| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Integration by substitution |
| Difficulty | Challenging +1.2 This is a moderately challenging C4 integration question requiring substitution with trigonometric functions and application to volumes of revolution. Part (a) involves standard trigonometric substitution technique with careful handling of limits, while part (b) requires recognizing the connection between the integral and the volume formula. The substitution is given, reducing problem-solving demand, but execution requires careful algebraic manipulation and exact value calculation, placing it above average difficulty. |
| Spec | 1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\frac{dx}{du} = -2\sin u\) then \(\int \frac{1}{x^2\sqrt{4-x^2}} \, dx = \int \frac{1}{(2\cos u)^2\sqrt{4-(2\cos u)^2}} \times (-2\sin u) \, du = \int \frac{-2\sin u}{4\cos^2 u\sqrt{4\sin^2 u}} \, du = -\frac{1}{4}\int \frac{1}{\cos^2 u} \, du\) | B1; M1; M1; Use of \(1 - \cos^2 u = \sin^2 u\) M1; \(\pm k \int \frac{1}{\cos^2 u} \, du\) M1 |
| \(= -\frac{1}{4}\tan u (+C)\) | \(\pm k\tan u\) M1 | |
| \(x = \sqrt{2} \Rightarrow \sqrt{2} = 2\cos u \Rightarrow u = \frac{\pi}{4}\); \(x = 1 \Rightarrow 1 = 2\cos u \Rightarrow u = \frac{\pi}{3}\) | M1 | |
| \(\left[-\frac{1}{4}\tan u\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}} = -\frac{1}{4}\left(\tan\frac{\pi}{4} - \tan\frac{\pi}{3}\right) = -\frac{1}{4}(1-\sqrt{3}) = \frac{\sqrt{3}-1}{4}\) | A1 | (7 marks) |
| (b) | \(V = \pi\int_1^{\sqrt{2}}\left(\frac{4}{x(4-x^2)^{\frac{1}{2}}}\right)^2 dx = 16\pi\int_1^{\sqrt{2}}\frac{1}{x^2\sqrt{4-x^2}} dx\) | M1; 16π × integral in (a) M1; A1fft |
| \(= 16\pi\left(\frac{\sqrt{3}-1}{4}\right)\) | [3] |
(a) | $\frac{dx}{du} = -2\sin u$ then $\int \frac{1}{x^2\sqrt{4-x^2}} \, dx = \int \frac{1}{(2\cos u)^2\sqrt{4-(2\cos u)^2}} \times (-2\sin u) \, du = \int \frac{-2\sin u}{4\cos^2 u\sqrt{4\sin^2 u}} \, du = -\frac{1}{4}\int \frac{1}{\cos^2 u} \, du$ | B1; M1; M1; Use of $1 - \cos^2 u = \sin^2 u$ M1; $\pm k \int \frac{1}{\cos^2 u} \, du$ M1 |
$= -\frac{1}{4}\tan u (+C)$ | $\pm k\tan u$ M1 |
$x = \sqrt{2} \Rightarrow \sqrt{2} = 2\cos u \Rightarrow u = \frac{\pi}{4}$; $x = 1 \Rightarrow 1 = 2\cos u \Rightarrow u = \frac{\pi}{3}$ | M1 |
$\left[-\frac{1}{4}\tan u\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}} = -\frac{1}{4}\left(\tan\frac{\pi}{4} - \tan\frac{\pi}{3}\right) = -\frac{1}{4}(1-\sqrt{3}) = \frac{\sqrt{3}-1}{4}$ | A1 | (7 marks)
(b) | $V = \pi\int_1^{\sqrt{2}}\left(\frac{4}{x(4-x^2)^{\frac{1}{2}}}\right)^2 dx = 16\pi\int_1^{\sqrt{2}}\frac{1}{x^2\sqrt{4-x^2}} dx$ | M1; 16π × integral in (a) M1; A1fft |
$= 16\pi\left(\frac{\sqrt{3}-1}{4}\right)$ | **[3]** |
**Total: [10]**8. (a) Using the substitution $x = 2 \cos u$, or otherwise, find the exact value of
$$\int _ { 1 } ^ { \sqrt { 2 } } \frac { 1 } { x ^ { 2 } \sqrt { } \left( 4 - x ^ { 2 } \right) } d x$$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ef3ae4a-a06d-48c1-8b79-7d7c3f95d120-14_680_1264_502_338}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation $y = \frac { 4 } { x \left( 4 - x ^ { 2 } \right) ^ { \frac { 1 } { 4 } } } , \quad 0 < x < 2$.
The shaded region $S$, shown in Figure 3, is bounded by the curve, the $x$-axis and the lines with equations $x = 1$ and $x = \sqrt { } 2$. The shaded region $S$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.\\
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.
\hfill \mbox{\textit{Edexcel C4 2010 Q8 [10]}}