| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| 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 +0.8 This is a Further Maths FP3 question requiring understanding of improper integrals and integration by parts with an exponential. Part (a) tests conceptual understanding (straightforward - infinite limit), while part (b) requires correct application of integration by parts, proper limit notation, and evaluation of limits involving exponentials. The technique is standard for FP3 but more sophisticated than typical A-level Core questions, placing it moderately above average difficulty. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| The upper limit of integration is infinity | B1 | Must mention infinite/unbounded limit |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_2^t (x-2)e^{-2x}\,dx\) with limit as \(t \to \infty\) stated | M1 | Limiting process must be shown |
| Integration by parts: \(u = (x-2)\), \(\frac{dv}{dx} = e^{-2x}\) | M1 | Correct choice for IBP |
| \(\left[{-\frac{1}{2}(x-2)e^{-2x}}\right]_2^t + \int_2^t \frac{1}{2}e^{-2x}\,dx\) | A1 | Correct expression after first IBP |
| \(= \left[-\frac{1}{2}(x-2)e^{-2x} - \frac{1}{4}e^{-2x}\right]_2^t\) | A1 | Correct integration of second term |
| \(= \left(-\frac{1}{2}(t-2)e^{-2t} - \frac{1}{4}e^{-2t}\right) - \left(0 - \frac{1}{4}e^{-4}\right)\) | A1 | Correct substitution of limits |
| As \(t \to \infty\), \(te^{-2t} \to 0\) and \(e^{-2t} \to 0\), so integral \(= \frac{1}{4}e^{-4}\) | A1 | Correct limit evaluated with justification |
# Question 4:
## Part (a)
| The upper limit of integration is infinity | B1 | Must mention infinite/unbounded limit |
## Part (b)
| $\int_2^t (x-2)e^{-2x}\,dx$ with limit as $t \to \infty$ stated | M1 | Limiting process must be shown |
| Integration by parts: $u = (x-2)$, $\frac{dv}{dx} = e^{-2x}$ | M1 | Correct choice for IBP |
| $\left[{-\frac{1}{2}(x-2)e^{-2x}}\right]_2^t + \int_2^t \frac{1}{2}e^{-2x}\,dx$ | A1 | Correct expression after first IBP |
| $= \left[-\frac{1}{2}(x-2)e^{-2x} - \frac{1}{4}e^{-2x}\right]_2^t$ | A1 | Correct integration of second term |
| $= \left(-\frac{1}{2}(t-2)e^{-2t} - \frac{1}{4}e^{-2t}\right) - \left(0 - \frac{1}{4}e^{-4}\right)$ | A1 | Correct substitution of limits |
| As $t \to \infty$, $te^{-2t} \to 0$ and $e^{-2t} \to 0$, so integral $= \frac{1}{4}e^{-4}$ | A1 | Correct limit evaluated with justification |
---4
\begin{enumerate}[label=(\alph*)]
\item Explain why $\int _ { 2 } ^ { \infty } ( x - 2 ) \mathrm { e } ^ { - 2 x } \mathrm {~d} x$ is an improper integral.
\item Evaluate $\int _ { 2 } ^ { \infty } ( x - 2 ) \mathrm { e } ^ { - 2 x } \mathrm {~d} x$, showing the limiting process used.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2015 Q4 [7]}}