| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Improper integral to infinity |
| Difficulty | Moderate -0.3 This is a straightforward improper integral question requiring standard exponential integration and taking a limit as a→∞. Part (i) uses direct integration of e^(-x) and e^(-3x) with reverse chain rule, while part (ii) simply requires evaluating the limit of the result. The techniques are routine for P2 level with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Integrate to obtain form \(pe^{-x} + qe^{-3x}\) where \(p \neq 1, q \neq 6\) | M1 | |
| Obtain \(-e^{-x} - 2e^{-3x}\) (allow unsimplified) | A1 | |
| Apply both limits to \(pe^{-x} + qe^{-3x}\) (allow \(p = 1, q = 6\)) | M1 | |
| Obtain \(3 - e^{-\pi} - 2e^{-3\pi}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| State 3 following a result of the form \(k + pe^{-x} + qe^{-3x}\) | B1 | [1] |
**(i)**
Integrate to obtain form $pe^{-x} + qe^{-3x}$ where $p \neq 1, q \neq 6$ | M1 |
Obtain $-e^{-x} - 2e^{-3x}$ (allow unsimplified) | A1 |
Apply both limits to $pe^{-x} + qe^{-3x}$ (allow $p = 1, q = 6$) | M1 |
Obtain $3 - e^{-\pi} - 2e^{-3\pi}$ | A1 | [4]
**(ii)**
State 3 following a result of the form $k + pe^{-x} + qe^{-3x}$ | B1 | [1]2 (i) Find $\int _ { 0 } ^ { a } \left( \mathrm { e } ^ { - x } + 6 \mathrm { e } ^ { - 3 x } \right) \mathrm { d } x$, where $a$ is a positive constant.\\
(ii) Deduce the value of $\int _ { 0 } ^ { \infty } \left( \mathrm { e } ^ { - x } + 6 \mathrm { e } ^ { - 3 x } \right) \mathrm { d } x$.
\hfill \mbox{\textit{CAIE P2 2014 Q2 [5]}}