| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Show that integral equals expression |
| Difficulty | Standard +0.3 This is a standard integration by parts question with ln(x-4), requiring the well-known technique of setting u=ln(x-4) and dv=dx. Part (a) is routine application of the formula, part (b) is trivial identification of an asymptote, and part (c) involves straightforward definite integration with some algebraic manipulation. While it requires multiple steps and careful algebra, it follows a predictable pattern with no novel insight needed, making it slightly easier than average. |
| Spec | 1.08f Area between two curves: using integration1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int 1 \cdot \ln(x-4)\,dx\), so \(u = \ln(x-4)\) and \(v' = 1\) | M1 | Attempt integration by parts, with correct parts. \(u\) and \(v'\) correctly allocated and correct formula used. M0 if \(v = x - 4\) from \(v' = 1\) |
| \(x\ln | x-4 | - \int\frac{x}{x-4}\,dx\) |
| \(\int\frac{x}{x-4}\,dx = \int 1 + \frac{4}{x-4}\,dx\) | M1 | Attempt to deal with improper fraction. Allow sign error ie \(1 - \frac{4}{x-4}\). Could use substitution of \(u = x - 4\) but must get as far as a proper fraction (ie \(1 \pm 4u^{-1}\)). Do not need to actually integrate for M1 |
| \(= x + 4\ln | x-4 | \) |
| \(\int\ln(x-4)\,dx = x\ln | x-4 | - x - 4\ln |
| [5] | NB differentiating given answer is 0/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(x = 4\) | B1 | Must be an equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left | \int_5^7 \ln(x-4)\,dx\right | + \left |
| \((3\ln 3 - 2) + (\frac{1}{2}\ln\frac{1}{2} + \frac{1}{2})\) | A1 | Correct expression for area. Any unsimplified equiv |
| \(3\ln 3 - 2 - \frac{1}{2}\ln 2 + \frac{1}{2}\) | M1 | Attempt to rearrange to required form. Use \(\ln\frac{1}{2} = -\ln 2\) and gather like terms. Could follow M0. Allow M1 (implied) for \(3\ln 3 + 0.5\ln 2 - 2.5\), even if \(-0.5\ln 0.5\) not seen first |
| \(3\ln 3 - \frac{1}{2}\ln 2 - \frac{3}{2}\) | A1 | Obtain \(3\ln 3 - \frac{1}{2}\ln 2 - \frac{3}{2}\). Or \(3\ln 3 - 0.5\ln 2 - 1.5\) |
# Question 11:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int 1 \cdot \ln(x-4)\,dx$, so $u = \ln(x-4)$ and $v' = 1$ | M1 | Attempt integration by parts, with correct parts. $u$ and $v'$ correctly allocated and correct formula used. M0 if $v = x - 4$ from $v' = 1$ |
| $x\ln|x-4| - \int\frac{x}{x-4}\,dx$ | A1 | Correct expression. Allow brackets not modulus. Allow $x \times \frac{1}{x-4}$, even if subsequently spoilt |
| $\int\frac{x}{x-4}\,dx = \int 1 + \frac{4}{x-4}\,dx$ | M1 | Attempt to deal with improper fraction. Allow sign error ie $1 - \frac{4}{x-4}$. Could use substitution of $u = x - 4$ but must get as far as a proper fraction (ie $1 \pm 4u^{-1}$). Do not need to actually integrate for M1 |
| $= x + 4\ln|x-4|$ | A1 | Correct integration of fraction. Allow brackets not modulus. Using a substitution gives $x - 4 + 4\ln|x-4|$; must be in terms of $x$ and not $u$ for A1 |
| $\int\ln(x-4)\,dx = x\ln|x-4| - x - 4\ln|x-4| + c = (x-4)\ln|x-4| - x + c$ **A.G.** | A1 | Show given answer with no errors seen. Modulus required in final answer, as well as $+c$. Can go from penultimate line in MS to given answer with no further detail needed. Answer from using substitution will need to justify changing $c$ eg $c + 4$ is a constant hence $c'$ is also a constant |
| | **[5]** | **NB** differentiating given answer is 0/5 |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $x = 4$ | B1 | Must be an equation |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left|\int_5^7 \ln(x-4)\,dx\right| + \left|\int_{4.5}^5 \ln(x-4)\,dx\right|$ | M1 | Attempt sum of absolute areas. Or integral $\int_5^7 \ln(x-4)\,dx - \int_{4.5}^5 \ln(x-4)\,dx$ |
| $(3\ln 3 - 2) + (\frac{1}{2}\ln\frac{1}{2} + \frac{1}{2})$ | A1 | Correct expression for area. Any unsimplified equiv |
| $3\ln 3 - 2 - \frac{1}{2}\ln 2 + \frac{1}{2}$ | M1 | Attempt to rearrange to required form. Use $\ln\frac{1}{2} = -\ln 2$ and gather like terms. Could follow M0. Allow M1 (implied) for $3\ln 3 + 0.5\ln 2 - 2.5$, even if $-0.5\ln 0.5$ not seen first |
| $3\ln 3 - \frac{1}{2}\ln 2 - \frac{3}{2}$ | A1 | Obtain $3\ln 3 - \frac{1}{2}\ln 2 - \frac{3}{2}$. Or $3\ln 3 - 0.5\ln 2 - 1.5$ |11\\
\begin{tikzpicture}[x=0.7cm, y=0.7cm]
% x-axis
\draw[-latex, semithick] (-1.5, 0) -- (12.5, 0) node[below right] {$x$};
% y-axis
\draw[-latex, semithick] (0, -4.5) -- (0, 3.2) node[above left] {$y$};
% Origin
\node[below left] at (0,0) {$O$};
% Curve: y = ln(x - 4)
\draw[thick, smooth, domain=-4.3:2.1, variable=\y, samples=100] plot ({4 + exp(\y)}, \y);
\end{tikzpicture}
The diagram shows part of the curve $y = \ln ( x - 4 )$.
\begin{enumerate}[label=(\alph*)]
\item Use integration by parts to show that $\int \ln ( x - 4 ) \mathrm { d } x = ( x - 4 ) \ln | x - 4 | - x + c$.
\item State the equation of the vertical asymptote to the curve $y = \ln ( x - 4 )$.
\item Find the total area enclosed by the curve $y = \ln ( x - 4 )$, the $x$-axis and the lines $x = 4.5$ and $x = 7$. Give your answer in the form $a \ln 3 + b \ln 2 + c$ where $a , b$ and $c$ are constants to be found.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2019 Q11 [10]}}