| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Improper integral with parts |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question combining standard integration by parts with a limit to evaluate an improper integral. Part (a) is routine IBP, part (b) tests knowledge of exponential dominance (a standard result), and part (c) applies the limit. While it's FP3 content, the execution is mechanical with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int xe^{-2x}\,dx = -\frac{1}{2}xe^{-2x} - \int -\frac{1}{2}e^{-2x}\,dx\) | M1 | Reasonable attempt at parts |
| A1 | ||
| \(= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \{+c\}\) | A1\(\checkmark\) | Condone absence of \(+c\) |
| \(\int_0^a xe^{-2x}\,dx = -\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} - (0 - \frac{1}{4})\) | M1 | \(F(a) - F(0)\) |
| \(= \frac{1}{4} - \frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a}\) | A1 | Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lim_{a\to\infty} a^k e^{-2a} = 0\) | B1 | Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^\infty xe^{-2x}\,dx = \lim_{a\to\infty}\left\{\frac{1}{4} - \frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a}\right\}\) | M1 | If this line is missing then 0/2 |
| \(= \frac{1}{4} - 0 - 0 = \frac{1}{4}\) | A1\(\checkmark\) | 2 marks; On candidate's "\(1/4\)" in part (a). B1 must have been earned |
# Question 2:
## Part 2(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int xe^{-2x}\,dx = -\frac{1}{2}xe^{-2x} - \int -\frac{1}{2}e^{-2x}\,dx$ | M1 | Reasonable attempt at parts |
| | A1 | |
| $= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \{+c\}$ | A1$\checkmark$ | Condone absence of $+c$ |
| $\int_0^a xe^{-2x}\,dx = -\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} - (0 - \frac{1}{4})$ | M1 | $F(a) - F(0)$ |
| $= \frac{1}{4} - \frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a}$ | A1 | Total: 5 marks |
## Part 2(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lim_{a\to\infty} a^k e^{-2a} = 0$ | B1 | Total: 1 mark |
## Part 2(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^\infty xe^{-2x}\,dx = \lim_{a\to\infty}\left\{\frac{1}{4} - \frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a}\right\}$ | M1 | If this line is missing then 0/2 |
| $= \frac{1}{4} - 0 - 0 = \frac{1}{4}$ | A1$\checkmark$ | 2 marks; On candidate's "$1/4$" in part (a). B1 must have been earned |
---2
\begin{enumerate}[label=(\alph*)]
\item Find $\int _ { 0 } ^ { a } x \mathrm { e } ^ { - 2 x } \mathrm {~d} x$, where $a > 0$.
\item Write down the value of $\lim _ { a \rightarrow \infty } a ^ { k } \mathrm { e } ^ { - 2 a }$, where $k$ is a positive constant.
\item Hence find $\int _ { 0 } ^ { \infty } x \mathrm { e } ^ { - 2 x } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2006 Q2 [8]}}