Year: 2022
Paper: 3
Question Number: 5
Course: LFM Pure
Section: Integration
One question was attempted by well over 90% of the candidates two others by about 90%, and a fourth by over 80%. Two questions were attempted by about half the candidates and a further three questions by about a third of the candidates. Even the other three received attempts from a sixth of the candidates or more, meaning that even the least popular questions were markedly more popular than their counterparts in previous years. Nearly 90% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Show that
\[ \int_{-a}^{a} \frac{1}{1+\mathrm{e}^x}\,\mathrm{d}x = a \quad \text{for all } a \geqslant 0. \]
\item Explain why, if $\mathrm{g}$ is a continuous function and
\[ \int_0^a \mathrm{g}(x)\,\mathrm{d}x = 0 \quad \text{for all } a \geqslant 0, \]
then $\mathrm{g}(x) = 0$ for all $x \geqslant 0$.
Let $\mathrm{f}$ be a continuous function with $\mathrm{f}(x) \geqslant 0$ for all $x$. Show that
\[ \int_{-a}^{a} \frac{1}{1+\mathrm{f}(x)}\,\mathrm{d}x = a \quad \text{for all } a \geqslant 0 \]
if and only if
\[ \frac{1}{1+\mathrm{f}(x)} + \frac{1}{1+\mathrm{f}(-x)} - 1 = 0 \quad \text{for all } x \geqslant 0, \]
and hence if and only if $\mathrm{f}(x)\mathrm{f}(-x) = 1$ for all $x$.
\item Let $\mathrm{f}$ be a continuous function such that, for all $x$, $\mathrm{f}(x) \geqslant 0$ and $\mathrm{f}(x)\mathrm{f}(-x) = 1$. Show that, if $\mathrm{h}$ is a continuous function with $\mathrm{h}(x) = \mathrm{h}(-x)$ for all $x$, then
\[ \int_{-a}^{a} \frac{\mathrm{h}(x)}{1+\mathrm{f}(x)}\,\mathrm{d}x = \int_0^a \mathrm{h}(x)\,\mathrm{d}x\,. \]
\item Hence find the exact value of
\[ \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{\mathrm{e}^{-x}\cos x}{\cosh x}\,\mathrm{d}x\,. \]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& I &= \int_{-a}^a \frac{1}{1+e^x} \d x \\
&&&= \int_{-a}^a \frac{e^{-x}}{e^{-x}+1} \d x \\
&&&= \left [ -\ln(1 + e^{-x} ) \right ]_{-a}^a \\
&&&= \ln(1 + e^a) - \ln(1 + e^{-a}) \\
&&&= \ln \left ( \frac{1+e^a}{1+e^{-a}} \right) \\
&&&= \ln \left ( e^a \frac{1+e^a}{e^a+1} \right) \\
&&& = a
\end{align*}
\item Suppose $g$ is continuous and $\int_0^a g(x) \d x = 0$ for all $a \geq 0$ then $g(x) = 0$ for all $x$.
Proof: Differentiate with respect to $a$ to obtain $g(a) = 0$ for all $a$ as required.
\begin{align*}
&& a &= \int_{-a}^a \frac{1}{1+ f(x)} \d x \\
\Leftrightarrow && 1 &= \frac{1}{1 + f(a)} + \frac{1}{1 + f(-a)} \tag{FTC} \\
\Leftrightarrow && (1+f(x))(1+f(-x)) &= 2+f(-x) + f(x) \\
\Leftrightarrow && f(x) f(-x) & = 1
\end{align*}
\item $\,$ \begin{align*}
&& J &= \int_{-a}^a \frac{h(x)}{1 + f(x)} \d x \\
y = - x: &&&=\int_{-a}^a \frac{h(-y)}{1 + f(-y)} \d y \\
&&&= \int_{-a}^a \frac{h(y)}{1 + f(-y)} \d y \\
\Rightarrow && 2J &= \int_{-a}^a h(x) \left ( \frac{1}{1+f(x)} + \frac{1}{1+f(-x)} \right) \d x \\
&&&= \int_{-a}^a h(x) \d x \\
&&&= \int_{-a}^0 h(x) \d x + \int_0^a h(x) \d x\\
&&&= \int_0^a h(-x) \d x + \int_0^a h(x) \d x \\
&&&= 2 \int_0^a h(x) \d x \\
\Rightarrow && J &= \int_0^a h(x) \d x
\end{align*}
\item First notice that $h(x) = \cos x = h(-x)$. Also notice that if $f(x) = e^{2x}$ then $f(x)f(-x) = 1$ so \begin{align*}
&& K &= \int_{-\frac12 \pi}^{\frac12\pi} \frac{e^{-x}\cos x}{\cosh x} \d x \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \frac{2h(x)}{1+f(x)} \d x \\
&&&= 2 \int_0^{\frac12 \pi} h(x) \d x \\
&&&= 2 \int_0^{\frac12 \pi} \cos x \d x \\
&&&= 2
\end{align*}
\end{questionparts}
This was only very slightly less popular than question 3, but it was the third most successful with a mean of just under half marks. Part (i) was well done, with a variety of methods used, the most common being by a substitution of e^x. In this part, the most frequent errors were showing insufficient working to fully justify the given result, not spotting how to simplify (1+e^a)/(1+e^{-a}), and incorrectly integrating ∫1/(1+e^x) dx to get ln(1+e^x). Part (ii) was generally found to be the hardest. There was a range of responses to the first requirement from concise use of the Fundamental theorem of Calculus to long, often imprecise, paragraphs of text. Candidates attempting proof by contradiction tended to be more successful if they used a sketch to back up their argument. The second result saw many different methods used. The most common mistakes were not showing enough working when using a u=-x substitution, not showing that the argument can be reversed, and using an incorrect argument such as ∫_{-a}^{a} g(x) = 0 → g(x) = 0 (to which g(x) = sin x is a counterexample). Many candidates did not see the link with the first requirement of the part. However, the final result of this part was usually done well. Candidates found part (iii) easier than part (ii), the most common mistake was not realising that h(x) = h(-x) holds, even though this was stated in the question. Part (iv) was generally done well, with the most common mistakes being neglecting to show that the functions satisfied the conditions in part 3 or omitting a factor of 2. A few candidates did not use the results from the previous parts, instead using other methods, which, as the question stated "hence", gained no credit for this part.