| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Volume of revolution with substitution |
| Difficulty | Standard +0.8 This C4 question requires executing a non-trivial substitution (u² = 1-x) with careful handling of limits and differentials, then computing a volume of revolution. Part (a) involves multiple algebraic steps after substitution, and part (b) requires integrating (1-x)x² which needs expansion and term-by-term integration. The substitution is more complex than routine u-substitutions, requiring students to work backwards from u² rather than u, making this moderately challenging but still within standard C4 scope. |
| Spec | 1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(u^2 = 1 - x \Rightarrow x = 1 - u^2\), \(\frac{dx}{du} = -2u\) | M1 | |
| \(x = 0 \Rightarrow u = 1, x = 1 \Rightarrow u = 0\) | B1 | |
| \(\text{area} = \int_0^1 x\sqrt{1-x}\,dx = \int_1^0 (1-u^2) \times u \times (-2u)\,du\) | M1 | |
| \(= \int_0^1 (2u^2 - 2u^4)\,du\) | A1 | |
| \(= [\frac{2}{3}u^3 - \frac{2}{5}u^5]_0^1\) | M1 A1 | |
| \(= (\frac{2}{3} - \frac{2}{5}) - (0) = \frac{4}{15}\) | M1 A1 | |
| \(= \pi\int_0^1 x^2(1-x)\,dx\) | M1 | |
| \(= \pi\int_0^1 (x^2 - x^3)\,dx\) | M1 | |
| \(= \pi[\frac{1}{3}x^3 - \frac{1}{4}x^4]_0^1\) | M1 A1 | |
| \(= \pi[(\frac{1}{3} - \frac{1}{4}) - (0)] = \frac{1}{12}\pi\) | M1 A1 | (13 marks) |
$u^2 = 1 - x \Rightarrow x = 1 - u^2$, $\frac{dx}{du} = -2u$ | M1 |
$x = 0 \Rightarrow u = 1, x = 1 \Rightarrow u = 0$ | B1 |
$\text{area} = \int_0^1 x\sqrt{1-x}\,dx = \int_1^0 (1-u^2) \times u \times (-2u)\,du$ | M1 |
$= \int_0^1 (2u^2 - 2u^4)\,du$ | A1 |
$= [\frac{2}{3}u^3 - \frac{2}{5}u^5]_0^1$ | M1 A1 |
$= (\frac{2}{3} - \frac{2}{5}) - (0) = \frac{4}{15}$ | M1 A1 |
$= \pi\int_0^1 x^2(1-x)\,dx$ | M1 |
$= \pi\int_0^1 (x^2 - x^3)\,dx$ | M1 |
$= \pi[\frac{1}{3}x^3 - \frac{1}{4}x^4]_0^1$ | M1 A1 |
$= \pi[(\frac{1}{3} - \frac{1}{4}) - (0)] = \frac{1}{12}\pi$ | M1 A1 | (13 marks)
---\includegraphics{figure_6}
Figure 1 shows the curve with equation $y = x\sqrt{1-x}$, $0 \leq x \leq 1$.
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $u^2 = 1 - x$ to show that the area of the region bounded by the curve and the $x$-axis is $\frac{8}{15}$. [8]
\item Find, in terms of $\pi$, the volume of the solid formed when the region bounded by the curve and the $x$-axis is rotated through $360°$ about the $x$-axis. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [13]}}