= Solution
For fixed $v>0$, the support inequalities become $v/2<u<v$. The <marginal density> is therefore
$$
\boxed{f_V(v)=\int_{v/2}^{v}e^{-u}\,du
=\begin{cases}e^{-v/2}-e^{-v},&v>0,\\0,&v\leq0.\end{cases}}
$$
It is nonnegative and integrates to $2-1=1$. Another interpretation is that $V=X+2Y$ sums <independent> <exponential random variables> of rates $1$ and $1/2$, so it has a <hypoexponential distribution>. The <mean> is $1+2=3$, which agrees with $\int_0^\infty v(e^{-v/2}-e^{-v})\,dv=4-1=3$.
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