| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Independent multi-part (different techniques) |
| Difficulty | Moderate -0.3 Part (a) is a standard integration by parts with a straightforward choice of u and dv (polynomial times trig function). Part (b) is a routine substitution requiring polynomial division after substitution. Both are textbook-style exercises with clear methods and no problem-solving insight required, making this slightly easier than average for A-level. |
| Spec | 1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = x\), \(\frac{dv}{dx} = \sin(2x-1)\) attempted | M1 | \(\int \sin f(x), \frac{d}{dx}(x)\) attempted |
| \(\frac{du}{dx} = 1\), \(v = -\frac{1}{2}\cos(2x-1)\) | A1 | All correct — condone omission of brackets |
| \((\int =) -\frac{x}{2}\cos(2x-1)\) | m1 | correct substitution of their terms into parts |
| \(= -\frac{x}{2}\cos(2x-1) + \frac{1}{2}\int\cos(2x-1)\,(dx)\) | A1 | All correct — condone omission of brackets |
| \(= -\frac{x}{2}\cos(2x-1) + \frac{1}{4}\sin(2x-1) + c\) | A1 | CSO; condone missing \(+c\) and \(dx\); condone missing brackets around \(2x-1\) if recovered in final line; ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 2x - 1\), \(du = 2\,dx\) | M1 | OE |
| \(\int\frac{x^2}{2x-1}\,dx = \int\frac{(u+1)^2}{4u}\cdot\frac{du}{2}\) | m1 A1 | All in terms of \(u\); all correct; PI from later working |
| \(= \left(\frac{1}{8}\right)\int\frac{u^2 + 2u + 1}{u}\,du\) | ||
| \(= \left(\frac{1}{8}\right)\int\left(u + 2 + \frac{1}{u}\right)du\) | A1 | |
| \(= \left(\frac{1}{8}\right)\left[\frac{u^2}{2} + 2u + \ln u\right]\) | B1 | or \(\left(\frac{1}{8}\right)\left[\frac{(u+2)^2}{2} + \ln u\right]\) |
| \(= \frac{1}{8}\left[\frac{(2x-1)^2}{2} + 2(2x-1) + \ln(2x-1)\right] + c\) | A1 | or \(= \frac{1}{8}\left[\frac{(2x+1)^2}{2} + \ln(2x-1)\right] + c\); CSO condone missing \(+c\) only; ISW |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x$, $\frac{dv}{dx} = \sin(2x-1)$ attempted | M1 | $\int \sin f(x), \frac{d}{dx}(x)$ attempted |
| $\frac{du}{dx} = 1$, $v = -\frac{1}{2}\cos(2x-1)$ | A1 | All correct — condone omission of brackets |
| $(\int =) -\frac{x}{2}\cos(2x-1)$ | m1 | correct substitution of their terms into parts |
| $= -\frac{x}{2}\cos(2x-1) + \frac{1}{2}\int\cos(2x-1)\,(dx)$ | A1 | All correct — condone omission of brackets |
| $= -\frac{x}{2}\cos(2x-1) + \frac{1}{4}\sin(2x-1) + c$ | A1 | CSO; condone missing $+c$ and $dx$; condone missing brackets around $2x-1$ if recovered in final line; ISW |
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 2x - 1$, $du = 2\,dx$ | M1 | OE |
| $\int\frac{x^2}{2x-1}\,dx = \int\frac{(u+1)^2}{4u}\cdot\frac{du}{2}$ | m1 A1 | All in terms of $u$; all correct; PI from later working |
| $= \left(\frac{1}{8}\right)\int\frac{u^2 + 2u + 1}{u}\,du$ | | |
| $= \left(\frac{1}{8}\right)\int\left(u + 2 + \frac{1}{u}\right)du$ | A1 | |
| $= \left(\frac{1}{8}\right)\left[\frac{u^2}{2} + 2u + \ln u\right]$ | B1 | or $\left(\frac{1}{8}\right)\left[\frac{(u+2)^2}{2} + \ln u\right]$ |
| $= \frac{1}{8}\left[\frac{(2x-1)^2}{2} + 2(2x-1) + \ln(2x-1)\right] + c$ | A1 | or $= \frac{1}{8}\left[\frac{(2x+1)^2}{2} + \ln(2x-1)\right] + c$; CSO condone missing $+c$ only; ISW |8
\begin{enumerate}[label=(\alph*)]
\item Using integration by parts, find $\int x \sin ( 2 x - 1 ) \mathrm { d } x$.
\item Use the substitution $u = 2 x - 1$ to find $\int \frac { x ^ { 2 } } { 2 x - 1 } \mathrm {~d} x$, giving your answer in terms of $x$.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q8 [11]}}