| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Independent multi-part (different techniques) |
| Difficulty | Moderate -0.3 Part (a) is trivial exponential integration, part (b) is a standard single-application integration by parts with no complications, and part (c) is straightforward substitution that reduces to a basic integral. All three parts are routine textbook exercises requiring only direct application of standard techniques with no problem-solving insight needed. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{4}e^{4x}\) | B1 | Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(u=2x+1\), \(dv = e^{4x}\), \(du=2\), \(v=\frac{1}{4}e^{4x}\) | M1 | By parts |
| \(= \frac{1}{4}(2x+1)e^{4x} - \frac{1}{2}\int e^{4x}\,dx\) | M1 | Their \(\left(uv - \int v\,du\right)\) |
| \(= \frac{1}{4}(2x+1)e^{4x} - \frac{1}{8}e^{4x}(+c)\) | A1 | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(u = 1+\ln x\); \(\frac{du}{dx} = \frac{1}{x}\) or \(\frac{dx}{du} = e^{u-1}\) | B1 | |
| \(\int = \int u\,du = \frac{u^2}{2}(+c)\) | M1, A1 | In terms of \(u\) only |
| \(= \frac{(1+\ln x)^2}{2}(+c)\) | A1 | Total: 4 marks |
## Question 3:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $\frac{1}{4}e^{4x}$ | B1 | Total: 1 mark |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|---------|
| $u=2x+1$, $dv = e^{4x}$, $du=2$, $v=\frac{1}{4}e^{4x}$ | M1 | By parts |
| $= \frac{1}{4}(2x+1)e^{4x} - \frac{1}{2}\int e^{4x}\,dx$ | M1 | Their $\left(uv - \int v\,du\right)$ |
| $= \frac{1}{4}(2x+1)e^{4x} - \frac{1}{8}e^{4x}(+c)$ | A1 | Total: 3 marks |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|---------|
| $u = 1+\ln x$; $\frac{du}{dx} = \frac{1}{x}$ or $\frac{dx}{du} = e^{u-1}$ | B1 | |
| $\int = \int u\,du = \frac{u^2}{2}(+c)$ | M1, A1 | In terms of $u$ only |
| $= \frac{(1+\ln x)^2}{2}(+c)$ | A1 | Total: 4 marks |
---3
\begin{enumerate}[label=(\alph*)]
\item Find $\int \mathrm { e } ^ { 4 x } \mathrm {~d} x$.
\item Use integration by parts to find $\int \mathrm { e } ^ { 4 x } ( 2 x + 1 ) \mathrm { d } x$.
\item By using the substitution $u = 1 + \ln x$, or otherwise, find $\int \frac { 1 + \ln x } { x } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q3 [8]}}