| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper integrals with infinite upper limit (exponential/IBP) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths improper integral question requiring standard integration by parts and a limit evaluation. Part (a) is definitional recall, part (b) is routine integration by parts, and part (c) applies a standard limiting process. While it's Further Maths content, the techniques are 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 |
|---|---|---|
| (a) The interval of integration is infinite | E1 | Total: 1 |
| (b) \(\int xe^{-3x}\,dx = -\frac{1}{3}xe^{-3x} - \int -\frac{1}{3}e^{-3x}\,dx\) | M1, A1 | |
| \(= -\frac{1}{3}xe^{-3x} - \frac{1}{9}e^{-3x} + c\) | A1/ | Total: 3 |
| (c) \(I = \int_1^{\infty} xe^{-3x}\,dx = \lim_{a \to \infty}\int_1^a xe^{-3x}\,dx\) | M1 | |
| \(\lim_{a \to \infty}\left[-\frac{1}{3}ae^{-3a} - \frac{1}{9}e^{-3a}\right] - \left[-\frac{4}{9}e^{-3}\right]\) | M1 | |
| \(\lim_{a \to \infty} ae^{-3a} = 0\) | M1 | |
| \(I = \frac{4}{9}e^{-3}\) | A1 | Total: 3 |
**(a)** The interval of integration is infinite | E1 | Total: 1 | OE
**(b)** $\int xe^{-3x}\,dx = -\frac{1}{3}xe^{-3x} - \int -\frac{1}{3}e^{-3x}\,dx$ | M1, A1 | | Reasonable attempt at parts
$= -\frac{1}{3}xe^{-3x} - \frac{1}{9}e^{-3x} + c$ | A1/ | Total: 3 | Condone absence of $+c$
**(c)** $I = \int_1^{\infty} xe^{-3x}\,dx = \lim_{a \to \infty}\int_1^a xe^{-3x}\,dx$ | M1 | |
$\lim_{a \to \infty}\left[-\frac{1}{3}ae^{-3a} - \frac{1}{9}e^{-3a}\right] - \left[-\frac{4}{9}e^{-3}\right]$ | M1 | | F(a) – F(1) with an indication of limit '$a \to \infty$'
$\lim_{a \to \infty} ae^{-3a} = 0$ | M1 | | For statement with limit/limiting process shown
$I = \frac{4}{9}e^{-3}$ | A1 | Total: 3 |4
\begin{enumerate}[label=(\alph*)]
\item Explain why $\int _ { 1 } ^ { \infty } x \mathrm { e } ^ { - 3 x } \mathrm {~d} x$ is an improper integral.
\item Find $\int x \mathrm { e } ^ { - 3 x } \mathrm {~d} x$.
\item Hence evaluate $\int _ { 1 } ^ { \infty } x \mathrm { e } ^ { - 3 x } \mathrm {~d} x$, showing the limiting process used.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q4 [7]}}