| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts with a definite integral. While it involves multiple steps (choosing u and dv, applying integration by parts, evaluating limits, and algebraic manipulation), the technique is routine for Further Maths students and follows a well-practiced pattern. The presence of the square root adds mild complexity but the approach is straightforward once the correct parts are identified. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
$I_n = \left[x^n \cdot \frac{1}{3}(2x+1)^{\frac{3}{2}}\right]_0^4 - \int_0^4 nx^{n-1} \cdot \frac{1}{3}(2x+1)^{\frac{3}{2}}\ \mathrm{d}x$ Use of parts **M1**
For $\int\sqrt{2x+1} = \frac{1}{3}(2x+1)^{\frac{3}{2}}$ at any stage **B1**
$3I_n = \left(4^n \cdot 27 - 0\right) - n\int_0^4 x^{n-1}(2x+1)\sqrt{2x+1}\ \mathrm{d}x$ Splitting the power of $2x+1$ **M1**
$-n(2I_n + I_{n-1})$ **A1**
$\Rightarrow (2n+3)I_n = 27 \times 4^n - nI_{n-1}$ Legitimately (**Answer Given**) **A1** [5]4 Let $I _ { n } = \int _ { 0 } ^ { 4 } x ^ { n } \sqrt { 2 x + 1 } \mathrm {~d} x$ for $n \geqslant 0$. Show that, for $n \geqslant 1$,
$$( 2 n + 3 ) I _ { n } = 27 \times 4 ^ { n } - n I _ { n - 1 }$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q4 [5]}}