| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts with a clear structure. Part (a) involves routine manipulation to derive the formula (finding p and q), while part (b) applies it twice with straightforward arithmetic. The technique is well-practiced in FP3, though the algebraic manipulation and exact value calculation require care. |
| Spec | 1.08i Integration by parts4.08f Integrate using partial fractions8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \frac{2}{3}x^n(x+8)^{\frac{3}{2}} - \int \frac{2}{3}nx^{n-1}(x+8)^{\frac{3}{2}} \, dx\) | M1 | Parts in the correct direction |
| (same line) | A1 | Correct expression |
| \(I_n = \ldots - \frac{2}{3}n\int x^{n-1}(x+8)(x+8)^{\frac{1}{2}} \, dx\) | M1 | Writes \((x+8)^{\frac{3}{2}}\) as \((x+8)(x+8)^{\frac{1}{2}}\) |
| \(I_n = \frac{2}{3}x^n(x+8)^{\frac{3}{2}} - \frac{2}{3}nI_n - \frac{16}{3}nI_{n-1}\) | dM1 | Substitutes \(I_n\) and \(I_{n-1}\) correctly. Dependent on previous M mark. |
| \(I_n + \frac{2}{3}nI_n = \frac{2}{3}x^n(x+8)^{\frac{3}{2}} - \frac{16}{3}nI_{n-1}\) | ddM1 | Collects \(I_n\) terms to lhs. Dependent on both previous M marks. |
| \(I_n = \frac{2x^n(x+8)^{\frac{3}{2}}}{2n+3} - \frac{16n}{2n+3}I_{n-1}\) | A1 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_0 = \int \sqrt{x+8} \, dx = \frac{2}{3}(x+8)^{\frac{3}{2}}(+c)\) | M1 | Attempts \(I_0\), must be of form \(k(x+8)^{\frac{3}{2}}\) |
| (same line) | A1 | Correct expression |
| \(I_2 = \frac{2x^2(x+8)^{\frac{3}{2}}}{7} - \frac{32}{7}I_1\) and/or \(I_1 = \frac{2x(x+8)^{\frac{3}{2}}}{5} - \frac{16}{5}I_0\) | M1 | Reduction formula applied at least once |
| \(I_2 = \frac{2x^2(x+8)^{\frac{3}{2}}}{7} - \frac{32}{7}\left(\frac{2x(x+8)^{\frac{3}{2}}}{5} - \frac{16}{5}I_0\right) = \ldots\) | ddM1 | Full complete and correct method with limits applied to obtain numerical value for \(I_2\). Dependent on both previous M marks. |
| \(\int_0^{10} x^2\sqrt{x+8} \, dx = \frac{97232}{105}\sqrt{2}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_2 = \frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{4}{3}\int x(8+x)^{\frac{3}{2}} \, dx\) | M1A1 | M1: Correct first application of parts on \(I_2\); A1: Correct expression |
| \(= \frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{4}{3}\left(\frac{2}{5}x(8+x)^{\frac{5}{2}} - \int \frac{2}{5}(8+x)^{\frac{5}{2}} \, dx\right)\) | M1 | Applies parts again |
| \(= \frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{8}{15}x(8+x)^{\frac{5}{2}} + \frac{16}{105}(8+x)^{\frac{7}{2}}\) | ||
| \(\left[\frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{8}{15}x(8+x)^{\frac{5}{2}} + \frac{16}{105}(8+x)^{\frac{7}{2}}\right]_0^{10}\) | ddM1 | Fully complete and correct method with limits. Dependent on both previous M marks. |
| \(= \frac{97232}{105}\sqrt{2}\) | A1 | cao |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \frac{2}{3}x^n(x+8)^{\frac{3}{2}} - \int \frac{2}{3}nx^{n-1}(x+8)^{\frac{3}{2}} \, dx$ | M1 | Parts in the correct direction |
