| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Exponential times polynomial |
| Difficulty | Standard +0.8 This is a standard Further Maths reduction formula question requiring integration by parts to establish the recurrence relation, then iterative application to find I₃. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in FP2 and follows a predictable pattern. The exponential-polynomial combination is a classic type, making it moderately challenging but not requiring novel insight. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + 4x + 8 = (x+2)^2 + 4\) | M1 | Complete the square in order to use standard form |
| \(\int_0^1 \frac{1}{\sqrt{x^2+4x+8}} dx = \int_0^1 \frac{1}{\sqrt{(x+2)^2+4}} dx\) | M1 | Use correct standard form in integration |
| \(= [\sinh^{-1} \frac{x+2}{2}] = \sinh^{-1}(3/2) - \sinh^{-1} 1\) | A1 | Answer in \(\sinh^{-1}\) form |
| \(= \ln(3 + \sqrt{1+9/4}) - \ln(1 + \sqrt{2}) = \ln(\frac{3+\sqrt{13}}{2+2\sqrt{2}})\) | M1 | Attempt to turn into log form |
| \(= \ln(\frac{3+\sqrt{13}}{2(1+\sqrt{2})})\) | A1 | www isw |
| ALT: \(\int_0^1 \frac{1}{\sqrt{(x+2)^2+4}} dx = [\ln((x+2) + \sqrt{(x+2)^2+4})]_0^1\) | M1 | Attempt to use Standard form |
| \(= \ln(3 + \sqrt{13}) - \ln(2 + \sqrt{8}) = \ln(\frac{3+\sqrt{13}}{2+2\sqrt{2}})\) | A1 | www isw |
| \(= \ln(\frac{3+\sqrt{13}}{2+2\sqrt{2}})\) | M1 | Limits |
| A1 | www isw | |
| ALT: \(x + 2 = 2\tan\theta \Rightarrow I = [\ln(\sec\theta + \tan\theta)]_{\tan^{-1}3/2}^{\tan^{-1}1/2}\) | M1 | Substitution |
| A1 | Indefinite integral | |
| M1 | Deal with limits | |
| A1 | www isw |
$x^2 + 4x + 8 = (x+2)^2 + 4$ | M1 | Complete the square in order to use standard form
$\int_0^1 \frac{1}{\sqrt{x^2+4x+8}} dx = \int_0^1 \frac{1}{\sqrt{(x+2)^2+4}} dx$ | M1 | Use correct standard form in integration
$= [\sinh^{-1} \frac{x+2}{2}] = \sinh^{-1}(3/2) - \sinh^{-1} 1$ | A1 | Answer in $\sinh^{-1}$ form
$= \ln(3 + \sqrt{1+9/4}) - \ln(1 + \sqrt{2}) = \ln(\frac{3+\sqrt{13}}{2+2\sqrt{2}})$ | M1 | Attempt to turn into log form
$= \ln(\frac{3+\sqrt{13}}{2(1+\sqrt{2})})$ | A1 | www isw
ALT: $\int_0^1 \frac{1}{\sqrt{(x+2)^2+4}} dx = [\ln((x+2) + \sqrt{(x+2)^2+4})]_0^1$ | M1 | Attempt to use Standard form
$= \ln(3 + \sqrt{13}) - \ln(2 + \sqrt{8}) = \ln(\frac{3+\sqrt{13}}{2+2\sqrt{2}})$ | A1 | www isw
$= \ln(\frac{3+\sqrt{13}}{2+2\sqrt{2}})$ | M1 | Limits
| A1 | www isw
ALT: $x + 2 = 2\tan\theta \Rightarrow I = [\ln(\sec\theta + \tan\theta)]_{\tan^{-1}3/2}^{\tan^{-1}1/2}$ | M1 | Substitution
| A1 | Indefinite integral
| M1 | Deal with limits
| A1 | www isw6 (i) It is given that, for non-negative integers $n$,
$$I _ { n } = \int _ { 0 } ^ { 1 } \mathrm { e } ^ { - x } x ^ { n } \mathrm {~d} x$$
Prove that, for $n \geqslant 1$,
$$I _ { n } = n I _ { n - 1 } - \mathrm { e } ^ { - 1 } .$$
(ii) Evaluate $I _ { 3 }$, giving the answer in terms of e.
\hfill \mbox{\textit{OCR FP2 Q6 [8]}}