2014 Paper 1 Q2
Year: 2014
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Integration
Problem
- Show that \(\int \ln (2-x) \d x = -(2-x)\ln (2-x) + (2-x) + c \,,\ \) where \(x<2\).
- Sketch the curve \(A\) given by \(y= \ln \vert x^2-4\vert\).
- Show that the area of the finite region enclosed by the positive \(x\)-axis, the \(y\)-axis and the curve \(A\) is \(4\ln(2+\sqrt3)-2\sqrt3\,\).
- The curve \(B\) is given by \(y= \vert \ln \vert x^2-4\vert \vert\,\). Find the area between the curve \(B\) and the \(x\)-axis with \(| x| <2\). Note: you may assume that \(t \ln t \to 0\) as \(t\to 0\).
Solution
- \(\,\) \begin{align*} && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] - \int -\frac{x-2}{2-x} \d x \\ && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] + \int 1 \d x \\ &&&= -(2-x) \ln (2-x) +(2-x) + C \end{align*}
- \(\,\)
- \begin{align*} && \text{Area} &= \int_0^{\sqrt{3}} \ln | x^2 - 4 | \d x \\ &&&= \int_0^\sqrt{3} \ln(4-x^2) \d x \\ &&&= \int_0^\sqrt{3} \left ( \ln(2-x) + \ln (2+x) \right) \d x \\ &&&= \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_0^{\sqrt{3}} \\ &&&= \left ( -(2-\sqrt{3}) \ln (2-\sqrt{3}) +(2-\sqrt{3}) +(2+\sqrt{3})\ln(2+\sqrt{3})-(2+\sqrt{3}) \right) - \\ &&&\quad \quad \left (- 2\ln (2)+2 +2\ln(2)-2 \right) \\ &&&=\left ( -(2-\sqrt{3}) \ln \left ( \frac{1}{2+\sqrt{3}} \right) -2\sqrt{3} +(2+\sqrt{3})\ln(2+\sqrt{3}) \right) \\ &&&= 4\ln(2 + \sqrt{3}) - 2 \sqrt{3} \end{align*}
- \begin{align*} && \text{Area} &= 2 \left ( \int_0^\sqrt{3} \ln (4-x^2) \d x - \lim_{t \to 2}\int_{\sqrt{3}}^t \ln(4-x^2) \d x \right) \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2}\int_{\sqrt{3}}^t \left ( \ln (2-x) + \ln (2+x) \right) \d x \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2} \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_{\sqrt{3}}^{t} \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} - 2 \lim_{t \to 2} \left(-(2-t) \ln (2-t) +(2-t) +(2+t)\ln(2+t)-(2+x) \right) \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} -2(4 \ln4-4) \\ &&&= 16 \ln(2 + \sqrt{3}) - 16 \ln 2 +8(1-\sqrt{3}) \end{align*}
Examiner's report
— 2014 STEP 1, Question 2More than 1800 candidates sat this paper, which represents another increase in uptake for this STEP paper. The impression given, however, is that many of these extra candidates are just not sufficiently well prepared for questions which are not structured in the same way as are the A-level questions that they are, perhaps, more accustomed to seeing. Although STEP questions try to give all able candidates "a bit of an intro." into each question, they are not intended to be easy, and (at some point) imagination and real flair (as well as determination) are required if one is to score well on them. In general, it is simply not possible to get very far into a question without making some attempt to think about what is actually going on in the situation presented therein; and those students who expect to be told exactly what to do at each stage of a process are in for a shock. Too many candidates only attempt the first parts of many questions, restricting themselves to 3-6 marks on each, rather than trying to get to grips with substantial portions of work – the readiness to give up and try to find something else that is "easy pickings" seldom allows such candidates to acquire more than 40 marks (as was the case with almost half of this year's candidature, in fact). Poor preparation was strongly in evidence – curve-sketching skills were weak, inequalities very poorly handled, algebraic capabilities (especially in non-standard settings) were often pretty poor, and the ability to get to grips with extended bits of working lacking in the extreme; also, an unwillingness to be imaginative and creative, allied with a lack of thoroughness and attention to detail, made this a disappointing (and, possibly, very uncomfortable) experience for many of those students who took the paper. On the other side of the coin, there was a very pleasing number of candidates who produced exceptional pieces of work on 5 or 6 questions, and thus scored very highly indeed on the paper overall. Around 100 of them scored 90+ marks of the 120 available, and they should be very proud of their performance – it is a significant and noteworthy achievement.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Show LaTeX source
Problem source
\begin{questionparts}
\item Show that $\int \ln (2-x) \d x = -(2-x)\ln (2-x) + (2-x) + c \,,\ $ where $x<2$.
\item Sketch the curve $A$ given by $y= \ln \vert x^2-4\vert$.
\item Show that the area of the finite region enclosed by the
positive $x$-axis, the $y$-axis and the curve $A$ is $4\ln(2+\sqrt3)-2\sqrt3\,$.
\item The curve $B$ is given by $y= \vert \ln \vert x^2-4\vert \vert\,$. Find the area between the curve $B$ and the $x$-axis with $| x| <2$.
