| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Independent multi-part (different techniques) |
| Difficulty | Standard +0.3 Part (a)(i) is trivial exponential integration; (a)(ii) is standard integration by parts with ln x; part (b) requires recognizing a trigonometric substitution (x = sin θ) and evaluating definite integral limits, which is moderately challenging but a well-practiced technique at A-level. The 8-mark allocation suggests extended working but no novel insight—slightly above average difficulty overall. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{e^{3x+5}}{3} + C\) | M1, A1 | (\(ke^{3x+5}\)) (\(k = -\frac{1}{3}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int x^2 \ln x \, dx\) | ||
| \(u = \ln x\), \(\frac{dv}{dx} = x^2\) | M1 | (Correct \(u\) and \(\frac{dv}{dx}\)) |
| \(\frac{du}{dx} = \frac{1}{x}\), \(v = \frac{x^3}{3}\) | ||
| \(\int x^2 \ln x \, dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx\) | A1, A1 | |
| \(= \frac{x^3}{3}\ln x - \frac{x^3}{9} + C\) | A1 | |
| (Penalise omission of C once only) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} dx\) | ||
| \(x = \sin\theta\), \(dx = \cos\theta d\theta\) | B1 | |
| \(x = 0, \theta = 0\); \(x = \frac{1}{2}, \theta = \frac{\pi}{6}\) | B1 | |
| \(\int_0^{\frac{\pi}{6}} \frac{\sin^2\theta \cos\theta d\theta}{\sqrt{1-\sin^2\theta}}\) | M1 | (attempt to substitute) |
| \(\int_0^{\frac{\pi}{6}} \frac{\sin^2\theta}{\cos\theta}\cos\theta d\theta\) | A1 | (Correct) |
| \(\int_0^{\frac{\pi}{6}} \sin^2\theta d\theta\) | A1 | |
| \(\int_0^{\frac{\pi}{6}} \frac{1-\cos 2\theta}{2} d\theta\) | m1 | |
| \(\left[\frac{\theta}{2} - \frac{\sin 2\theta}{4}\right]_0^{\frac{\pi}{6}}\) | A1 | |
| \(= \frac{\pi}{12} - \frac{\sin\frac{\pi}{3}}{4} - 0 + 0 = \frac{\pi}{12} - \frac{\sqrt{3}}{8}\) | A1 | (both correct) |
### Part (a)(i):
$\frac{e^{3x+5}}{3} + C$ | M1, A1 | ($ke^{3x+5}$) ($k = -\frac{1}{3}$)
### Part (a)(ii):
$\int x^2 \ln x \, dx$ | |
$u = \ln x$, $\frac{dv}{dx} = x^2$ | M1 | (Correct $u$ and $\frac{dv}{dx}$)
$\frac{du}{dx} = \frac{1}{x}$, $v = \frac{x^3}{3}$ | |
$\int x^2 \ln x \, dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx$ | A1, A1 |
$= \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$ | A1 |
(Penalise omission of C once only) | |
### Part (b):
$\int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$ | |
$x = \sin\theta$, $dx = \cos\theta d\theta$ | B1 |
$x = 0, \theta = 0$; $x = \frac{1}{2}, \theta = \frac{\pi}{6}$ | B1 |
$\int_0^{\frac{\pi}{6}} \frac{\sin^2\theta \cos\theta d\theta}{\sqrt{1-\sin^2\theta}}$ | M1 | (attempt to substitute)
$\int_0^{\frac{\pi}{6}} \frac{\sin^2\theta}{\cos\theta}\cos\theta d\theta$ | A1 | (Correct)
$\int_0^{\frac{\pi}{6}} \sin^2\theta d\theta$ | A1 |
$\int_0^{\frac{\pi}{6}} \frac{1-\cos 2\theta}{2} d\theta$ | m1 |
$\left[\frac{\theta}{2} - \frac{\sin 2\theta}{4}\right]_0^{\frac{\pi}{6}}$ | A1 |
$= \frac{\pi}{12} - \frac{\sin\frac{\pi}{3}}{4} - 0 + 0 = \frac{\pi}{12} - \frac{\sqrt{3}}{8}$ | A1 | (both correct)
**Total: [14]**\begin{enumerate}[label=(\alph*)]
\item Integrate
\begin{enumerate}[label=(\roman*)]
\item $e^{-3x+5}$ [2]
\item $x^2 \ln x$ [4]
\end{enumerate}
\item Use an appropriate substitution to show that
$$\int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} dx = \frac{\pi}{12} - \frac{\sqrt{3}}{8}.$$ [8]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q8 [14]}}