| (same line) | A1 | Correct expression |
| $I_n = \ldots - \frac{2}{3}n\int x^{n-1}(x+8)(x+8)^{\frac{1}{2}} \, dx$ | M1 | Writes $(x+8)^{\frac{3}{2}}$ as $(x+8)(x+8)^{\frac{1}{2}}$ |
| $I_n = \frac{2}{3}x^n(x+8)^{\frac{3}{2}} - \frac{2}{3}nI_n - \frac{16}{3}nI_{n-1}$ | dM1 | Substitutes $I_n$ and $I_{n-1}$ correctly. **Dependent on previous M mark.** |
| $I_n + \frac{2}{3}nI_n = \frac{2}{3}x^n(x+8)^{\frac{3}{2}} - \frac{16}{3}nI_{n-1}$ | ddM1 | Collects $I_n$ terms to lhs. **Dependent on both previous M marks.** |
| $I_n = \frac{2x^n(x+8)^{\frac{3}{2}}}{2n+3} - \frac{16n}{2n+3}I_{n-1}$ | A1 | All correct |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_0 = \int \sqrt{x+8} \, dx = \frac{2}{3}(x+8)^{\frac{3}{2}}(+c)$ | M1 | Attempts $I_0$, must be of form $k(x+8)^{\frac{3}{2}}$ |
| (same line) | A1 | Correct expression |
| $I_2 = \frac{2x^2(x+8)^{\frac{3}{2}}}{7} - \frac{32}{7}I_1$ and/or $I_1 = \frac{2x(x+8)^{\frac{3}{2}}}{5} - \frac{16}{5}I_0$ | M1 | Reduction formula applied at least once |
| $I_2 = \frac{2x^2(x+8)^{\frac{3}{2}}}{7} - \frac{32}{7}\left(\frac{2x(x+8)^{\frac{3}{2}}}{5} - \frac{16}{5}I_0\right) = \ldots$ | ddM1 | Full complete and correct method with limits applied to obtain numerical value for $I_2$. **Dependent on both previous M marks.** |
| $\int_0^{10} x^2\sqrt{x+8} \, dx = \frac{97232}{105}\sqrt{2}$ | A1 | cao |
**Alternative by parts from scratch:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_2 = \frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{4}{3}\int x(8+x)^{\frac{3}{2}} \, dx$ | M1A1 | M1: Correct first application of parts on $I_2$; A1: Correct expression |
| $= \frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{4}{3}\left(\frac{2}{5}x(8+x)^{\frac{5}{2}} - \int \frac{2}{5}(8+x)^{\frac{5}{2}} \, dx\right)$ | M1 | Applies parts again |
| $= \frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{8}{15}x(8+x)^{\frac{5}{2}} + \frac{16}{105}(8+x)^{\frac{7}{2}}$ | | |
| $\left[\frac{2}{3}x^2(8+x)^{\frac{3}{2}} - \frac{8}{15}x(8+x)^{\frac{5}{2}} + \frac{16}{105}(8+x)^{\frac{7}{2}}\right]_0^{10}$ | ddM1 | Fully complete and correct method with limits. **Dependent on both previous M marks.** |
| $= \frac{97232}{105}\sqrt{2}$ | A1 | cao |5. Given that
$$I _ { n } = \int x ^ { n } \sqrt { ( x + 8 ) } \mathrm { d } x , \quad n \geqslant 0 , x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item show that, for $n \geqslant 1$
$$I _ { n } = \frac { p x ^ { n } ( x + 8 ) ^ { \frac { 3 } { 2 } } } { 2 n + 3 } - \frac { q n } { 2 n + 3 } I _ { n - 1 }$$
where $p$ and $q$ are constants to be found.
\item Use part (a) to find the exact value of
$$\int _ { 0 } ^ { 10 } x ^ { 2 } \sqrt { ( x + 8 ) } d x$$
giving your answer in the form $k \sqrt { 2 }$, where $k$ is rational.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2018 Q5 [11]}}