or \(x^4y = \displaystyle\int 2x^5\mathrm{e}^{-x^2}\,\mathrm{d}x\) Multiply through by their IF | M1 |
| \(x^4y = -x^4\mathrm{e}^{-x^2} - 2x^2\mathrm{e}^{-x^2} - 2\mathrm{e}^{-x^2}\ (+C)\) Use their answer for (a), which must be a function of \(x\), to integrate RHS | A1ft |
| \(y = -\mathrm{e}^{-x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^4} + \dfrac{C}{x^4}\) Complete to \(y = \ldots\) Include the constant and deal with it correctly Not follow through | A1 |
| (4) |
Notes
ALT:
| Scheme | Marks |
|---|---|
| Use the same substitution as in (a) Following work uses the work shown in (a) rather than just the final answer. No marks until a first order exact equation in \(y\) and \(t\) reached and an attempt is made to solve this. | |
| \(t = x^2 \Rightarrow \mathrm{d}t = 2x\,\mathrm{d}x\) or \(\mathrm{d}x = \dfrac{1}{2}t^{-\frac{1}{2}}\mathrm{d}t\), | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t}\times\dfrac{\mathrm{d}t}{\mathrm{d}x} \qquad x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2t\dfrac{\mathrm{d}y}{\mathrm{d}t} \qquad\) Equation becomes \(2t\dfrac{\mathrm{d}y}{\mathrm{d}t} + 4y = 2t\mathrm{e}^{-t}\) | |
| Integrating Factor \(\mathrm{e}^{\int\frac{2}{t}\mathrm{d}t} = t^2\) Use of \(t^2\) seen | B1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(t^2y\right) = t^2\mathrm{e}^{-t}\) or \(t^2y = \displaystyle\int t^2\mathrm{e}^{-t}\,\mathrm{d}t\) Multiply through by IF | M1 |
| \(t^2y = -t^2\mathrm{e}^{-t} - 2t\mathrm{e}^{-t} - 2\mathrm{e}^{-t}\ (+C)\) oe Use their work in (a) to integrate RHS | A1ft |
| \(y = -\mathrm{e}^{-x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^4} + \dfrac{C}{x^4}\) Reverse the substitution Complete to \(y = \ldots\) Include the constant and deal with it correctly Not follow through | A1 |
| (4) |
(Corrected from the printed mark scheme: the transformed equation is printed as \(2t\dfrac{\mathrm{d}t}{\mathrm{d}x} + 4y = 2t\mathrm{e}^{-t}\); it is \(2t\dfrac{\mathrm{d}y}{\mathrm{d}t} + 4y = 2t\mathrm{e}^{-t}\).)
Some common alternative forms for the answers:
1) \(y = -\mathrm{e}^{-x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^4} + \dfrac{C}{x^4}\)
2) \(y = \mathrm{e}^{-x^2}\left(-1 - \dfrac{2}{x^2} - \dfrac{2}{x^4}\right) + \dfrac{C}{x^4}\)
3) \(y = -\mathrm{e}^{-x^2}\left(1 + \dfrac{2}{x^2} + \dfrac{2}{x^4}\right) + \dfrac{C}{x^4}\)
4) \(y = \dfrac{-\left(x^4 + 2x^2 + 2\right)}{x^4\mathrm{e}^{x^2}} + \dfrac{C}{x^4}\)
| Scheme | Marks |
|---|---|
| \(0 = -\mathrm{e}^{-1} - 2\mathrm{e}^{-1} - 2\mathrm{e}^{-1} + C\) Attempt to substitute \(x = 1,\ y = 0\) into their \(y\) provided it includes a constant | M1 |
| \(\Rightarrow C = 5\mathrm{e}^{-1}\) oe NB: Not ft so must have been obtained using a correct expression for \(y\) | A1 |
| \(y = -\mathrm{e}^{-x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^4} + \dfrac{5\mathrm{e}^{-1}}{x^4}\) Must start \(y = \ldots\) Follow through their \(C\) and expression for \(y\) | A1ft |
| (3) | |
| (13 marks) |
Notes
Some common alternative forms for the answers:
1) \(y = -\mathrm{e}^{-x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^2} - \dfrac{2\mathrm{e}^{-x^2}}{x^4} + \dfrac{5\mathrm{e}^{-1}}{x^4}\)
2) \(y = \mathrm{e}^{-x^2}\left(-1 - \dfrac{2}{x^2} - \dfrac{2}{x^4}\right) + \dfrac{5\mathrm{e}^{-1}}{x^4}\)
3) \(y = -\mathrm{e}^{-x^2}\left(1 + \dfrac{2}{x^2} + \dfrac{2}{x^4}\right) + \dfrac{5\mathrm{e}^{-1}}{x^4}\)
4) \(y = \dfrac{-\left(x^4 + 2x^2 + 2\right)}{x^4\mathrm{e}^{x^2}} + \dfrac{5\mathrm{e}^{-1}}{x^4}\)