| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Improper integral with parts |
| Difficulty | Challenging +1.8 This FP3 question requires multiple sophisticated techniques: recognizing a limit involving exponential dominance, executing a non-obvious substitution with implicit differentiation, and carefully handling an improper integral with proper limiting arguments. While the individual steps are methodical once the approach is identified, the substitution is not standard and the connection between parts (b) and (c) requires algebraic manipulation and insight beyond typical A-level integration questions. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = xe^{-x} + 1 \Rightarrow du = (e^{-x} - xe^{-x})dx\) | M1 | Attempts to find \(du\) |
| \(\int \frac{e^{-x}(1-x)}{xe^{-x}+1}dx = \int \frac{1}{u}du = \ln u + c\) | ||
| \(= \ln(xe^{-x}+1)\{+c\}\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{1-x}{x+e^x}dx = \int \frac{e^{-x}(1-x)}{xe^{-x}+1}dx\) | B1 | |
| \(\int_1^\infty \frac{1-x}{x+e^x}dx = \lim_{a\to\infty}\left[\ln(xe^{-x}+1)\right]_1^a\) | ||
| \(= \lim_{a\to\infty}\left\{\ln(ae^{-a}+1)\right\} - \ln(e^{-1}+1)\) | M1 | For using part (b) and \(F(B)-F(A)\) |
| \(= \ln\left\{\lim_{a\to\infty}(ae^{-a}+1)\right\} - \ln(e^{-1}+1)\) | ||
| \(= \ln 1 - \ln(e^{-1}+1) = -\ln(e^{-1}+1)\) | M1, A1 | 4 marks total |
# Question 7:
## Part 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0$ | B1 | 1 mark total |
## Part 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = xe^{-x} + 1 \Rightarrow du = (e^{-x} - xe^{-x})dx$ | M1 | Attempts to find $du$ |
| $\int \frac{e^{-x}(1-x)}{xe^{-x}+1}dx = \int \frac{1}{u}du = \ln u + c$ | | |
| $= \ln(xe^{-x}+1)\{+c\}$ | A1 | 2 marks total | Condone missing $c$ |
## Part 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{1-x}{x+e^x}dx = \int \frac{e^{-x}(1-x)}{xe^{-x}+1}dx$ | B1 | |
| $\int_1^\infty \frac{1-x}{x+e^x}dx = \lim_{a\to\infty}\left[\ln(xe^{-x}+1)\right]_1^a$ | | |
| $= \lim_{a\to\infty}\left\{\ln(ae^{-a}+1)\right\} - \ln(e^{-1}+1)$ | M1 | For using part (b) and $F(B)-F(A)$ |
| $= \ln\left\{\lim_{a\to\infty}(ae^{-a}+1)\right\} - \ln(e^{-1}+1)$ | | |
| $= \ln 1 - \ln(e^{-1}+1) = -\ln(e^{-1}+1)$ | M1, A1 | 4 marks total | For using limiting process |7
\begin{enumerate}[label=(\alph*)]
\item Write down the value of
$$\lim _ { x \rightarrow \infty } x \mathrm { e } ^ { - x }$$
\item Use the substitution $u = x \mathrm { e } ^ { - x } + 1$ to find $\int \frac { \mathrm { e } ^ { - x } ( 1 - x ) } { x \mathrm { e } ^ { - x } + 1 } \mathrm {~d} x$.
\item Hence evaluate $\int _ { 1 } ^ { \infty } \frac { 1 - x } { x + \mathrm { e } ^ { x } } \mathrm {~d} x$, showing the limiting process used.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q7 [7]}}