Taking the typical approach I consider events of the form $E_s:=\{X\leq s\}$ for $s\geq 0$ and would like to write $$\mathbb{E}[X\wedge Y~\mathbb{1}_{E_s}]=\lambda^2\int_0^s\left(\int_0^\infty x\wedge y~e^{-\lambda y}~\mathrm{d}y\right) e^{-\lambda x}~\mathrm{d}x$$ in a way that suggests a solution like e.g. $\frac{1}{2}X$ (very naive), which would correspond to the result $$\frac{\lambda}{2}\int_0^s x~e^{-\lambda x}~\mathrm{d}x.$$ However, through writing $x\wedge y=x\mathbb{1}_{\{x\leq y\}}+y\mathbb{1}_{\{y< x\}}$ and explicit computation I seem to obtain $$\mathbb{E}[X\wedge Y~\mathbb{1}_{E_s}]=\int_0^s e^{-\lambda x}+e^{-2\lambda x}~\mathrm{d}x$$ and, provided this is correct (?), I don't feel a solution suggests itself.
I would like to know if my approach is sensible here and if you can confirm the computation. If so, what would be the solution and if not, what should I do instead?
An approach which might be computationally easier (and which I would call the typical one...) is often called the functional approach, and is based on the fact that $$E(X\wedge Y\mid X)=g(X)$$ if and only if, for every measurable and bounded function $u$,
Assuming that $X$ and $Y$ are independent, $$E((X\wedge Y)u(X))=E(Xu(X);Y>X)+E(Yu(X);Y<X)$$ hence $$g(x)=E(x\wedge Y)=xP(Y>x)+E(Y;Y<x)$$ In the case at hand, $Y$ has PDF $\lambda e^{-\lambda y}$ on $y>0$ hence $$g(x)=xe^{-\lambda x}+\int_0^xy\lambda e^{-\lambda y}dy=\lambda^{-1}(1-e^{-\lambda x})$$ that is,
One sees that the distribution of $X$ is irrelevant, and that this approach shows the much more general result that, if $Z$ and $W$ are independent and $h(Z,W)$ is integrable, then $$E(h(Z,W)\mid W)=g(W)$$ where $$g(w)=E(h(Z,w))$$