| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Integration of x^n·ln(x) |
| Difficulty | Standard +0.3 This is a straightforward integration by parts question with clear guidance. Part (i) provides a helpful hint by asking for the derivative first, and part (ii) explicitly tells students to use integration by parts with the answer given to verify. The technique is standard C3 material requiring quotient rule differentiation and one application of integration by parts, making it slightly easier than average but still requiring proper execution of the method. |
| Spec | 1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{x^4}\) | M1 | Quotient rule with \(u=\ln x\) and \(v=x^2\); consistent with their derivatives. \(udv \pm vdu\) in quotient rule is M0 |
| B1 | \(d/dx(\ln x) = 1/x\) soi | |
| \(= \frac{x-2x\ln x}{x^4}\) | A1 | Correct expression (o.e.). Condone \(\ln x \cdot 2x = \ln 2x^2\) for this A1 (provided \(\ln x \cdot 2x\) is shown) |
| \(= \frac{1-2\ln x}{x^3}\) | A1 | o.e. cao, mark final answer, but must have divided top and bottom by \(x\) |
| [4] | e.g. \(\frac{1}{x^3}-\frac{2\ln x}{x^3}\), \(x^{-3}-2x^{-3}\ln x\) | |
| *Or*: \(\frac{dy}{dx} = -2x^{-3}\ln x + x^{-2}\cdot\frac{1}{x}\) | M1 | Product rule with \(u=x^{-2}\) and \(v=\ln x\) |
| B1 | \(d/dx(\ln x)=1/x\) soi | |
| \(= -2x^{-3}\ln x + x^{-3}\) | A1 | Correct expression |
| A1 | o.e. cao, mark final answer, must simplify the \(x^{-2}\cdot(1/x)\) term | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{\ln x}{x^2}\,dx\); let \(u=\ln x\), \(du/dx = 1/x\); \(dv/dx = 1/x^2\), \(v=-x^{-1}\) | M1 | Integration by parts with \(u=\ln x\), \(du/dx=1/x\), \(dv/dx=1/x^2\), \(v=-x^{-1}\). Must be correct. |
| \(= -\frac{1}{x}\ln x + \int \frac{1}{x}\cdot\frac{1}{x}\,dx\) | A1 | Must be correct, condone \(+c\) |
| \(= -\frac{1}{x}\ln x + \int \frac{1}{x^2}\,dx\) | At this stage need to see \(1/x^2\) | |
| \(= -\frac{1}{x}\ln x - \frac{1}{x} + c\) | A1 | Condone missing \(c\) |
| \(= -\frac{1}{x}(\ln x + 1) + c\) | A1 | NB AG must have \(c\) shown in final answer |
| [4] |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{x^4}$ | M1 | Quotient rule with $u=\ln x$ and $v=x^2$; consistent with their derivatives. $udv \pm vdu$ in quotient rule is M0 |
| | B1 | $d/dx(\ln x) = 1/x$ soi |
| $= \frac{x-2x\ln x}{x^4}$ | A1 | Correct expression (o.e.). Condone $\ln x \cdot 2x = \ln 2x^2$ for this A1 (provided $\ln x \cdot 2x$ is shown) |
| $= \frac{1-2\ln x}{x^3}$ | A1 | o.e. cao, mark final answer, but must have divided top and bottom by $x$ |
| **[4]** | | e.g. $\frac{1}{x^3}-\frac{2\ln x}{x^3}$, $x^{-3}-2x^{-3}\ln x$ |
| *Or*: $\frac{dy}{dx} = -2x^{-3}\ln x + x^{-2}\cdot\frac{1}{x}$ | M1 | Product rule with $u=x^{-2}$ and $v=\ln x$ |
| | B1 | $d/dx(\ln x)=1/x$ soi |
| $= -2x^{-3}\ln x + x^{-3}$ | A1 | Correct expression |
| | A1 | o.e. cao, mark final answer, must simplify the $x^{-2}\cdot(1/x)$ term |
| **[4]** | | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{\ln x}{x^2}\,dx$; let $u=\ln x$, $du/dx = 1/x$; $dv/dx = 1/x^2$, $v=-x^{-1}$ | M1 | Integration by parts with $u=\ln x$, $du/dx=1/x$, $dv/dx=1/x^2$, $v=-x^{-1}$. Must be correct. |
| $= -\frac{1}{x}\ln x + \int \frac{1}{x}\cdot\frac{1}{x}\,dx$ | A1 | Must be correct, condone $+c$ |
| $= -\frac{1}{x}\ln x + \int \frac{1}{x^2}\,dx$ | | At this stage need to see $1/x^2$ |
| $= -\frac{1}{x}\ln x - \frac{1}{x} + c$ | A1 | Condone missing $c$ |
| $= -\frac{1}{x}(\ln x + 1) + c$ | A1 | **NB AG** must have $c$ shown in final answer |
| **[4]** | | |4 (i) Differentiate $\frac { \ln x } { x ^ { 2 } }$, simplifying your answer.\\
(ii) Using integration by parts, show that $\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c$.
\hfill \mbox{\textit{OCR MEI C3 Q4 [8]}}