| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Compound expressions with binomial expansion |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts with clear substitution choices. Part (a) follows a well-practiced technique (letting u=(3-x²)^n, dv=dx, then manipulating), and part (b) is straightforward recursive calculation starting from I₀=√3. While it requires careful algebra and is from FP3, the method is entirely routine for students at this level with no novel insight needed. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int_0^{\sqrt{3}}(3-x^2)^n dx = \left[x(3-x^2)^n\right]_0^{\sqrt{3}}+\int_0^{\sqrt{3}}2x^2 n(3-x^2)^{n-1}dx\) | M1A1 | M1: Integration by parts in correct direction; A1: Correct expression (ignore limits) |
| \(= 0-2n\displaystyle\int_0^{\sqrt{3}}(3-x^2-3)(3-x^2)^{n-1}dx\) | dM1 | Substitutes limits and uses \(x^2 = x^2-3+3\); dependent on first M |
| \(= -2n\displaystyle\int_0^{\sqrt{3}}(3-x^2)^n dx + 6n\int_0^{\sqrt{3}}(3-x^2)^{n-1}dx\) | A1 | Correct expressions |
| \(= 6nI_{n-1}-2nI_n\) | ddM1 | Substitutes for \(I_{n-1}\) and \(I_n\); dependent on both M's |
| \(I_n = \dfrac{6n}{2n+1}I_{n-1}\;*\) | A1* | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \displaystyle\int_0^{\sqrt{3}}(3-x^2)^{n-1}(3-x^2)\,dx = 3\int_0^{\sqrt{3}}(3-x^2)^{n-1}dx - \int_0^{\sqrt{3}}x^2(3-x^2)^{n-1}dx\) | dM1 | M1: Writes bracket as product and separates; this is the second M and depends on the first M below |
| \(= 3I_{n-1}+\left\{\left[\dfrac{x(3-x^2)^n}{-2n}\right]_0^{\sqrt{3}}-\displaystyle\int_0^{\sqrt{3}}\dfrac{(3-x^2)^n}{-2n}dx\right\}\) | M1A1 | M1: Parts in correct direction (First M1); A1: Correct expression (First A1) |
| \(= 3I_{n-1} - \displaystyle\int_0^{\sqrt{3}}\dfrac{(3-x^2)^n}{2n}\,dx\) | A1 | Correct expression with no errors |
| \(= 3I_{n-1} - \dfrac{1}{2n}I_{n-1}\) ... wait, \(= 3I_{n-1}-\dfrac{1}{2n}I_n\) | ddM1 | Substitutes for \(I_{n-1}\) and \(I_n\); dependent on both M's |
| \(I_n = \dfrac{6n}{2n+1}I_{n-1}\;*\) | A1 | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_0=\sqrt{3}\) or \(I_1=2\sqrt{3}\) | B1 | |
| \(I_4 = \dfrac{24}{9}I_3\) | M1 | Attempt \(I_4\) in terms of \(I_3\) |
| \(I_4 = \dfrac{24}{9}\cdot\dfrac{18}{7}I_2 = \dfrac{24}{9}\cdot\dfrac{18}{7}\cdot\dfrac{12}{5}I_1\) | M1A1 | M1: Attempt \(I_4\) in terms of \(I_1\); A1: Correct expression for \(I_4\) as shown or correct numerical expression |
| \(I_4 = \dfrac{24}{9}\cdot\dfrac{18}{7}\cdot\dfrac{12}{5}\cdot\dfrac{6}{3}\cdot\sqrt{3}\) | ||
| \(I_4 = \dfrac{1152}{35}\sqrt{3}\) | A1 |
# Question 4: $\displaystyle\int_0^{\sqrt{3}}(3-x^2)^n\,dx$
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int_0^{\sqrt{3}}(3-x^2)^n dx = \left[x(3-x^2)^n\right]_0^{\sqrt{3}}+\int_0^{\sqrt{3}}2x^2 n(3-x^2)^{n-1}dx$ | M1A1 | M1: Integration by parts in correct direction; A1: Correct expression (ignore limits) |
| $= 0-2n\displaystyle\int_0^{\sqrt{3}}(3-x^2-3)(3-x^2)^{n-1}dx$ | dM1 | Substitutes limits and uses $x^2 = x^2-3+3$; **dependent on first M** |
| $= -2n\displaystyle\int_0^{\sqrt{3}}(3-x^2)^n dx + 6n\int_0^{\sqrt{3}}(3-x^2)^{n-1}dx$ | A1 | Correct expressions |
| $= 6nI_{n-1}-2nI_n$ | ddM1 | Substitutes for $I_{n-1}$ and $I_n$; **dependent on both M's** |
| $I_n = \dfrac{6n}{2n+1}I_{n-1}\;*$ | A1* | Correct completion with **no** errors |
**Part (a) Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \displaystyle\int_0^{\sqrt{3}}(3-x^2)^{n-1}(3-x^2)\,dx = 3\int_0^{\sqrt{3}}(3-x^2)^{n-1}dx - \int_0^{\sqrt{3}}x^2(3-x^2)^{n-1}dx$ | dM1 | M1: Writes bracket as product and separates; **this is the second M and depends on the first M below** |
| $= 3I_{n-1}+\left\{\left[\dfrac{x(3-x^2)^n}{-2n}\right]_0^{\sqrt{3}}-\displaystyle\int_0^{\sqrt{3}}\dfrac{(3-x^2)^n}{-2n}dx\right\}$ | M1A1 | M1: Parts in correct direction (First M1); A1: Correct expression (First A1) |
| $= 3I_{n-1} - \displaystyle\int_0^{\sqrt{3}}\dfrac{(3-x^2)^n}{2n}\,dx$ | A1 | Correct expression with no errors |
| $= 3I_{n-1} - \dfrac{1}{2n}I_{n-1}$ ... wait, $= 3I_{n-1}-\dfrac{1}{2n}I_n$ | ddM1 | Substitutes for $I_{n-1}$ and $I_n$; **dependent on both M's** |
| $I_n = \dfrac{6n}{2n+1}I_{n-1}\;*$ | A1 | Correct completion with no errors |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_0=\sqrt{3}$ or $I_1=2\sqrt{3}$ | B1 | |
| $I_4 = \dfrac{24}{9}I_3$ | M1 | Attempt $I_4$ in terms of $I_3$ |
| $I_4 = \dfrac{24}{9}\cdot\dfrac{18}{7}I_2 = \dfrac{24}{9}\cdot\dfrac{18}{7}\cdot\dfrac{12}{5}I_1$ | M1A1 | M1: Attempt $I_4$ in terms of $I_1$; A1: Correct expression for $I_4$ as shown or correct numerical expression |
| $I_4 = \dfrac{24}{9}\cdot\dfrac{18}{7}\cdot\dfrac{12}{5}\cdot\dfrac{6}{3}\cdot\sqrt{3}$ | | |
| $I_4 = \dfrac{1152}{35}\sqrt{3}$ | A1 | |
---4.
$$I _ { n } = \int _ { 0 } ^ { \sqrt { 3 } } \left( 3 - x ^ { 2 } \right) ^ { n } \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 1$
$$I _ { n } = \frac { 6 n } { 2 n + 1 } I _ { n - 1 }$$
\item Hence find the exact value of $I _ { 4 }$, giving your answer in the form $k \sqrt { 3 }$ where $k$ is a rational number to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2014 Q4 [11]}}