4 It is given that
$$I _ { n } = \int _ { 0 } ^ { 1 } \frac { x ^ { n } } { \sqrt { } ( 1 + 2 x ) } \mathrm { d } x$$
Show that, for \(n \geqslant 1\),
$$( 2 n + 1 ) I _ { n } = \sqrt { } 3 - n I _ { n - 1 }$$
Show that
$$I _ { 3 } = \frac { 2 } { 35 } ( \sqrt { } 3 + 1 )$$
Show mark scheme
Show mark scheme source
Question 4:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(I_n\left[x^n(1+2x)^{\frac{1}{2}}\right]_0^1 - \int_0^1 \frac{nx^{n-1}(1+2x)}{(1+2x)^{\frac{1}{2}}}\,dx\) M1A1
Integrates by parts
\(= \sqrt{3} - n\int_0^1 \frac{x^{n-1}}{(1+2x)^{\frac{1}{2}}}\,dx - 2n\int_0^1\frac{x^n}{(1+2x)^{\frac{1}{2}}}\,dx\) \(\Rightarrow (2n+1)I_n = \sqrt{3} - nI_{n-1}\) (AG) A1
Obtains result
Alternatively: \(\frac{d}{dx}\left\{x^n(1+2x)^{\frac{1}{2}}\right\} = (1+2x)^{\frac{1}{2}}\cdot nx^{n-1} + x^n(1+2x)^{-\frac{1}{2}}\)(M1)
\(\Rightarrow \left[x^n(1+2x)^{\frac{1}{2}}\right]_0^1 = \frac{n(1+2x)x^{n-1}}{\sqrt{1+2x}} + I_n\) (A1)
\(\Rightarrow (2n+1)I_n = \sqrt{3} - nI_{n-1}\) (AG) (A1)
\(I_0 = \left[\sqrt{1+2x}\right]_0^1 = \sqrt{3}-1\) B1
Finds \(I_0\) (or \(I_1\))
\(3I_1 = \sqrt{3}-(\sqrt{3}-1) \Rightarrow I_1 = \frac{1}{3}\) M1
Uses reduction formula
\(\Rightarrow I_2 = \frac{\sqrt{3}}{5} - \frac{2}{15} \Rightarrow I_3 = \frac{2}{35}(\sqrt{3}+1)\) (AG) A1A1
Finds \(I_2\) and \(I_3\)
Copy
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n\left[x^n(1+2x)^{\frac{1}{2}}\right]_0^1 - \int_0^1 \frac{nx^{n-1}(1+2x)}{(1+2x)^{\frac{1}{2}}}\,dx$ | M1A1 | Integrates by parts |
| $= \sqrt{3} - n\int_0^1 \frac{x^{n-1}}{(1+2x)^{\frac{1}{2}}}\,dx - 2n\int_0^1\frac{x^n}{(1+2x)^{\frac{1}{2}}}\,dx$ | | |
| $\Rightarrow (2n+1)I_n = \sqrt{3} - nI_{n-1}$ (AG) | A1 | Obtains result |
| **Alternatively:** $\frac{d}{dx}\left\{x^n(1+2x)^{\frac{1}{2}}\right\} = (1+2x)^{\frac{1}{2}}\cdot nx^{n-1} + x^n(1+2x)^{-\frac{1}{2}}$ | (M1) | |
| $\Rightarrow \left[x^n(1+2x)^{\frac{1}{2}}\right]_0^1 = \frac{n(1+2x)x^{n-1}}{\sqrt{1+2x}} + I_n$ | (A1) | |
| $\Rightarrow (2n+1)I_n = \sqrt{3} - nI_{n-1}$ (AG) | (A1) | |
| $I_0 = \left[\sqrt{1+2x}\right]_0^1 = \sqrt{3}-1$ | B1 | Finds $I_0$ (or $I_1$) |
| $3I_1 = \sqrt{3}-(\sqrt{3}-1) \Rightarrow I_1 = \frac{1}{3}$ | M1 | Uses reduction formula |
| $\Rightarrow I_2 = \frac{\sqrt{3}}{5} - \frac{2}{15} \Rightarrow I_3 = \frac{2}{35}(\sqrt{3}+1)$ (AG) | A1A1 | Finds $I_2$ and $I_3$ |
---
Show LaTeX source
Copy
4 It is given that
$$I _ { n } = \int _ { 0 } ^ { 1 } \frac { x ^ { n } } { \sqrt { } ( 1 + 2 x ) } \mathrm { d } x$$
Show that, for $n \geqslant 1$,
$$( 2 n + 1 ) I _ { n } = \sqrt { } 3 - n I _ { n - 1 }$$
Show that
$$I _ { 3 } = \frac { 2 } { 35 } ( \sqrt { } 3 + 1 )$$
\hfill \mbox{\textit{CAIE FP1 2013 Q4 [7]}}