\textit{Note: you may assume that $t \ln t \to 0$ as $t\to 0$.}
\end{questionparts}Solution source
\begin{questionparts}
\item $\,$ \begin{align*}
&& \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] - \int -\frac{x-2}{2-x} \d x \\
&& \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] + \int 1 \d x \\
&&&= -(2-x) \ln (2-x) +(2-x) + C
\end{align*}
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){ln(abs((#1)^2-4))};
\def\xl{-6};
\def\xu{6};
\def\yl{-3};
\def\yu{3};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
% \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, blue, smooth, domain=\xl:-2.01, samples=301]
plot (\x, {\functionf(\x)});
\draw[thick, blue, smooth, domain=-1.999:1.999, samples=301]
plot (\x, {\functionf(\x)});
\draw[thick, blue, smooth, domain=2.01:\xu, samples=100]
plot (\x, {\functionf(\x)});
\draw[thin, red, dashed] (-2, \yl) -- (-2, \yu) node[sloped, above, pos=0.8] {\tiny $x = -2$};
\draw[thin, red, dashed] (2, \yl) -- (2, \yu) node[sloped, above, pos=0.8] {\tiny $x = 2$};
\node[blue, above, rotate=40] at (-1, {\functionf(-1)}) {\tiny $y = \ln|x^2-4|$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
\item \begin{align*}
&& \text{Area} &= \int_0^{\sqrt{3}} \ln | x^2 - 4 | \d x \\
&&&= \int_0^\sqrt{3} \ln(4-x^2) \d x \\
&&&= \int_0^\sqrt{3} \left ( \ln(2-x) + \ln (2+x) \right) \d x \\
&&&= \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_0^{\sqrt{3}} \\
&&&= \left ( -(2-\sqrt{3}) \ln (2-\sqrt{3}) +(2-\sqrt{3}) +(2+\sqrt{3})\ln(2+\sqrt{3})-(2+\sqrt{3}) \right) - \\
&&&\quad \quad \left (- 2\ln (2)+2 +2\ln(2)-2 \right) \\
&&&=\left ( -(2-\sqrt{3}) \ln \left ( \frac{1}{2+\sqrt{3}} \right) -2\sqrt{3} +(2+\sqrt{3})\ln(2+\sqrt{3}) \right) \\
&&&= 4\ln(2 + \sqrt{3}) - 2 \sqrt{3}
\end{align*}
\item
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){abs(ln(abs((#1)^2-4)))};
\def\xl{-6};
\def\xu{6};
\def\yl{-3};
\def\yu{3};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
% \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, blue, smooth, domain=\xl:-2.01, samples=301]
plot (\x, {\functionf(\x)});
\draw[thick, blue, smooth, domain=-1.999:1.999, samples=301]
plot (\x, {\functionf(\x)});
\draw[thick, blue, smooth, domain=2.01:\xu, samples=301]
plot (\x, {\functionf(\x)});
\draw[thin, red, dashed] (-2, \yl) -- (-2, \yu) node[sloped, above, pos=0.8] {\tiny $x = -2$};
\draw[thin, red, dashed] (2, \yl) -- (2, \yu) node[sloped, above, pos=0.8] {\tiny $x = 2$};
\node[blue, above, rotate=40] at (-1, {\functionf(-1)}) {\tiny $y = |\ln|x^2-4||$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
\begin{align*}
&& \text{Area} &= 2 \left ( \int_0^\sqrt{3} \ln (4-x^2) \d x - \lim_{t \to 2}\int_{\sqrt{3}}^t \ln(4-x^2) \d x \right) \\
&&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2}\int_{\sqrt{3}}^t \left ( \ln (2-x) + \ln (2+x) \right) \d x \\
&&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2} \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_{\sqrt{3}}^{t} \\
&&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} - 2 \lim_{t \to 2} \left(-(2-t) \ln (2-t) +(2-t) +(2+t)\ln(2+t)-(2+x) \right) \\
&&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} -2(4 \ln4-4) \\
&&&= 16 \ln(2 + \sqrt{3}) - 16 \ln 2 +8(1-\sqrt{3})
\end{align*}
\end{questionparts}
This was another very popular question, attempted by around 80% of candidates, and producing much greater success, although it has to be noted that this usually extended to work up to the end of part (iii), after which the integration efforts had become so prolonged and involved that many candidates simply moved on elsewhere rather than plough on into (iv). This all led to a mean score on the question of almost exactly 10/20. There were two perfectly acceptable approaches to part (i); integrating the LHS by parts, after writing ln(2 – x) as the product ln(2 – x).1 or, instead, differentiating the RHS in order to verify the result. The first method is probably that which is most "in the spirit of the game", though the second is, perhaps, the shrewder tactic. For those adopting the first approach, some difficulties were encountered when the extra "2 – x" failed to appear naturally, indicating a failure to appreciate the nature of arbitrary constants (namely, that "– x + C" could equally well have been written as "– x + 2 + C"). The curve-sketching was handled in the usual mixed way, with many candidates clearly well-prepared and dealing admirably with crossing-points on the axes and asymptotes, while others at the other end of the ability range seemed capable only of plotting points and "joining the dots". There were also many who thought that, towards asymptotes, the graph of a function should only approach positive infinity, and this led to a ∪-shaped central portion. However, there were very few sketch-attempts that failed to pick up at least 3 or 4 point-scoring features of the graph at hand. Following this, attempts at part (iii) again generally picked up quite a few of the available marks by using the correct limits, separating ln(4 − x²) as ln(2 – x) + ln(2 + x), and then adapting the given result of (i) to the second term (although this frequently took far longer than should have been the case). However, when it came to part (iv), most candidates decided they'd had enough and went elsewhere. Many who started (iv) were clearly thrown by the extra set of modulus-brackets (even though they were intended to make life easy by rendering every part of the first curve positive). The extra bits of area then came simply from the use of a new pair of limits and the given result for evaluating the integrand in the vicinity of the asymptote.