| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Standard +0.3 This is a straightforward integration by parts question with standard setup. Part (a) is a simple substitution (2 marks), part (b)(i) applies integration by parts with clear choice of u and dv, and part (b)(ii) is routine evaluation of definite integral using the result from (b)(i). The question scaffolds well and requires no novel insight beyond applying the standard technique. |
| Spec | 1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\int \sqrt{5-x}\,dx = \int (5-x)^{\frac{1}{2}}\,dx = \frac{(5-x)^{\frac{3}{2}}}{-\frac{3}{2}}\) \((+C)\) | M1 | |
| \(= -\frac{2}{3}(5-x)^{\frac{3}{2}} + C\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\int (x-1)\sqrt{5-x}\,dx = -\frac{2}{3}(x-1)(5-x)^{\frac{3}{2}} + \frac{2}{3}\int (5-x)^{\frac{3}{2}}\,dx\) | M1 A1ft | |
| \(= \ldots + \frac{2}{3} \times \frac{(5-x)^{\frac{5}{2}}}{-\frac{5}{2}}\) \((+C)\) | M1 | |
| \(= -\frac{2}{3}(x-1)(5-x)^{\frac{3}{2}} - \frac{4}{15}(5-x)^{\frac{5}{2}}\) \((+C)\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\left[-\frac{2}{3}(x-1)(5-x)^{\frac{3}{2}} - \frac{4}{15}(5-x)^{\frac{5}{2}}\right]_1^5 = (0-0) - \left(0 - \frac{4}{15}\times 4^{\frac{5}{2}}\right)\) | M1 | |
| \(= \frac{128}{15}\left(= 8\frac{8}{15} \approx 8.53\right)\) | A1 | awrt 8.53 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(u^2 = 5-x \Rightarrow 2u\frac{du}{dx} = -1 \Rightarrow \frac{dx}{du} = -2u\) | ||
| \(\int (x-1)\sqrt{5-x}\,dx = \int (4-u^2)u(-2u)\,du\) | M1 A1 | |
| \(= \int (2u^4 - 8u^2)\,du = \frac{2}{5}u^5 - \frac{8}{3}u^3\) \((+C)\) | M1 | |
| \(= \frac{2}{5}(5-x)^{\frac{5}{2}} - \frac{8}{3}(5-x)^{\frac{3}{2}}\) \((+C)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(x=1 \Rightarrow u=2,\quad x=5 \Rightarrow u=0\) | ||
| \(\left[\frac{2}{5}u^5 - \frac{8}{3}u^3\right]_2^0 = (0-0) - \left(\frac{64}{5} - \frac{64}{3}\right)\) | M1 | |
| \(= \frac{128}{15}\left(= 8\frac{8}{15} \approx 8.53\right)\) | A1 | awrt 8.53 (2) |
## Question 6:
**Part (a):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\int \sqrt{5-x}\,dx = \int (5-x)^{\frac{1}{2}}\,dx = \frac{(5-x)^{\frac{3}{2}}}{-\frac{3}{2}}$ $(+C)$ | M1 | |
| $= -\frac{2}{3}(5-x)^{\frac{3}{2}} + C$ | A1 | (2) |
**Part (b)(i):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\int (x-1)\sqrt{5-x}\,dx = -\frac{2}{3}(x-1)(5-x)^{\frac{3}{2}} + \frac{2}{3}\int (5-x)^{\frac{3}{2}}\,dx$ | M1 A1ft | |
| $= \ldots + \frac{2}{3} \times \frac{(5-x)^{\frac{5}{2}}}{-\frac{5}{2}}$ $(+C)$ | M1 | |
| $= -\frac{2}{3}(x-1)(5-x)^{\frac{3}{2}} - \frac{4}{15}(5-x)^{\frac{5}{2}}$ $(+C)$ | A1 | (4) |
**Part (b)(ii):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\left[-\frac{2}{3}(x-1)(5-x)^{\frac{3}{2}} - \frac{4}{15}(5-x)^{\frac{5}{2}}\right]_1^5 = (0-0) - \left(0 - \frac{4}{15}\times 4^{\frac{5}{2}}\right)$ | M1 | |
| $= \frac{128}{15}\left(= 8\frac{8}{15} \approx 8.53\right)$ | A1 | awrt 8.53 (2) |
**Alternative for (b) — substitution $u^2 = 5-x$:**
| Working | Marks | Notes |
|---------|-------|-------|
| $u^2 = 5-x \Rightarrow 2u\frac{du}{dx} = -1 \Rightarrow \frac{dx}{du} = -2u$ | | |
| $\int (x-1)\sqrt{5-x}\,dx = \int (4-u^2)u(-2u)\,du$ | M1 A1 | |
| $= \int (2u^4 - 8u^2)\,du = \frac{2}{5}u^5 - \frac{8}{3}u^3$ $(+C)$ | M1 | |
| $= \frac{2}{5}(5-x)^{\frac{5}{2}} - \frac{8}{3}(5-x)^{\frac{3}{2}}$ $(+C)$ | A1 | |
**Alternative for (c):**
| Working | Marks | Notes |
|---------|-------|-------|
| $x=1 \Rightarrow u=2,\quad x=5 \Rightarrow u=0$ | | |
| $\left[\frac{2}{5}u^5 - \frac{8}{3}u^3\right]_2^0 = (0-0) - \left(\frac{64}{5} - \frac{64}{3}\right)$ | M1 | |
| $= \frac{128}{15}\left(= 8\frac{8}{15} \approx 8.53\right)$ | A1 | awrt 8.53 (2) |
---6. (a) Find $\int \sqrt { } ( 5 - x ) \mathrm { d } x$.\\
(2)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-11_503_1270_370_335}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve with equation
$$y = ( x - 1 ) \sqrt { } ( 5 - x ) , \quad 1 \leqslant x \leqslant 5$$
(b) (i) Using integration by parts, or otherwise, find
$$\int ( x - 1 ) \sqrt { } ( 5 - x ) \mathrm { d } x$$
(ii) Hence find $\int _ { 1 } ^ { 5 } ( x - 1 ) \sqrt { } ( 5 - x ) \mathrm { d } x$.\\
\hfill \mbox{\textit{Edexcel C4 2009 Q6 [8]}}