= Solution
Set $a_n=1-1/(2n)$, $b_n=1+1/(2n)$. The finite pulse has <Laplace transform> $\bar S_n(p)=n(e^{-pa_n}-e^{-pb_n})/p$. Since $f(0)=0$, transforming the equation gives
$$
\bar f(p)=\frac{n(e^{-pa_n}-e^{-pb_n})}{p(p-1)}.
$$
The <partial fraction decomposition> $1/[p(p-1)]=1/(p-1)-1/p$ has inverse $e^t-1$. By the <Laplace transform time-shift rule>,
$$
f(t)=n[(e^{t-a_n}-1)H(t-a_n)-(e^{t-b_n}-1)H(t-b_n)].
$$
This <finite-pulse response of a first-order linear equation> gives the requested outer regions:
$$
\boxed{f(t)=0\quad(t<a_n),\qquad f(t)=2n\sinh\left(\frac1{2n}\right)e^{t-1}\quad(t>b_n).}
$$
For completeness, while the pulse is active the solution is $n(e^{t-a_n}-1)$. It is continuous at both switching times. After the forcing stops it grows as $e^t$, because the homogeneous equation is $f'=f$. The apparent pole at $p=0$ in the full transform is removable; the <Bromwich contour> may be taken with $c>1$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2-pulse-response.png]
{title=A unit-area pulse excites an exponentially growing response}
{description=Original calculated responses for two pulse widths and their limiting impulse response. The lower panel shows the prescribed rectangular inputs; their areas are both one.}
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