| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Derivative then integrate by parts |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through numerical integration, product rule differentiation, and integration by parts. Part (b)(i) is routine product rule, part (b)(ii) is a standard 'reverse' integration by parts heavily hinted by part (i), and part (b)(iii) is straightforward evaluation. The mid-ordinate rule in (a) is also standard C3 content. While it requires multiple techniques, each step is well-signposted and represents typical textbook exercises, making it slightly easier than average overall. |
| Spec | 1.07l Derivative of ln(x): and related functions1.08d Evaluate definite integrals: between limits1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore \ln x = \ln(1\ln 1.5 + \ln 2.5 + \ln 3.5 + \ln 4.5) = 4.08\) | M1 A1 A1 | use of 1.5, 2.5, ..., 3 or 4 correct \(x\) values; AWFW 4 to 4.2; CAO; 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x\ln x\); \(\frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x = \ln x + 1\) | M1 A1 | use of product rule (only differentiating, 2 terms with + sign); 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(\ln x + 1)dx = x\ln x\); \(\int \ln x dx = x\ln x - x(+c)\) | M1 A1 | OE; attempt at parts with \(u = \ln x\); 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^5 \ln x\, dx = [x\ln x - x]_1^5 = (5\ln 5 - 5) - (\ln 1 - 1) = 5\ln 5 - 4\) | M1 A1 | correct substitution of limits into their (ii) provided ln \(x\) is involved; ISW; 2 marks total |
### 6(a)
$\therefore \ln x = \ln(1\ln 1.5 + \ln 2.5 + \ln 3.5 + \ln 4.5) = 4.08$ | M1 A1 A1 | use of 1.5, 2.5, ..., 3 or 4 correct $x$ values; AWFW 4 to 4.2; CAO; 3 marks total
### 6(b)(i)
$y = x\ln x$; $\frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x = \ln x + 1$ | M1 A1 | use of product rule (only differentiating, 2 terms with + sign); 2 marks total
### 6(b)(ii)
$\int(\ln x + 1)dx = x\ln x$; $\int \ln x dx = x\ln x - x(+c)$ | M1 A1 | OE; attempt at parts with $u = \ln x$; 2 marks total
### 6(b)(iii)
$\int_1^5 \ln x\, dx = [x\ln x - x]_1^5 = (5\ln 5 - 5) - (\ln 1 - 1) = 5\ln 5 - 4$ | M1 A1 | correct substitution of limits into their (ii) provided ln $x$ is involved; ISW; 2 marks total
**Total for Question 6:** 9 marks6
\begin{enumerate}[label=(\alph*)]
\item Use the mid-ordinate rule with four strips to find an estimate for $\int _ { 1 } ^ { 5 } \ln x \mathrm {~d} x$, giving your answer to three significant figures.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = x \ln x$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence, or otherwise, find $\int \ln x \mathrm {~d} x$.
\item Find the exact value of $\int _ { 1 } ^ { 5 } \ln x \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q6 [9]}}